Write the Expression as the Sine, Cosine, or Tangent of an Angle. Sin 8x Cos X - Cos 8x Sin X
TRIGONOMETRY
- Formulary
- Definitions and basics
- Trigonometric circle and angles
- Trigonometric numbers of a real number t
- Basic formulas
- Related values
- The difference between the two values is an integer multiple of 2.pi
- supplementary values
- complementary values
- Opposite values
- Anti-supplementary values
- The difference between the two values is an integer multiple of 2.pi
- The right-angled triangle
- Other properties in a right-angled triangle
- Other properties in a right-angled triangle
- Area of a triangle
- Sine rule
- Homogeneous expression in a, b and c
- Homogeneous expression in a, b and c
- Cosine rule
- Trigonometric circle and angles
- Special values
- pi/3
- pi/4
- pi/6
- pi/3
- Solving Triangles
- Case SSS
- Case ASA or AAS
- Case SAS
- Case SSA
- Case SSS
- Trigonometric functions
- The sine function
- The cosine function
- The tangent function
- The cotangent function
- Related functions and period
- Period of a sum of two functions
- The sine function
- Inverse Trigonometric Functions = Cyclometric functions
- The arcsin function
- The arccos function
- The arctan function
- The arccot function
- No period
- Transformations
- The arcsin function
- Sum formulas
- cos(u - v)
- cos(u + v)
- sin(u - v)
- sin(u + v)
- tan(u + v)
- tan(u - v)
- cos(u - v)
- Doubling formulas
- sin(2u)
- cos(2u)
- tan(2u)
- sin(2u)
- Carnot formulas
- t-Formulas or Half-Angle Formulas
- Simpson formulas
- Factoring trigonometric forms
- Period of the product of two related functions
- The general sine function
- Trigonometric equations
- Base equations
- cos(u) = cos(v)
- sin(u) = sin(v)
- tan(u) = tan(v)
- cot(u) = cot(v)
- cos(u) = cos(v)
- Reducing to base equations
- Using an additional unknown
- Using factoring
- The equation a.sin(u)+b.cos(u) = c
- First Method
- Second Method
- First Method
- Homogeneous equations
- Other equations
- Base equations
- Trigonometric inequalities
- Conventions
- Examples
- Conventions
- Cyclometric equations
- Calculations with cyclometric functions
- A Formulary
- Solved problems about trigonometry
- Other trigonometry tutorials
Formulary
Click here to find basic formulas.Definitions and basics
Trigonometric circle and angles
Choose an x-axis and a y-axis (orthonormal) and let O be the origin.
A circle of radius one centered at O is called 'the' trigonometric circle or 'the' unit circle.
Turning counterclockwise is the positive orientation in trigonometry.
Angles are measured starting from the x-axis.
The units used to measure an angle are 'degree' and 'radian'.
A right angle is an angle whose measure is exactly 90 degrees or pi/2 radians.
In this theory we use mainly radians.
Each real number t corresponds to exactly one angle, and to exactly one point P on the unit circle.
We call that point the 'image point' of t.
Examples:
- pi/6 corresponds to the angle t and to point P on the circle.
- -pi/2 corresponds to the angle u and to point Q on the circle.
Trigonometric numbers of a real number t
The real number t corresponds to exactly one point P on the unit circle.- The x-coordinate of P is called the cosine of t. We write cos(t).
- The y-coordinate of P is called the sine of t. We write sin(t).
- The number sin(t)/cos(t) is called the tangent of t. We write tan(t).
- The number cos(t)/sin(t) is called the cotangent of t. We write cot(t).
- The number 1/cos(t) is called the secant of t. We write sec(t)
- The number 1/sin(t) is called the cosecant of t. We write csc(t) or cosec(t)
contains the origin and point P(cos(t),sin(t)). So this line is OP.
On this line we take the intersection point S(1,?) with the line x = 1.
It is easy to see that ? = tan(t).
So tan(t) is the y-coordinate of the point S.
In an analogous manner we find that cotan(t) is the x-coordinate of the intersection point S' of the line OP with the line y = 1.
Basic formulas
With t radians corresponds exactly one point P(cos(t),sin(t)) on the unit circle. The square of the distance [OP] = 1. Calculating |OP|2, using the coordinates of P, we find for each t :cos2(t) + sin2(t) = 1 sin2(t) 1 + tan2(t) = 1 + ---------- cos2(t) cos2(t)+sin2(t) = ----------------- cos2(t) 1 = ----------- = sec2(t) cos2(t) In the same way : 1 + cotan2(t) = 1/ sin2(t) = csc2(t)
cos2(t) + sin2(t) = 1 1 + tan2(t) = sec2(t) 1 + cot2(t) = csc2(t)
Usage examples:
sin2(t) = 1 - cos2(t) cos2(4t) = 1 - sin2(4t) 1 + tan2(t/2) = sec2(t/2) csc2(t2) - cot2(t2) = 1Exercise:
If cos(t)=0.5 then sin2(t) = ... If cos(t)=0.1 then tan2(t) = ... If cot(t)=0.2 then sin2(t) = ...
Related values
The difference between the two values is an integer multiple of 2.pi
If the difference between t an t' is an integer multiple of 2.pi, the corresponding points on the unit circle coincide. So
If t - t' = 2.k.pi (k is an integer) then sin(t) = sin(t')
cos(t) = cos(t')
tan(t) = tan(t')
cot(t) = cot(t')
supplementary values
t and t' are supplementary values <=> t+t' = pi.
With the help of a unit circle we see that the corresponding image points are symmetric relative to the y-axis. Hence, we have :
sin(t) = sin(pi - t)
cos(t) = -cos(pi - t)
tan(t) = -tan(pi - t)
cot(t) = -cot(pi - t)
Usage examples:
sin(t + pi/2) = sin(pi/2 - t) tan(2t + 0.2) = - tan(pi -0.2 - 2t) - tan(pi -t) = tan(t) sin(pi-t) + cos(3pi-t) - sin(t+4pi) + cos(t) = sin(t) + cos(pi-t) - sin(t) + cos(t) = sin(t) - cos(t) - sin(t) + cos(t) = 0
complementary values
t and t' are complementary values <=> t+t' = pi/2.
The corresponding image points on a unit circle are symmetric relative to the line y = x . Hence, we have :
sin(t) = cos(pi/2 - t)
cos(t) = sin(pi/2 - t)
tan(t) = cot(pi/2 - t)
cot(t) = tan(pi/2 - t)
Usage examples:
tan(pi/4 +3t) = cot(pi/4 -3t) cos(3pi/2 -t) = sin( t - pi) = sin(-t + 2pi) = sin(-t) cot(3x - pi/2) = tan(-3x + pi ) = - tan(3x) - cos(pi/2 - 2x) + sin(-2x - pi) - cos(3pi - 2x) = - sin(2x) + sin(pi - 2x) - cos(pi - 2x) = - sin(2x) + sin(2x) + cos(2x) = cos(2x)
Opposite values
t and t' are opposite values <=> t+t' = 0.
Now, the corresponding image points are symmetric relative to the x-axis. Hence, we have :
sin(t) = -sin(-t)
cos(t) = cos(-t)
tan(t) = -tan(-t)
cot(t) = -cot(-t)
Usage examples:
cos(-pi/2 + x) = cos(pi/2 - x) = sin (x) sin(6x - pi) = - sin(pi - 6x) = - sin(6x) cot(-x + 4pi) = cot(-x) = - cot(x)
Anti-supplementary values
t and t' are anti-supplementary values <=> ( t-t' = pi or t'-t = pi )
The corresponding image points are symmetric relative to the origin O . Hence, we have :
sin(t) = -sin(t+pi)
cos(t) = -cos(t+pi)
tan(t) = tan(t+pi)
cot(t) = cot(t+pi)
Usage examples:
tan(5a + 3pi) = tan(5a + pi) = tan(5a) cot(t/2 + pi/2) = cot(t/2 - pi/2) = - cot(pi/2 - t/2) = - tan(t/2) sin(x + 3 pi) + sin(x) = -sin(x) + sin(x) = 0
The right-angled triangle
Let the right angle of a triangle ABC be labelled A. The distances |AB|, |BC| and |CA| are usually denoted by c, a and b. Choose point B in a suitable way as center of a trigonometric circle (see figure).Now sin(B),cos(B) and 1 are directly proportional with b, c and a.
sin(B) cos(B) 1 ------ = ------ = --- b c a => sin(B) = b/a cos(B) = c/a tan(B) = b/c and since the angles B and C are complementary angles cos(C) = b/a sin(C) = c/a tan(C) = c/b
sin(B) = b/a cos(B) = c/a tan(B) = b/c cos(C) = b/a sin(C) = c/a tan(C) = c/b
Other properties in a right-angled triangle
- The square of the hypotenuse is equal to the sum of the squares of the other two sides. a2 = b2 + c2
- The altitude to the hypotenuse of a right triangle is the mean proportional between the segments into which it divides the hypotenuse. h2 = x.y
- Each leg of a right triangle is the mean proportional between the hypotenuse and the its orthogonal projection on the hypotenuse. c2 = a.x and b2 = a.y
Applications:
The tangent lines, in the points A and B of a circle with center O and radius r, intersect in point P. The chord AB and the line OP intersect in point S. Let a = |OP| and k = |AB|.
Express k as a function of r and a.
In the right-angled triangle AOP : The leg of a right triangle is the mean proportional between the hypotenuse and the its orthogonal projection on the hypotenuse.
|OA|2 = |OP| |OS|. => |OS| = r2/aIn the right-angled triangle OAS : The square of the hypotenuse is equal to the sum of the squares of the other two sides.
|OA|2 = |OS|2 + |AS|2 => r2 = |OS|2 + k2/4 => r2 = r4/a2 + k2/4 => .... 2r ___________ => k = ---- \/ a2 - r2 a
Divide a given segment BC by the construction of a point D. The two parts BD and DC must have a suitable length x and y such that the product x.y is equal to a given value m2.
x+y=|BC| en m2 = x.y
m is the mean proportional between x and y
We know that the altitude to the hypotenuse BC of a right triangle ABC is the mean proportional between the segments into which it divides the hypotenuse.
We are looking for a right triangle ABC with base the hypotenuse BC and with m as the height.
The vertex A of the right-angled triangle is located on the circle with diameter BC.
Construction steps:
- Draw a semicircle with diameter BC
- Draw a parallel to BC at distance m from BC.
- This parallel gives us point A.
- Draw the altitude AD from A to BC.
Area of a triangle
The area of the triangle is a.h/2 .
But in triangle BAH, we have sin(B) = h/c .
Hence the area of the triangle is a.c.sin(B) /2.
Similarly, we have that the area of the triangle
= b.c.sin(A) /2 = a.b.sin(C) /2
The area of a triangle ABC =(1/2) a.c.sin(B) = (1/2) b.c.sin(A) = (1/2) a.b.sin(C)
You can also use Heron's formula to calculate the area of a triangle.
Let s = half the circumference of the triangle = (a +b + c)/2.The area of a triangle ABC = ______________________________ V s (s - a) (s - b) (s - c)
Exercise:
A triangle has sides with length 5, 4 and 7.
Draw accurately the triangle.
Calculate the area of the triangle with the formula of Heron and check the result by measuring height of the triangle and calculating the area with this height. Now measure an angle of the triangle and calculate the area a third time. Conversely, you can calculate the angles of the triangle if you know the area of the triangle.
Sine rule
The area of a triangle ABC = a.c.sin(B)/2 = b.c.sin(A)/2 = a.b.sin(C)/2 => a.c.sin(B) = b.c.sin(A) = a.b.sin(C)dividing through by a.b.c, we get
a b c ------ = ------ = ------ sin(A) sin(B) sin(C)This formula is called the sine rule in a triangle ABC.
Let R be the radius of the circle with center O through the points A,B and C. Let B' be the second intersection point of BO with the circle. The angle B' in triangle BB'C is equal to, or supplementary with, A.
In the right-angled triangle BB'C we see that a = 2R sin(B') = 2R sin(A).
Thus, the fractions in the sinus rule are all equal to 2R.
a b c ------ = ------ = ------ = 2R sin(A) sin(B) sin(C)
Exercise:
A triangle has sides with length a = 5, b = 4 and c = 7.
Draw accurately the triangle. Calculate the area of the triangle with the formula of Heron.
Now calculate the angle A with the area formula (1/2).b.c.sin (A).
Now, use the angle A to find the radius R of the circumscribed circle.
Check the result on your sketch.
Homogeneous expression in a, b and c
Note:A relation is called homogeneous in a, b and c if and only if this relation remains valid when we replace a, b and c by a multiple r.a, r.b and r.c (r not 0).
If an expression between the sides of a triangle is homogeneous in a, b and c, we obtain an equivalent expression by replacing a,b and c by sin(A), sin(B) and sin(C).
Example:
In a triangle
b.sin(A-C) = 3.c.cos(A+C) <=> sin(B).sin(A-C) = 3.sin(C).cos(A+C)
Cosine rule
a2 = b2 + c2 - 2 b c cos(A) b2 = c2 + a2 - 2 c a cos(B) c2 = a2 + b2 - 2 a b cos(C)
Proof:
We'll prove that a2 = b2 + c2 - 2 b c cos(A)
If the angle A is a right angle, then the proof is obvious.
Now, suppose the angle A is an acute angle.
Now, suppose the angle A is an obtuse angle.a2 = h2 + p2 (*) b2 = h2 + q2 = h2 + (c - p)2 so, h2 = b2 - (c - p)2 (**) From (*) and (**) a2 = b2 - (c - p)2 + p2 = b2 - (c2 - 2 p c + p2) + p2 = b2 - c2 + 2 p c = b2 + c2 + 2 p c - 2 c2 = b2 + c2 + 2 c (p - c) = b2 + c2 - 2 c (c - p) = b2 + c2 - 2 c q = b2 + c2 - 2 c b cos(A)
The proof proceeds in the same manner as above.
Draw a new picture and work this out as an exercise.
This cosine rule can also be proved using the dot product of vectors.
See Proof cosine rule
Special values
pi/3
Let V be the image point corresponding with the angle pi/3 on the unit circle and let E the intersection point of that circle with the positive X-axis.The triangle 'OVE' is equilateral. Hence cos(pi/3) = 1/2.
sin2 (pi/3) = sqrt( 1 - cos2 (pi/3)) = sqrt(3)/2 So, sin(pi/3) = sqrt(3)/2 and cos(pi/3) = 1/2. tan(pi/3) = sqrt(3)
pi/4
Let V be the image point corresponding with the angle pi/4 on the unit circle. From this, it is obvious that cos(pi/4) = sin(pi/4) and tan(pi/4) = 1.cos2(pi/4)+sin2(pi/4) = 1 => 2cos2(pi/4) = 1 => cos (pi/4) = sqrt(1/2) So, cos (pi/4) = sin(pi/4) = sqrt(1/2) tan(pi/4) = 1
pi/6
From properties for complementary angles we have: cos (pi/6) = sqrt(3)/2 and sin(pi/6) = 1/2.tan(pi/6) = 1/sqrt(3).
Solving Triangles
Case SSS
Given : The three sides.Substitute all the sides in de Cosine Rule to compute the angles.
Example: a=4 b=5 c=7
The Cosine Rule gives
58 = 70 cos(A) 40 = 56 cos(B) -8 = 40 cos(C) A = 34.05� B = 44.41� C = 101.53�Test : A + B + C = ...
Case ASA or AAS
Given : Two angles and a side.Calculate the third angle and then the sides with the Sine Rule.
Example: a=4 A=34� B=45�
The third angle is C =101�
From the Sine Rule
4 sin(45�) b = -------------- = 5.06 sin(34�) 4 sin(101�) c = ------------- = 7.02 sin(34�)Test : draw a sketch of the triangle
Case SAS
Given : Two sides and an included angle.Use the Cosine Rule.
Example: b=5 c=7 A=34.05�
From the Cosine Rule
a2 = 25 + 49 - 70 cos(34.05�) => a = 4 The other two formulas of the cosine rule give 40 = 56 cos(B) -8 = 40 cos(C) B = 44.41� C = 101.53�Test : A + B + C = ...
Case SSA
Given : Two sides and a non-included angle. Draw a sketch. There are three cases.
1) no solutions
2) one solution
3) two solutions
- A=60� b=5 a=1
From a sketch we see that there are no solutions.
- A=60� b=5 a=7
From a sketch we see that there is one solution. We use the Sine Rule.
7 5 c --------- = -------- = --------- sin(60�) sin(B) sin(C)
So, sin(B)= 0.6186 and this gives us two supplementary solutions for B.
But from our sketch, we know what value to choose. B = 38.21�.
Then C = 180� - 38.21� - 60� = 81.79�
and c = 8 - A=60� b=5 a=4.5
From a sketch we see that there are two solutions for B. We use the Sine Rule.
4.5 5 c --------- = -------- = --------- sin(60�) sin(B) sin(C)
So, sin(B)= 0.96225 and this gives us two supplementary solutions for B.
B = 74.2� of 105.8�
First choose B = 74.2� and first compute C and then c with the Sine Rule.
Then choose B = 105.8� and first compute C and then c with the Sine Rule.
Check the results using your sketch.
Trigonometric functions
The sine function
The function defined by :sin : R -> R : x -> sin(x)is called, the sine function.
The images are bounded in [-1,1] and the period is 2.pi .
We see that the range of the function is [-1,1].
The cosine function
The function defined by :cos : R -> R : x -> cos(x)is called, the cosine function.
The images are bounded in [-1,1] and the period is 2.pi .
The range of the function is [-1,1].
The tangent function
The function defined by :tan : R -> R : x -> tan(x)is called, the tangent function.
Now, the period is pi and the images are not defined in x = (pi/2) + k.pi
The range or image is R.
The cotangent function
The function defined by :cot : R -> R : x -> cot(x)is called, the cotangent function.
The period is pi and that the images are not defined in x = k.pi
The range or image is R.
Related functions and period
We can submit previous functions to all kinds of transformations. We obtain related functions. ( see Influence of a transformation on the graph of a function )Example 1
y = sin(4x)
The graph of this function arises from the graph of sin(x) when we compress the graph of sin(x) towards the y-axis with a factor 4.
From this it follows that the period of sin(4x) is pi/2. The function y = sin(ax) has a period 2.pi/a if a > 0.
Similar rules apply to the other trigonometric functions. Thus the period of tan(x/3) is 3.pi.
Example 2
y = sin(x+5)
The graph of this function comes about by moving the graph of sin(x) five units to the left. The period does not change.
Example 3
y = tan(x)+5
The graph of this function is obtained by moving the graph of tan(x) five units upwards. The period does not change.
Example 4
We start with y = tan(x). We compress the graph towards the y-axis with a factor 3. The new function is y = tan(3x). We move the graph two units to the right. The new function is y = tan(3(x-2)) . Finally, we move the last graph two units downwards. We obtain y = tan(3x -6)-2. The period is pi/3.
Generalization:
The period of A sin(a x + b ) is 2 pi/|a|
The period of A cos(a x + b ) is 2 pi/|a|
The period of A tan(a x + b ) is pi/|a|
The period of A cot(a x + b ) is pi/|a|
The period of A / sin(a x + b ) is 2 pi/|a|
The period of A / cos(a x + b ) is 2 pi/|a|
The period of A / tan(a x + b ) is pi/|a|
The period of A / cot(a x + b ) is pi/|a|
Period of a sum of two functions
If f(x) is a function with period = a g(x) is a function with period = b Then f(x)+g(x) is a function with period = c <=> There are strictly positive and relatively prime integers m and n such that c = m.a = n.bExamples
sin(2x) has pi as period and cos(3x) has 2pi/3 as period .
Now, c = 2.(pi) = 3.(2pi/3). So, 2 pi is a period of sin(2x) + cos(3x)
sin(pi x) has 2 as period and tan(2 pi x/7) has 7/2 as period.
Now, c = 7.(2) = 4.(7/2). So, 14 is a period of sin(pi x) + tan(2 pi x/7)
sin(sqrt(2) x) has pi.sqrt(2) as period and cos(2x) has pi as period .
There are no strictly positive integers m and n such that
m.(pi.sqrt(2)) = n.(pi). So, sin(sqrt(2) x) + cos(2x) has NO period!
sin(x) has 2pi as period and cos(pi x) has 2 as period.
There are no strictly positive integers m and n such that
m.(2pi) = n.(2). So, sin(x) + cos(pi x) has NO period!
Inverse Trigonometric Functions = Cyclometric functions
The arcsin function
We restrict the domain of the sine function to [-pi/2 , pi/2].Now this restriction is invertible because each image value in [-1,1] corresponds to exactly one original value in [-pi/2 , pi/2].
The inverse function of that restricted sine function is called the arcsine function.
We write arcsin(x) or asin(x).
The graph y = arcsin(x) is the mirror image of the restricted sine graph relative to the line y = x.
The domain is [-1,1] and the range is [-pi/2 , pi/2].
The arccos function
We restrict the domain of the cosine function to [0 , pi].Now this restriction is invertible because each image value in [-1,1] corresponds to exactly one original value in [0 , pi].
The inverse function of that restricted cosine function is called the arccosine function.
We write arccos(x) or acos(x) .
The graph y = arccos(x) is the mirror image of the restricted cosine graph relative to the line y = x.
The domain is [-1,1] and the range is [0 , pi].
The arctan function
We restrict the domain of the tangent function to [-pi/2 , pi/2].The inverse function of that restricted tangent function is called the arctangent function. We write arctan(x) or atan(x) . The graph y = arctan(x) is the mirror image of the restricted tangent graph relative to the line y = x.
The domain is R and the range is [-pi/2 , pi/2].
The arccot function
We restrict the domain of the cotangent function to [0 , pi].The inverse function of that restricted cotangent function is called the arccotangent function.
We write arccot(x) or acot(x) .
The graph y = arccot(x) is the mirror image of the restricted cotangent graph relative to the line y = x.
The domain is R and the range is [0 , pi].
No period
The inverse trigonometric functions have no period!Transformations
As with the trigonometric functions, we can create related functions using simple transformations. Example:
y = 2.arcsin(x-1) comes about by moving the graph of arcsin(x) one unit to the right, and then by multiplying all the images by two. The domain is [0,2] and the range is [-pi,pi].
Sum formulas
cos(u - v)
We prove this formula using the concept of dot product of two vectors. (See theory about vectors)With u corresponds one point P(cos(u),sin(u)) on the unit circle
With v corresponds one point Q(cos(v),sin(v)) on the unit circle
The angle, corresponding with the arc QP on the circle, has a value u - v .
The dot product P.Q = 1.1.cos(u-v) .
But using the coordinates we also have P.Q = cos(u).cos(v)+sin(u).sin(v).
Hence,
cos(u-v) = cos(u).cos(v)+sin(u).sin(v)
Example:
cos(pi/3-2x) = cos(pi/3)cos(2x) + sin(pi/3)sin(2x) = 0.5 cos(2x) + 0.5 sqrt(3) sin(2x)
cos(u + v)
cos(u + v) = cos(u - (-v)) = cos(u).cos(-v)+sin(u).sin(-v)
cos(u + v) = cos(u).cos(v)-sin(u).sin(v)
Example:
cos(x + x/2) + cos(x - x/2) = cos(x)cos(x/2) + sin(x)sin(x/2) + cos(x)cos(x/2) - sin(x)sin(x/2) = 2 cos(x)cos(x/2)
sin(u - v)
sin(u - v) = cos(pi/2-(u-v)) = cos( (pi/2-u) +v )= cos(pi/2 - u).cos(v)-sin(pi/2 - u).sin(v)
sin(u - v) = sin(u).cos(v)-cos(u).sin(v)
Example:
sin(x - pi/4) = sin(x) cos(pi/4) - cos(x) sin(pi/4) = (sin(x)-cos(x))/sqrt(2)
sin(u + v)
sin(u + v) = cos(pi/2-(u+v)) = cos( (pi/2-u) -v )= cos(pi/2 - u).cos(v)+sin(pi/2 - u).sin(v)
sin(u + v) = sin(u).cos(v)+cos(u).sin(v)
tan(u + v)
sin(u + v) sin(u).cos(v)+cos(u).sin(v) tan(u+v) = ------------ = --------------------------- cos(u + v) cos(u).cos(v)-sin(u).sin(v)Dividing the dominator and denominator by cos(u).cos(v) we have
tan(u) + tan(v) tan(u+v) = ----------------- 1 - tan(u).tan(v)
Example:
tan(u) + tan(pi/4) tan(u) + 1 1 + tan(u) tan(u+pi/4) = -------------------- = -------------- = ------------- 1 - tan(u).tan(pi/4) 1 - tan(u) 1 - tan(u)
tan(u - v)
In the same way, we have
tan(u) - tan(v) tan(u-v) = ----------------- 1 + tan(u).tan(v)
Doubling formulas
sin(2u)
sin(2u) = sin(u + u) = sin(u).cos(u)+cos(u).sin(u) = 2sin(u).cos(u)
Examples
sin(x) = 2 sin(x/2).cos(x/2) sin(4x) = 2 sin(2x).cos(2x) = 4 sin(x) cos(x) cos(2x) 12 sin(8x) cos(8x) = 6 sin(16x)
cos(2u)
cos(2u) = cos(u+u) = cos(u).cos(u)-sin(u).sin(u) = cos2 (u) - sin2 (u)
cos(2u) = cos2 (u) - sin2 (u)
tan(2u)
tan(u) + tan(u) 2 tan(u) tan(2u) = ------------------ = --------------- 1 - tan(u).tan(u) 1- tan(u)tan(u)
2 tan(u) tan(2u) = ----------- 1- tan2(u)
Example:
1 cot(2x) = -------- tan(2x) 1 - tan2(x) = ------------- 2 tan(x)
Carnot formulas
1 + cos(2u) = 1+cos2 (u)-sin2 (u) = 2 cos2 (u) 1 - cos(2u) = 1-cos2 (u)+sin2 (u) = 2 sin2 (u)
1 + cos(2u) = 2 cos2 (u) 1 - cos(2u) = 2 sin2 (u)
Applications:
- From the carnot formulas, it follows that :
The period of cos(2u) = the period of cos2 (u) = the period of sin2 (u) - Factor the expression 1 + 2 cos(x) + cos(2x)
1 + 2 cos(x) + cos(2x) = 2 cos(x) + ( 1 + cos(2x)) = 2 cos(x) + 2 cos2 (x) = 2 cos(x) (1 + cos(x)) = 2 cos(x) 2 cos2 (x/2) = 4 cos(x) cos2 (x/2)
Since 2 pi is the period of (1 + 2 cos(x) + cos(2x)), it follows that the period of cos(x) cos2 (x/2) is 2pi. - Find the period of tan2(4x)
The period of tan2(4x) is equal to the period of 1+tan2(4x).
The period of 1+tan2(4x) is equal to the period of 1/ cos2(4x).
The period of 1/ cos2(4x) is equal to the period of cos2(4x).
The period of cos2(4x) is equal to the period of 0.5(1+cos(8x)).
The period of 0.5(1+cos(8x)) is equal to the period of cos(8x).
And this period is pi/4. - In triangle ABC the sides a, b, c are such that 3a = 7c en 3b = 8c.
Find tan2(A/2) without calculating A or A/2.Solution:
About the three sides we know :
a b c --- = --- = --- 7 8 3 Since similar triangles have the same angles, we can use a = 7 , b = 8 and c = 3 as sides of the triangle. From the cosine rule we can write b2 + c2 - a2 cos(A) = ------------------ = 1/2 2 b c Now we use the Carnot formulas 1 - cos(A) 2 sin2(A/2) ---------- = -------------- = tan2(A/2) = 1/3 1 + cos(A) 2 cos2(A/2)
- Find the exact value of cos(pi/12)
We know : 1 + cos(2u) = 2 cos2 (u)
Now take 2u = pi/6 radians , then u = pi/12 radians.Now the exact value of cos(pi/6) is sqrt(3)/2. We can use the Carnot formula to calculate cos(pi/12).
cos2(pi/12) = (1 + cos(pi/6))/2 = (1 + sqrt(3)/2)/2 = (2 + sqrt(3))/4 So, cos(pi/12) = (1/2). sqrt(2 + sqrt(3))
t-Formulas or Half-Angle Formulas
From the Carnot formulas we havecos(2u) = 2 cos2(u) -1 2 = ------------ - 1 1 + tan2 (u) 1 - tan2(u) = ------------- 1 + tan2 (u) We know: 2 tan(u) tan(2u)= ------------- 1 - tan2 (u) Hence, 2 tan(u) sin(2u) = ----------- 1 + tan2 (u)
Let t = tan(u) , then 1 - t2 cos(2u) = --------- 1 + t2 or 1 - tan2(u) cos(2u) = ------------- 1 + tan2 (u) 2t sin(2u) = -------- 1 + t2 or 2 tan(u) sin(2u) = ----------- 1 + tan2 (u) 2t tan(2u) = ------- 1 - t2 or 2 tan(u) tan(2u)= ------------- 1 - tan2 (u)
These three formulas are called the t-formulas or the Half-Angle Formulas
Application:
The equation of a line d is y = 3 x + 4 .
u = the angle fom the x-axis to this line.
We know that tan(u) = 3.
Line d' is the reflection of the the line y = 3 in d.
The the angle fom the x-axis to the line d' is 2u.
The slope of line d' is tan(2u).
2t 2 tan(u) 2 . 3 tan(2u) = ------- = ------------ = ---------- = - 0.75 1 - t2 1 - tan2(u) 1 - 9
Simpson formulas
We know thatcos(u + v) = cos(u).cos(v)-sin(u).sin(v) cos(u - v) = cos(u).cos(v)+sin(u).sin(v) sin(u + v) = sin(u).cos(v)+cos(u).sin(v) sin(u - v) = sin(u).cos(v)-cos(u).sin(v)and from this, we have
cos(u + v) + cos(u - v) = 2.cos(u).cos(v) cos(u + v) - cos(u - v) = -2.sin(u).sin(v) sin(u + v) + sin(u - v) = 2. sin(u).cos(v) sin(u + v) - sin(u - v) = 2. cos(u).sin(v)Let x = u + v and y = u - v
then u = (1/2)(x + y) and v = (1/2)(x - y)
Now we have
cos(x) + cos(y) = 2 cos((1/2)(x + y)) cos((1/2)(x - y)) cos(x) - cos(y) = -2 sin((1/2)(x + y)) sin((1/2)(x - y)) sin(x) + sin(y) = 2 sin((1/2)(x + y)) cos((1/2)(x - y)) sin(x) - sin(y) = 2 cos((1/2)(x + y)) sin((1/2)(x - y))
x + y x - y cos(x) + cos(y) = 2 cos ------ cos ------- 2 2 x + y x - y cos(x) - cos(y) = -2 sin ------ sin ------- 2 2 x + y x - y sin(x) + sin(y) = 2 sin ------ cos ------- 2 2 x + y x - y sin(x) - sin(y) = 2 cos ------ sin ------- 2 2
Example: The formulas can be used to factor trigonometric expressions.
cos(2x) - cos(2y) ----------------- cos(2x) + cos(2y) -2 sin(x+y) sin(x-y) = -------------------- 2 cos(x+y) cos(x-y) = - tan(x+y) tan(x-y) = tan(y+x) tan(y-x)
Factoring trigonometric forms
Many of the previous formulas can be used to factor trigonometric forms. This factoring can be done in many ways. With some examples, we'll show various methods for factorization.-
sin(2a).(1 + tan2(a)) = 2 sin(a) cos(a) . (1/cos2(a)) = 2 sin(a) / cos(a) = 2 tan(a)
-
2 sin(2a) + sin(4a) = 2 sin(2a) + 2 sin(2a).cos(2a) = 2 sin(2a). (1 + cos(2a)) = 4 sin(a) cos(a) .2 cos2(a) = 8 sin(a) cos3(a)
-
1 - tan4(a) = (1 - tan2(a)).(1 + tan2(a)) sin2(a) 1 = ( 1 - ------- ) -------- cos2(a) cos2(a) cos(2a) = -------- cos4(a)
-
cos2(a) - sin2(a) - 2 cos(a) + 1 = cos2(a) - 2 cos(a) + cos2(a) = 2 cos2(a) - 2 cos(a) = 2 cos(a) (cos(a) - 1) = -2 cos(a) (1 - cos(a)) = -2 cos(a) 2 sin2(a/2) = -4 cos(a) sin2(a/2)
-
sin(2a) (1 + 2 cos(2a) ) + 2 sin(3a) = sin(2a) + 2 sin(2a) cos(2a) + 2 sin(3a) = sin(2a) + sin(4a) + 2 sin(3a) = 2 sin(3a) cos(a) + 2 sin(3a) = 2 sin(3a) (1+ cos(a)) = 4 sin(3a) cos2(a/2)
-
cos2(a) -2 cos(a) + cos(2a) + sin2(a) = 1 + cos(2a) - 2 cos(a) = 2 cos2(a) - 2 cos(a) = 2 cos(a) (cos(a) - 1) = -2 cos(a) (1 - cos(a)) = -2 cos(a) 2 sin2(a/2) = -4 cos(a) sin2(a/2)
-
(1 - cos(4a))2 ----------------- (1 - cos2(4a)) (1 - cos(4a))2 = ---------------------------- (1 - cos(4a))(1 + cos(4a)) (1 - cos(4a)) = ------------------- (1 + cos(4a)) 2 sin2(2a) = ----------------- 2 cos2(2a) = tan2(2a)
-
cos4(a) - sin4(a) = (cos2(a) - sin2(a)) (cos2(a) + sin2(a)) = (cos2(a) - sin2(a)) = cos(2a)
-
sin(a) + sin(b) + sin(c) - sin(a+b+c) = 2 sin( (a+b)/2 ) cos( (a-b)/2 ) + 2 cos( (a+b+2c)/2 ) sin( (-a-b)/2 ) = 2 sin( (a+b)/2 ) ( cos( (a-b)/2 ) - cos( (a+b+2c)/2 ) = -4 sin( (a+b)/2 ) sin( (a+c)/2 ) sin( (-b-c)/2 ) ) = 4 sin( (a+b)/2 ) sin( (a+c)/2 ) sin( (b+c)/2 ) )
-
sin2(a) - sin2(b) - sin2(a+b) = sin2(a) - sin2(b) - (sin(a) cos(b) + cos(a) sin(b))2 = sin2(a) - sin2(b) - sin2(a) cos2(b) - cos2(a) sin2(b) - 2 sin(a) sin(b) cos(a) cos(b) = sin2(a) (1- cos2(b)) - sin2(b) (1 + cos2(a)) - 2 sin(a) sin(b) cos(a) cos(b) = sin2(a) sin2(b) - sin2(b) (1 + cos2(a)) - 2 sin(a) sin(b) cos(a) cos(b) = sin2(b) ( sin2(a) -1 - cos2(a)) - 2 sin(a) sin(b) cos(a) cos(b) = -2 sin2(b) cos2(a) - 2 sin(a) sin(b) cos(a) cos(b) = -2 sin(b) cos(a) ( sin(b) cos(a) sin(a) cos(b)) = -2 cos(a) sin(b) sin(a+b)
-
sin(2b+2c) (cos(2b) + cos(2c)) - sin(2b) - sin(2c) = sin(2b+2c) 2 cos(b+c) cos(b-c) - 2 sin(b+c)cos(b-c) = 4 sin(b+c) cos(b+c) cos(b+c) cos(b-c) - 2 sin(b+c)cos(b-c) = 2 sin(b+c) cos(b-c) ( 2 cos2(b+c) - 1) = 2 sin(b+c) cos(b-c) ( 2 cos2(b+c) - cos2(b+c) - sin2(b+c) ) = 2 sin(b+c) cos(b-c) (cos2(b+c) - sin2(b+c) ) = 2 sin(b+c) cos(b-c) cos(2b +2c)
Period of the product of two related functions
We know thatcos(u + v) = cos(u).cos(v)-sin(u).sin(v) cos(u - v) = cos(u).cos(v)+sin(u).sin(v) sin(u + v) = sin(u).cos(v)+cos(u).sin(v) sin(u - v) = sin(u).cos(v)-cos(u).sin(v) Thus cos(u + v) + cos(u - v) = 2.cos(u).cos(v) cos(u + v) - cos(u - v) = -2.sin(u).sin(v) sin(u + v) + sin(u - v) = 2. sin(u).cos(v) sin(u + v) - sin(u - v) = 2. cos(u).sin(v) or 2.cos(u).cos(v) = cos(u + v) + cos(u - v) -2.sin(u).sin(v) = cos(u + v) - cos(u - v) 2. sin(u).cos(v) = sin(u + v) + sin(u - v) 2. cos(u).sin(v) = sin(u + v) - sin(u - v)The period of cos(u).cos(v) is equal to the period of cos(u + v) + cos(u - v)
The period of sin(u).sin(v) is equal to the period of cos(u + v) - cos(u - v)
The period of sin(u).cos(v) is equal to the period of sin(u + v) + sin(u - v)
The period of cos(u).sin(v) is equal to the period of sin(u + v) - sin(u - v)
Examples:
The period of cos(2x).sin(x+3) is equal to the period of sin(3x+3) - sin(x-3)
and this period is 2 pi.
The period of cos(4x).cos(x/2) is equal to the period of cos(9x/2) + cos(7x/2)
and this period is 4pi
The general sine function
The general sine function has y = a sin(b(x-c)) + d as equation,with a, b, c not negative and a and b not zero.
We can transform many equations of trigonometric functions to the form of a general sine function by using previous formulas.
We give some examples of such transformations.
-
y = -3 sin(x) <=> y = 3 sin(x - pi)
-
y = sin(-2x) <=> y = - sin(2x) <=> y = sin(2x - pi) <=> y = sin 2(x - pi/2)
-
y = -4 sin(-3x) <=> y = 4 sin(3x)
-
y = 2 cos(3x) <=> y = 2 sin(pi/2 - 3x) <=> y = -2 sin(3x -pi/2) <=> y = 2 sin(3x - 3pi/2) <=> y = 2 sin 3(x - pi/2)
-
y = -2 cos(3x-1) <=> y = -2 sin(pi/2 -3x + 1) <=> y = 2 sin(3x - pi/2 -1) <=> y = 2 sin 3(x - (pi/6 + 1/3))
-
y = cos(3x+4) - cos(3x-4) <=> y = -2 sin(3x) sin(4) <=> y = (-2sin(4)) sin(3x)
-
y = sin(4x-3).cos(4x-3) <=> y = (1/2) sin(8x-6) <=> y = (1/2) sin(8(x-(3/4))
This transformation is somewhat more difficult than in previous examples.
a.sin(u)+b.cos(u) can be brought in the form A.sin(u-uo).
Then, the transformation to the general sine function is easy.
a.sin(u) + b.cos(u) = a( sin(u) + (b/a) cos(u) ) Take uo such that tan(uo) = - b/a = a( sin(u) - tan(uo) cos(u) ) = (a/cos(uo)) . ( sin(u).cos(uo) - sin(uo).cos(u) ) Let A = (a/cos(uo)) = A . sin(u - uo)Example
3 sin(x) - 2 cos(x) = 3( sin(x) - (2/3) cos(x) ) Let tan(uo) = 2/3 ; take uo = 0.588 = 3( sin(x) - tan(uo) cos(x) ) = (3/cos(uo)) ( sin(x) cos(uo) - cos(x) sin(uo) ) = 3.6055 sin( x - 0.588)
Trigonometric equations
Base equations
cos(u) = cos(v)
With the help of the unit circle it is easy to see thatcos(u) = cos(v) <=> (u = v + k.2pi) or (u = -v + k.2pi)
sin(u) = sin(v)
With the help of the unit circle it is easy to see thatsin(u) = sin(v) <=> (u = v + 2.k.pi) or (u = pi - v + 2.k.pi)
tan(u) = tan(v)
With the help of the unit circle it is easy to see thattan(u) = tan(v) <=> (u = v + k.pi) on condition that tan(u) and tan(v) exist
cot(u) = cot(v)
With the help of the unit circle it is easy to see thatcot(u) = cot(v) <=> (u = v + k.pi) on condition that cot(u) and cot(v) exist
Reducing to base equations
Example 1cos(2x) = cos(pi-3x) <=> 2x = (pi-3x) + 2.k.pi or 2x = -(pi-3x) + 2.k'.pi <=> 5x = pi + 2.k.pi or -x = -pi + 2.k'.pi <=> x = pi/5 + 2.k.pi/5 or x = pi - 2.k'.piExample 2
tan(x-pi/2) = tan(2x) <=> (x-pi/2) = 2x + k.pi <=> -x = pi/2 + k.pi <=> x = -pi/2 - k.pi ( for these values tan(x-pi/2) and tan(2x)) exist)Example 3
cos(x) = -1/3 <=> cos(x) = cos(1.91) <=> x = 1.91 +2.k.pi or x = -1.91 - 2.k.piExample 4
sin(2x) = cos(x-pi/3) <=> cos(pi/2 - 2x) = cos(x-pi/3) <=> pi/2 - 2x = x - pi/3 + 2.k.pi or pi/2 - 2x = - x + pi/3 + 2.k'.pi <=> -3x = - pi/2 - pi/3 + 2.k.pi or -x = -pi/2 + pi/3 + 2.k'.pi <=> x = pi/6 + pi/9 + 2.k.pi/3 or x = pi/2 - pi/3 - 2.k'.pi <=> x = 5pi/18 + 2.k.pi/3 or x = pi/6 - 2.k'.piExample 5
3 sin(2x) = cos(2x) <=> 3 tan(2x) = 1 <=> tan(2x) = 1/3 <=> tan(2x) = tan(0.32) <=> 2x = 0.32 + k pi <=> x = 0.16 + k pi/2Example 6
tan(2x) . cot( x + pi/2) = 1 <=> tan(2x) = tan( x + pi/2) <=> 2x = x + pi/2 + k.pi <=> x = pi/2 + k.pi (on condition that tan(2x) and cot( x + pi/2) exist)But cot( x + pi/2) does not exist for x = pi/2 + k.pi !!!!!
So, tan(2x) . cot( x + pi/2) = 1 has no solutions !
The expression " on condition that ...." is not redundant!
Using an additional unknown
Example 12sin2 (2x)+sin(2x)-1=0 <=> (let t = sin(2x) ) 2t2 + t - 1 = 0 <=> t = 0.5 or t = -1 <=> sin(2x) = 0.5 or sin(2x) = -1 <=> sin(2x) = sin(pi/6) or sin(2x) = sin(-pi/2) <=> 2x = pi/6 +2.k.pi or 2x = pi - pi/6 +2.k.pi or 2x = -pi/2 +2.k.pi or 2x = pi + pi/2 +2.k.pi <=> x = pi/12 + k.pi or x = 5pi/12 + k.pi or x = -pi/4 + k.pi or x = 3pi/4 + k.piSometimes it is convenient to view these solutions on the unit circle.
Example 2
cos 10x + 7 = 8 cos 5x <=> cos 10x - 8 cos 5x + 7 =0 <=> 1 + cos 10x - 8 cos 5x + 6 =0 <=> 2 cos2 5x - 8 cos 5x + 6 =0 <=> cos2 5x - 4 cos 5x + 3 = 0 Let t = cos 5x t2 - 4t + 3 = 0 <=> t = 3 or t = 1 <=> cos 5x = 1 <=> cos 5x = cos 0 <=> 5x = 2kpi <=> x = 2kpi / 5Examples
In the same way, the following equations can be solved using an additional unknown.
tan2 (3x)+tan(3x)=0 sin2 (x)(sin(x)+1)-0.25(sin(x)+1) = 0 cos(2x)+sin2 (x) = 0.5 tan(2x)-cot(2x) = 1Check your results by plotting graphs.
Using factoring
Example 13.sin(2x)-2.sin(x) = 0 <=> 6sin(x)cos(x)-2.sin(x) = 0 <=> 2.sin(x).(3cos()-1) = 0 <=> sin(x) = 0 or cos(x) = 1/3 <=> x = k.pi or x = 1.23 + 2.k.pi or x = -1.23 + 2.k'.piExamples
In the same way, the following equations can be solved using factoring.
tan(x)tan(4x)+tan2 (x) = 0 sin(7x)-sin(x) = sin(3x) cos(4x) + cos(2x) + cos(x) = 0 sin(5x)+sin(3x) = cos(2x)-cos(6x)Check your results by plotting graphs.
The equation a.sin(u)+b.cos(u) = c
First Method
First we'll show that a.sin(u)+b.cos(u) can be transformed in the formA.sin(u-uo) or in the form A.cos(u-uo) .
a.sin(u) + b.cos(u) = a( sin(u) + (b/a) cos(u) ) Take uo such that tan(uo) = - b/a = a( sin(u) - tan(uo) cos(u) ) = (a/cos(uo)) . ( sin(u).cos(uo) - sin(uo).cos(u) ) Let A = (a/cos(uo)) = A . sin(u - uo) = A . cos(pi/2 - u + uo) = A . cos(u - uo')Example
3 sin(x) - 2 cos(x) = 3( sin(x) - (2/3) cos(x) ) Let tan(uo) = 2/3 ; take uo = 0.588 = 3( sin(x) - tan(uo) cos(x) ) = (3/cos(uo)) ( sin(x) cos(uo) - cos(x) sin(uo) ) = 3.6055 sin( x - 0.588) or = 3.6055 cos( x - 2.1598)Plot the graph of 3 sin(x) - 2 cos(x) and the graph of 3.6055 sin( x - 0.588)
With this method we can solve the equation
a.sin(u)+b.cos(u) = c
Example
3.sin(2x)+4.cos(2x) = 2 <=> sin(2x) + 4/3 .cos(2x) = 2/3 Let tan(t) = 4/3 <=> sin(2x) + tan(t) .cos(2x) = 2/3 <=> sin(2x)cos(t)+cos(2x)sin(t) = 2/3.cos(t) <=> sin(2x+t) = 2/3.cos(t) since 2/3.cos(t) = 0.4 <=> sin(2x+0.927) = sin(0.39) <=> 2x + 0.927 = 0.39 +2.k.pi or 2x + 0.927 = pi - 0.39 +2.k'.pi <=> ....
Second Method
Using the t-formulasExample
3 sin(2x) + 4 cos(2x) = 2 Let tan(x) = t <=> 2 t 1 - t2 3 ------- + 4 -------- = 2 1 + t2 1 + t2 <=> 6 t + 4 - 4 t2 = 2 + 2 t2 <=> 6 t2 - 6 t - 2 = 0 <=> 3 t2 - 3 t -1 = 0 <=> t = 1.26 or t = -0.26 <=> tan(x) = 1.26 or tan(x) = -0.26 <=> x = 0.9 + k pi or x = -0.26 + k pi
Homogeneous equations
An equation is homogeneous in a and b if and only if we obtain an equivalent equation when we replace a and b by ra and rb (r is not 0). Example: a3 x2 +5 a.b2 x +3 a2.b = 0 is an equation in x which is homogeneous in a en b. Now, we have in view the equations which are homogeneous in sin(u) and cos(u).
Procedure
- Reduce the equation to the form F = 0. If possible, use factoring to the left hand side and solve the simple parts.
- Divide the remaining equation through by a suitable power of cos(u), such that tan(u) appears everywhere.
- Let t = tan(u) and solve the algebraic equation.
- Return to tan(u)
2.cos3 (x)+2.sin2 (x)cos(x) = 5.sin(x)cos2 (x) <=> cos(x).(2.cos2 (x)+2.sin2 (x) - 5.sin(x)cos(x)) = 0 <=> The simple part cos(x) = 0 gives us x = pi/2 + k.pi In the second part, we divide both sides by cos2 (x). Then we have 2.tan2 (x) - 5.tan(x) +2 = 0 Let t = tan(x) <=> 2.t2 - 5 t + 2 = 0 <=> t = 0.5 or t = 2 <=> tan(x) = 0.5 or tan(x) = 2 <=> x = 0.464 +k.pi or x = 1.107 +k.pi
Other equations
Some equations can be solved, in an appropriate manner, by combining various discussed methods. In addition, some trigonometric formulas are often used to transform the equation into a suitable form. Moreover, experience and understanding play an important role by solving difficult equations. We give a not obvious example.
1/sin(x) + 1/cos(x) = sqrt(3)
1/sin(x) + 1/cos(x) = sqrt(3) sin(x) + cos(x) <=> ------------------- = sqrt(3) sin(x) cos(x) <=> sin(x) + cos(x) = sqrt(3) sin(x) cos(x) (1) --------------------------------------------------------If we square both sides of the equation (1), only the product sin(x)cos(x) occurs. Then, we can find the value of sin(x)cos(x) and with this we'll simplify the equation (1).
From (1) it follows ( sin(x) + cos(x))2 = 3 (sin(x) cos(x))2 <=> 1 + 2 sin(x) cos(x) = 3 (sin(x) cos(x))2 Let y = sin(x) cos(x) <=> 3 y2 - 2 y -1 = 0 <=> y = 1 or -1/3 <=> sin(x) cos(x) = 1 or sin(x) cos(x) = -1/3 If sin(x) cos(x) = 1 then sin(2x) = 2 and this is impossible. Conclusion: From (1) it follows that sin(x) cos(x) = -1/3 (2) ---------------------------------------------------------- Now, we use this result. We bring (2) in (1) sin(x) + cos(x) = - sqrt(3)/3 (3) If (1) is true, then (2) is true and thereby (3) is true. Now, we show that the reverse is true. We start with (3). sin(x) + cos(x) = - sqrt(3)/3 => (sin(x) + cos(x)2 = 1/3 => 1 + 2 sin(x) cos(x) = 1/3 => sin(x) cos(x) = -1/3 and this is (2) So, If (3) is true , then (2) is true and thereby (1) is true. Conclusion:A second method
(1) and (3) are equivalent equations. We'll solve (3) now. ------------------------------------------------------------- sin(x) + cos(x) = - sqrt(3)/3 <=> cos(pi/2 -x) + cos(x) = - sqrt(3)/3 <=> 2 cos(pi/4) cos (pi/4-x) = - sqrt(3)/3 <=> sqrt(2) cos (pi/4-x) = - sqrt(3)/3 <=> cos (pi/4-x) = -1/ sqrt(6) Let a = arccos(-1/sqrt(6)) <=> cos (pi/4-x) = cos(a) <=> pi/4 - x = a + 2 k pi or pi/4 - x = -a + 2 k pi <=> x = pi/4 - a + 2 k pi or x = pi/4 + a + 2 k pi The solutions of the given equation are the values pi/4 - a + 2 k pi en pi/4 + a + 2 k pi with a = arccos(-1/sqrt(6))
Many equations can be solved in different ways. We'll show an alternative way to solve the equation
1/sin(x) + 1/cos(x) = sqrt(3)The period of the function on the left-hand side is 2 pi. If we have the solutions to the equation in [0, 2pi], then we know all solutions.
First, we calculate the solutions in [0, 2pi]. Since the right-hand side of the equation is positive, solutions are only possible when the left-hand side is positive too.
By plotting the function 1/sin(x) + 1/cos(x), we see that the image is positive in the intervals (0,pi/2) ; (3pi/4, pi) and (3pi/2, 7pi/4).
The solutions can only occur in these intervals. If we restrict the values of x to these intervals, then both sides of the equation are positive and we can write
1/sin(x) + 1/cos(x) = sqrt(3) <=> (1/sin(x) + 1/cos(x))2 = 3 sin(x) + cos(x) <=> (------------------)2 = 3 sin(x) cos(x) <=> 1 + 2 sin(x) cos(x) = 3 sin2(x) cos2(x) Let y = sin(x) cos(x) <=> 3 y2 - 2 y - 1 = 0 <=> y = 1 or y = -1/3 Case y = 1 sin(x) cos(x) = 1 <=> 2 sin(x) cos(x) = 2 <=> sin(2x) = 2 In this case, there are no solutions Case y = -1/3 2 sin(x) cos(x) = -2/3 <=> sin(2x) = -2/3 let b = arcsin(-2/3) ; b = -0.7297 <=> sin(2x) = sin(b) <=> 2x = b + 2 k pi or 2x = (pi-b) + 2kpi <=> x = b/2 + k pi or x = pi/2 - b/2 + k pi Now, we will take only the x-values located in the intervals (0,pi/2) ; (3pi/4, pi) en (3pi/2, 7pi/4). There are 2 solutions : x = b/2 + pi = 2.7767 and x = pi/2 - b/2 + pi = 3pi/2 - b/2 = 5.077 All solutions to the equation 1/sin(x) + 1/cos(x) = sqrt(3) are b/2 + pi + 2 k pi en 3pi/2 - b/2 + 2 k pi
Trigonometric inequalities
Conventions
k is an integer.'=<' means equal or less than
'>=' means equal or greater than
Examples
- sin(x/2) > 1/2
We draw the solutions for (x/2) on the unit circle.
Now, we see that:
sin(x/2) > 1/2 <=> pi/6 + 2 k pi < x/2 < 5pi/6 + 2 k pi <=> pi/3 +4 k pi < x < 5pi/3 + 4 k pi
For each k, we have an open interval with solutions.
The solution set V is the union of all these open intervals.V = { U (pi/3 +4 k pi , 5 pi/3 + 4 k pi) | k in Z } k - tan(2x) < 1/3
We draw the solutions for (2x) on the unit circle.
Now, we see that:
tan(2x) < 1/3 <=> -pi/2 + k pi < 2x < 0.32 + k pi <=> -pi/4 + k pi/2 < x < 0.16 + k pi/2
For each k, we have an open interval with solutions.
The solution set V is the union of all these open intervals.V = { U (-pi/4 + k pi/2 , 0.16 + k pi/2) | k in Z } k - tan(2x + pi/5 ) < 1/3
This is a variation on the previous example.
The figure is the same as the previous one, but now it gives the solutions for (2x + pi/5).
Now we have :tan(2x + pi/5 ) < 1/3 <=> -pi/2 + k pi < 2x + pi/5 < 0.32 + k pi <=> -pi/2 -pi/5 + k pi < 2x < 0.32 - pi/5 + k pi <=> -7 pi/20 + k pi/2 < x < 0.16 -pi/10 + k pi/2
The solution set V is:V = { U (-7 pi/20 + k pi/2 , 0.16 - pi/10 + k pi/2) | k in Z } k - 2 sin2(x) - 3 sin(x) + 1 = < 0
Let t = sin(x) .
2 t2 - 3 t + 1 < 0 <=> 2 (t - 1)(t - 1/2) < 0 A sign study of the left hand side gives <=> 1/2 =< t =< 1 <=> 1/2 =< sin(x) =< 1 Draw the solutions for x on the unit circle. We see that: <=> pi/6 + 2 k pi =< x =< 5pi/6 + 2 k pi
The solution set isV = { U [pi/6 + 2 k pi ; 5pi/6 + 2 k pi] | k in Z } k - Another method to solve 2 sin2(x) - 3 sin(x) + 1 = < 0
One can also factor the left hand side directly and investigate the sign in a period-interval.
2 sin2(x) - 3 sin(x) + 1 <=> 2 (sin(x) - 1)(sin(x) - 1/2)
We take a simple period-interval [0,2pi). We investigate the sign of each factor.x 0 pi/6 pi/2 5pi/6 pi 2pi --------------------------------------------------------------- sin(x)-1 - - - - - - 0 - - - - - - - - - - - - - - --------------------------------------------------------------- sin(x)-1/2 - - 0 + + + + + + 0 - - - - - - - - - - --------------------------------------------------------------- product + + 0 - - - 0 - - 0 + + + + + + + + + + ---------------------------------------------------------------
The solutions in the period-interval are
pi/6 =< x =< 5pi/6The solution set is :
V = { U [pi/6 + 2 k pi ; 5pi/6 + 2 k pi] | k in Z } k - cosec(x) < sec(x)
We investigate first the inequality in the period-interval [0,2pi).
The values 0 ; pi/2 ; pi ; 3pi/2 are no solutions . We investigate the other values of x in each quadrant.
- First quadrant
cosec(x) < sec(x) <=> 1/ sin(x) < 1 / cos(x) now we have sin(x).cos(x) > 0 <=> cos(x) < sin(x)
The solution set is ( pi/4, pi/2 ). - Second quadrant
Now we have cos(x) < 0 and sin(x) > 0. There are no solutions.
- Third quadrant
cosec(x) < sec(x) <=> 1/ sin(x) < 1 / cos(x) now we have sin(x).cos(x) > 0 <=> cos(x) < sin(x)
The solution set is ( pi , 5 pi/4 ) - Fourth quadrant Now we have cos(x) > 0 and sin(x) < 0.
The solution set is ( 3 pi/2 , 2pi)
(pi/4 + 2 k pi , pi/2 + 2 k pi ) (pi + 2 k pi , 5pi/4 + 2 k pi ) (3pi/2 + 2 k pi , 2 pi + 2 k pi ) with k in Z
- First quadrant
- First transform, then solve.
cot(x) + 1 ------------- > 0 sin (x) cot(x) + 1 <=> ------------- > 0 and sin (x) not 0 sin (x) cos(x) + sin(x) <=> ------------------- > 0 and sin (x) not 0 sin2(x) <=> cos(x) + sin(x) > 0 and sin (x) not 0 <=> sin(x) + sin(pi/2 -x) > 0 and sin (x) not 0 with Simpson's formulas <=> 2 sin(pi/4) cos( x - pi/4) > 0 <=> cos( x - pi/4) > 0 and sin (x) not 0
Using the unit circle we see that<=> -pi/2 + 2k pi < x - pi/4 < pi/2 + 2k pi and sin (x) not 0 <=> -pi/4 + 2k pi < x < 3pi/4 + 2k pi and sin (x) not 0
The solution set V is the union of the open intervalsV = { U (-pi/4 + 2k pi , 3pi/4 + 2k pi ) | k in Z } \ { k pi | k in Z } k
Cyclometric equations
All equations are solved using the same method. We replace the equation successively by a necessary condition. This means that the values that we find are not necessary solutions. Afterwards, we have to test these values against the initial equation. The false or parasitic values must be deleted.Example 1
arcsin(2x) = pi/4 + arcsin(x) <=> / arcsin(2x) = a | arcsin(x) = b (1) \ a = pi/4 + b / sin(a) = 2x => | sin(b) = x \ a = pi/4 + b => / sin(pi/4 + b) = 2x \ sin(b) = x sum formulas => / cos(b) + sin(b) = 2x.sqrt(2) \ sin(b) = x => / cos(b) = 2x.sqrt(2) - x \ sin(b) = x => (2x.sqrt(2) - x)2 + x2 = 1 => .... => x = +0.4798 or x = -0.4798We test these values against the initial equation. The only solution is 0.4798.
The other x-value is false or parasitic.
Example 2
arctan(x+1) = 3.arctan(x-1) <=> / arctan(x+1) = a | arctan(x-1) = b \ a = 3 b => / tan(a) = x + 1 | tan(b) = x - 1 \ a = 3 b => / tan(3b) = x+ 1 \ tan(b) = x - 1 3 tan(b) - tan3(b) but tan(3b) = -------------------- 1 - 3.tan2(b) 3(x-1) - (x-1)3 => x+1 = -------------------- 1 - 3 (x-1)2 => (x+1) (1 - 3 (x-1)2) = 3(x-1) - (x-1)3 => ... => x = 0 or x = sqrt(2) or x = -sqrt(2)We test these values against the initial equation. The only solution is sqrt(2).
The other x-values are false or parasitic.
Example 3
arctan(x) + arctan(2x) = pi/4 <=> / arctan(x) = a | arctan(2x) = b \ a + b = pi/4 => / x = tan(a) | 2x = tan(b) \ a + b = pi/4 => / x = tan(a) \ 2x = tan(pi/4-a) 1 - tan(a) but tan(pi/4-a) = ---------------- since tan(pi/4) = 1 1 + tan(a) 1 - x => 2x = ---------- 1 + x => ... => x = (-3+sqrt(17))/4 or x = (-3-sqrt(17))/4We test these values against the initial equation. The only solution is (-3+sqrt(17))/4.
The other x-value is false or parasitic.
Example 4
arctan( (x+1)/(x+2) ) - arctan ( (x-1)/(x-2) ) = arccos( 3/sqrt(13) ) / arctan( (x+1)/(x+2) ) = a <=> | arctan( (x-1)/(x-2) ) = b | arccos( 3/sqrt(13) ) = c \ a - b = c / tan(a) = (x+1)/(x+2) => | tan(b) = (x-1)/(x-2) | cos(c) = 3/sqrt(13) \ a - b = c tan(a) - tan(b) but tan(a-b) = ------------------ and after calculation one finds 1 + tan(a) tan(b) -2 x / tan(c) = ---------------- => | 2 x2 - 5 | \ cos(c) = 3/sqrt(13) From the last equation it follows 1 + tan2(c) = 1/cos2(c) = 13/9 => tan2(c) = 4/9 There are now two cases First case : -2 x / tan(c) = ---------------- => | 2 x2 - 5 | \ tan(c) = 2/3 => .... => x = 1 or x = -5/2 Second case: -2 x / tan(c) = ---------------- => | 2 x2 - 5 | \ tan(c) = - 2/3 => .... => x = -1 or x = 5/2We test these values against the initial equation. The only solutions are 1 and -5/2.
The other x-values are false or parasitic.
Calculations with cyclometric functions
In the following examples, 'less or equal' is written as ' =< '.Example 1
________ | 2 \| 1 - p Show that cot(arcsin(p)) = ----------- p
let b = arcsin(p) ,then sin(b) = p with b in [-pi/2 , pi/2]. So, cos(b) = sqrt( 1 - p2) and ________ | 2 \| 1 - p cot(arcsin(p)) = cot(b) = --------- pExample 2
cos(arctan(x)) + x sin(arctan(x)) = x
Let u = arctan(x) ; Since x > 0 is u in (0, pi/2) and tan(u) = x 1 + tan2(u) = 1 + x2 <=> 1/ cos2(u) = 1 + x2 <=> cos(u) = 1/ sqrt(1+x2) - - - - - - - - - - - - - - - - sin(u) = tan(u) . cos(u) <=> sin(u) = x / sqrt(1+x2) - - - - - - - - - - - - - - - - cos(arctan(x)) + x sin(arctan(x)) = x <=> cos(u) + x sin(u) = x <=> 1/ sqrt(1+x2) + x2 / sqrt(1+x2) = x <=> 1 + x2 = x . sqrt(1+x2) <=> (1 + x2)2 = x2 . (1+x2) <=> 1 + x2 = x2 and this equation has no solutions.
Example 3
Find the domain of arccos(arcsin(x))
x belongs to the domain of arccos(arcsin(x)) <=> - 1 =< arcsin(x) =< 1 and since the sine function increases in [-1,1] <=> sin(-1) =< x =< sin(1)Conclusion : The domain of arccos(arcsin(x)) is [ -sin(1), sin(1) ]
Example 4
Find the domain of arcsin(arccos(x))
x belongs to the domain of arcsin(arccos(x)) <=> - 1 =< arccos(x) =< 1 <=> 0 =< arccos(x) =< 1 and since the cosine function decreases in [0,1] <=> cos(0) >= x >= cos(1)Conclusion : The domain of arcsin(arccos(x)) is [cos(1),1]
Example 5
Find the values of a such that the domain D of f(x) is as large as possible.
Calculate the value of a such that the domain D = [-0.5 ; 0.5]
x belongs to the domain of arccos(a. arcsin(x)) <=> - 1 =< a . arcsin(x) =< 1 <=> -1/a =< arcsin(x) =< 1/a (*) First case: 1/a =< pi/2 Now (*) is equivalent with sin(-1/a) =< x =< sin(1/a) (**) In this case the domain D is maximum if and only if sin(1/a)is maximum. Then a = 2/pi. Second case: 1/a > pi/2 Now (*) is equivalent with -pi/2 =< arcsin(x) =< pi/2 In this case the domain D is always [-1,1]. Conclusion: If a =< 2/pi , then the domain D is as large as possible. From (**) it follows that domain D is [-1/2,1/2] <=> sin(1/a) = 1/2 <=> a = 6/piRemark : you can illustrate and explore many of the previous steps with a function plotter.
Example 6
1 arctan -------------- = arctan(1/n) - arctan(1/(n+1)) n2 + n + 1
We note first that
1 1 1 0 < ------------- < -------- < --- =< 1 n2 + n + 1 n+1 n and therefore 1 1 1 0 < arctan ------------- < arctan -------- < arctan --- =< pi/4 n2 + n + 1 n+1 n This means that the expressions 1 arctan ------------- and arctan(1/n) - arctan(1/(n+1)) n2 + n + 1 are in (0, pi/4 ] for all n. Therefore, the expression 1 arctan -------------- = arctan(1/n) - arctan(1/(n+1)) n2 + n + 1 is equivalent with 1 ------------- = tan ( arctan(1/n) - arctan(1/(n+1)) ) n2 + n + 1 We simplify the right side tan ( arctan(1/n) - arctan(1/(n+1)) ) tan (arctan(1/n)) - tan (arctan(1/(n+1))) = ---------------------------------------------- 1 + tan (arctan(1/n)) . tan (arctan(1/(n+1))) 1/n - 1/(n+1) = ------------------------- 1 + (1/n)(1/(n+1)) 1 = ------------- n2 + n + 1Remark : you can illustrate and explore many of the previous steps with a function plotter.
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