Basic Trigonometric Identities
1. $sin^2 x + cos^x =1$ for all $x \in R$
2. $sec^2x =1 + tan^2x , x \neq (2n+1) \frac {\pi}{2}$
3. $cosec^2 x =1 + cot^2 x, x \neq n \pi$
4. $ sin (2n \pi + x) = sin x$
5. $ sin (n \pi +x) =(-1)^n sin x$
6. $ cos(2n \pi + x) = cos x$
7. $ tan (n \pi + x) = tan x$
8. $ tan (n \pi - x) = -tan x$
Trigonometric Identities of Sum and difference of angles
Sin and cos function
- $cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$
- $cos(A-B)=cos(A)cos(B)+sin(A)sin(B)$
- $cos(\frac {\pi}{2} -A)=sin(A)$
- $sin(\frac {\pi}{2} -A)=cos(A)$
- $sin(A+B)=sin(A)cos(B)+sin(B)cos(A)$
- $sin(A-B)=sin(A)cos(B)-sin(B)cos(A)$
Similarly we can have defined other sin and cos sum and differences
Tan and cot functions
- If none of the angles x, y and (x + y) is an odd multiple of π/2
$tan(A+B)=\frac{tan(A)+tan(B)}{1-tan(A)tan(B)}$
$tan(A-B)=\frac{tan(A)-tan(B)}{1+tan(A)tan(B)}$
- If none of the angles x, y and (x + y) is an multiple of π
$cot(A+B)=\frac{cot(A)cot(B)-1}{cot(A)+cot(B)}$
$cot(A-B)=\frac{cot(A)cot(B)+1}{cot(B)-cot(A)}$
Some other Important functions
- $cos(A)+cos(B)=2cos\frac{A+B}{2}cos\frac{A-B}{2}$
- $cos(A)-cos(B)=-2sin\frac{A+B}{2}sin\frac{A-B}{2}$
- $sin(A)+sin(B)=2sin\frac{A+B}{2}cos\frac{A-B}{2}$
- $sin(A)-sin(B)=2cos\frac{A+B}{2}sin\frac{A-B}{2}$
Trigonometric Identities of Multiple Angles
Now lets explore the multiple of x. These all can be proved from above equations
Double of x
$cos2x=cos^{^{2}}x-sin^{^{2}}x=2cos^{^{2}}x-1=1-2sin^{^{2}}x=\frac{1-tan^{^{2}}x}{1+tan^{^{2}}x}$
$sin2x=2cos(x)sin(x)=\frac{2tan(x)}{1+tan^{^{2}}x}$
$tan2x=\frac{2tan(x)}{1-tan^{^{2}}x}$
Triple of x
$sin3x=3sin(x)-4sin^{3}x$
$cos3x=4cos^{3}x-3cos(x)$
$tan(3x)=\frac{3tanx-tan^{^{3}}x}{1-3tan^{^{2}}x}$
Conditional Identities
if $A + B + C=180^{\circ}$ and A,B and C are positive angles then
1. $ sin 2A + sin 2B + sin 2C= 4 Sin(A) Sin (B) Sin (C)$
2. $ cos 2A + cos 2B + cos 2C= -1- 4 cos(A) cos (B) cos (C)$
3. $ sin A + sin B + sin C= 4 Sin \frac {A}{2} Sin \frac {B}{2} Sin \frac {C}{2}$
4. for no right angle in A,B,C $ tan A + tan B + tan C=tan(A) tan (B) tan (C)$
5. $ tan(\frac {A}{2}) tan (\frac {B}{2}) +tan(\frac {B}{2}) tan (\frac {C}{2}) + tan(\frac {A}{2}) tan (\frac {C}{2})=1$
6. $ cot A cot B + cot B cot C + cot C cot A=1$
Half Angle Identities
The ones for sine and cosine take the positive or negative square root depending on the quadrant of the angle $\frac {\theta}{2}$. For example, if $\frac {\theta}{2}$ is an acute angle, then the positive root would be used.
1. $sin \frac {\theta}{2}=\pm \sqrt {\frac {1-cos \theta }{2}}$
2. $cos \frac {\theta}{2}=\pm \sqrt {\frac {1+cos \theta }{2}}$
3. $tan \frac {\theta}{2}= \frac { sin \theta}{1 + cos \theta } = \frac { 1- cos \theta}{sin \theta }$
4. $ sin \theta = \frac {2 tan (\theta /2)}{1 + tan^2 (\theta /2)}$
5. $ cos \theta = \frac {1 - tan^2 (\theta /2)}{1 + tan^2 (\theta /2)}$
6. $ tan \theta = \frac {2 tan (\theta /2))}{1 - tan^2 (\theta /2)}$
Trigonometric Ratios for Some Non-common angles
1. $ sin 15^{\circ}=\frac {\sqrt {3} -1}{2 \sqrt {2}} =cos 75^{\circ}$
2. $ cos 15^{\circ}=\frac {\sqrt {3} +1}{2 \sqrt {2}} =sin 75^{\circ}$
3. $ tan 15^{\circ}= \frac {\sqrt {3} -1}{\sqrt {3} +1}$
4. $sin 18^{\circ}=\frac {\sqrt {5} -1}{4}=cos 72^{\circ}$
5. $ tan 18^{\circ}= 2 - \sqrt {3}$
6. $ sin 22 \frac {1}{2}^{\circ}=\frac {\sqrt {2 -\sqrt {2}}}{2}$
7. $ cos 22 \frac {1}{2}^{\circ}=\frac {\sqrt {2 +\sqrt {2}}}{2}$
8. $ tan 22 \frac {1}{2}^{\circ}= \sqrt {2} -1$
Solved Examples of trigonometric identities
1. Prove that $tan 70^{\circ} = 2tan 50^{\circ} + tan 20^{\circ} $
Solution
$tan 70^{\circ}= tan (50 + 20)$
$= \frac {tan 50 + tan 20}{1- tan 50 tan 20}$
or
$ tan 70(1- tan (50) tan(20)) = tan (50) + tan (20)$
$tan 70 = tan(70) tan(50) tan(20) +tan 50 + tan 20$
Now tan 70= tan (90-20) =cot 20
$tan 70 = cot 20 tan 50 tan 20 +tan 50 + tan 20$
$tan 70^{\circ} = 2 tan 50^{\circ} + tan 20^{\circ} $
2.Prove that $cot^4 x + cot^2 x = cosec^4 x -cosec^2x$
Solution
LHS = $cot^4 x + cot^2 x$
$=(cot^2x)^2 + cot^2x$
$=(cosec^2 x -1)^2 + cosec^2 x -1$
$=cosec^4 x -cosec^2x$
3. Prove that $ \frac {tan(A+B)}{cot(A-B)} = \frac {sin^2 A - Sin^2 B}{cos^2 A - cos^2 B}$
Solution
LHS= $ \frac {tan(A+B)}{cot(A-B)}$
$=\frac {sin(A+B) sin(A-B)}{cos(A+B)cos(A-B}$
$=\frac {sin^2 A - Sin^2 B}{cos^2 A - cos^2 B}$
4.Prove that
$(sin3A + sin A)sin A + (cos 3A -cos A)Cos A=0$
Solution
LHS=$(sin3A + sin A)sin A + (cos 3A -cos A)Cos A$
$=(2 sin \frac {3A+A}{2} cos \frac {3A-A}{2})sin A + (-2sin \frac {3A+A}{2} sin \frac {3A-A}{2})cos A$
$=2 sin 2A cos A sin A - 2 Sin2A sin A cos A =0$
Practice Questions
1.Prove that $cos (2\pi/15) cos (4\pi/15) cos (8\pi/15) cos (16\pi/15) = \frac {1}{16}$
2.Prove that $cos 4x + cos 8x - 2cos6x cos2x=0$
3.Prove that $tan20^{\circ}tan40 ^{\circ}tan80^{\circ}=tan60^{\circ}$
4.Prove that $sin^A =cos^2(A-B) + cos^2B - 2cos(A-B) cosA cos B$
5.Prove that $sin 3x + sin 2x – sin x = 4sin x cos(x/2)cos(3x/2)$
Related Topics
Related Links on Trigonometry
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Class 11 Maths
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