Trigonometric Identities | IIT JEE Trigonometric Series Formula

What is an Identity?

As the name suggests, an equation is said to be an identity if it holds good in every case. A trigonometric equation is hence an identity if it is true for all values of the angle or angles involved. A given identity may be established by reducing either side to the other one, or reducing each side to the same expression, or any other convenient modification.

Students must learn all the standard trigonometric identities as the examiners often frame questions based on the application of these identities. We discuss here some of the typical trigonometric formulae and identities:

For any angles A, B, C, following results hold true:

• sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C

• cos (A + B +C) = cos A cos B cos C- cos A sin B sin C – sin A cos B sin C – sin A sin B cos C

• tan (A + B + C) = (tan A + tan B + tan C – tan A tan B tan C) / (1- tan A tan B- tan B tan C – tan A tan C)/    

• cot (A + B + C) = (cot A cot B cot C – cot A – cot B – cot C) / (cot A cot B + cot B cot C + cot A cot C)
   

Watch this Video for more reference
 

If A, B, C are angles of a triangle (or A + B + C = π):

• sin A cos B cos C + cos A sin B cos C + cos A cos B sinC = sin A sin B sin C

• cos A sin B sin C + sin A cos B sin C + sin A sin B cos C = 1 + cos A cos B cos C

• tan A + tan B + tan C = tan A tan B tan C

• cot B cot C + cot C cot A + cot A cot B = 1

• tan B/2 tan C/2 + tan C/2 tan A/2 + tan A/2 tan B/2 = 1

• cot A/2 +  cot B/2 + cot C/2 = cot A/2 cot B/2 cot C/2

• sin 2A + sin 2B + sin 2C = 4sin A sin B sin C

• cos 2A + cos 2B + cos 2C = -1 - 4cos A cos B cos C

• cos2A + cos2B + cos2C = 1 - 2cos A cos B cos C

• sin A + sin B + sin C = 4 cos A/2 cos B/2 cos C/2

• cos A + cos B + cos C = 1 + 4 sin A/2 sin B/2 sin C/2
   

Illustration: If x + y + z = xyz, prove that

x/ (1-x²) + y/ (1-y²)  + z/ (1-z²) = 4xyz/ (1-x²)(1-y²)(1-z²)

Solution: Let x = tan A, y = tan B, z = tan C

Then tan A + tan B + tan C = tan A. tan B. tan C.

⇒ A + B + C = π  

Hence, we have tan (2A + 2B) = tan (2π – 2C)

or tan (2A + 2B) = - tan2C

This gives (tan 2A + tan2B) / (1- tan 2A tan 2B) = -tan2C

or tan2A + tan2B + tan2C = tan 2A.tan 2B.tan 2C

Also since tan 2A = 2 tan A / (1-tan²A), using this the above equation becomes

2x/ (1-x²) + 2y/ (1-y²) + 2z/ (1-z²) = 8xyz / (1-x²) (1-y²) (1-z²)

This gives x/ (1-x²) + y/ (1-y²) + z/ (1-z²) = 4xyz / (1-x²) (1-y²) (1-z²).

Illustration: If A + B + C = 180⁰, then prove that

sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C.        

Solution: sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C

Taking the LHS first we have

sin (B + C – A) + sin (C + A – B) + sin (A + B – C)

= sin (π – A – A) + sin (π – B – B) + sin (π – C – C) (Since A + B + C = π)

= sin 2A + sin 2B + sin 2C

= 4 sin A sin B sin C (using the above identities).
 

Trigonometric Series

Sometimes questions are also framed on trigonometric series like finding the product of n terms of sine or cosine series. There are particular methods for dealing with such questions. Like if we have a cosine series in its product form where the angles are in G.P. with common ratio 2 then both the numerator and denominator should be multiplied by 2 sin (least angle). We illustrate some of the examples based on these concepts:

Illustration: Simplify the product cos A. cos 2A. cos 2²A …. Cos 2^n–1A.

Solution: cos A× cos 2A….cos 2^n–1A

= 1/2 sin A. (sin 2A . cos 2A) . cos 2A….cos 2^n–1A

= 1/2² sin A. (sin 2A . cos 2A) . cos 2A….cos 2^n–1A

Continuing like this taking all the factors one by one we get the final product as

sin 2ⁿA / 2ⁿ sin A

In general, this can be written as

∏ cos 2^r A = sin 2ⁿA/ 2ⁿ sin A , where r varies from 0 to (n-1).

Similarly, if we have a sine or a cosine series in the sum form where the angles are in A.P., then multiply both numerator and denominator with 2sin (common difference /2).

Illustration: Prove that cos 2π/7 + cos 4π/7 + cos 6π/7 = -1/2.

Solution: The given term is cos 2π/7 + cos 4π/7 + cos 6π/7

= (2 sin π/7)/ (2 sin π/7) (cos 2π/7 + cos 4π/7 + cos 6π/7)

= (sin 3π/7 - sin π/7 + sin 5π/7 - sin 3π/7 + sin π - sin 5π/7) / (2 sin π/7)  

= -1/2
 

Some Important Results

1. -√a² + b² ≤ a sin x + b cos x ≤ √a² + b²

2.  sin² x + cosec² x ≥ 2

3. cos² x + sec² x ≥ 2

4.  tan² x + cot² x ≥ 2

5. cos A + cos (A + B) + cos (A + 2B) + …… cos (A + (n-1)B) = sin (nB/2)/ sin (B/2) 

6. tan a = cot A – 2 cot 2A

Illustration: Sum to n-terms of the series

sin α – sin (α + β) + sin (α + 2β) - (α + 3β) + …..

Solution:  Since, sin (π + θ) = - sin θ and sin (2π + θ) = sin θ

Therefore, sin (α + β) = sin (π + (α + β))

Hence, sin (α + 2β) = sin (2π + (α + 2β)) – sin (α + 3β)

= sin (3π + (α + 3β)) and so on.

Hence, the required sum is given by

S = sin α + sin (π + (α + β)) + sin (2π + (α + 2β)) + sin (3π + (α + 3β))…. to n    terms 

= sin α + sin (α + (π + β)) + sin (α + (2π + β)) sin (α + (3π + β))…. to n terms

Since, we know that sin α + sin (α + β) + sin (α + 2β) + … to n terms 

= (sin (nβ/ 2))/ (sin (β/ 2)). sin (α + (n-1)β/2)

Hence, the above equation gets reduced to 

= sin n. ((π + β)/ 2)/ sin ((π + β)/ 2). sin (α + (n-1)(π + β/ 2))

Illustration: Given α + β + γ = π, prove that

sin² α + sin² β - sin² γ =  2 sin α sin β cos γ

Solution: Considering the LHS,

sin² α + sin² β - sin² γ

= sin² α + (sin² β - sin² γ)

= sin² α + sin (β - γ) sin (β + γ)

= sin² α + sin (β - γ) sin (π - α)

= sin² α + sin α sin (β - γ)

= sin α [sin α + sin (β - γ)]

= sin α [sin (π – ((β + γ)) + sin (β - γ)]

= sin α [sin (β + γ) + sin (β - γ)]

= sin α [2 sin β cos γ]

= 2 sin α sin β cos γ

= RHS.
 

Related Resources

• Look into the Revision Notes on Trigonometry for a quick revision.

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