# Identities Involving Squares of Sines and Cosines

Identities involving squares of sines and cosines of multiples or submultiples of the angles involved.

To prove the identities involving squares sines and cosines we use the following algorithm.

Step I: Arrange the terms on the on the L.H.S. of the identity so that either sin$$^{2}$$ A - sin$$^{2}$$  B = sin (A + B) sin (A - B) or cos$$^{2}$$ A - sin$$^{2}$$  B = cos (A + B) cos (A - B) can be used.

Step II: Take the common factor outside.

Step III: Express the trigonometric ratio of a single angle inside the brackets into that of the sum of the angles.

Step IV: Use the formulas to convert the sum into product.

Examples on Identities involving squares of sines and cosines:

1.  If A + B + C = π, prove that,

sin$$^{2}$$ A + sin$$^{2}$$ B + sin$$^{2}$$ C = 2 + 2 cos A cos B cos C.

Solution:

L.H.S. = sin$$^{2}$$ A + sin$$^{2}$$ B + sin$$^{2}$$ C

= $$\frac{1}{2}$$(1 - cos$$^{2}$$ A) + $$\frac{1}{2}$$( 1- cos$$^{2}$$ B) + 1- cos$$^{2}$$ C

[Since, 2 sin$$^{2}$$ A = 1 - cos 2A

⇒ sin$$^{2}$$ A = $$\frac{1}{2}$$(1 - cos 2A)

Similarly, sin$$^{2}$$ B = $$\frac{1}{2}$$(1 - cos 2B) ]

= 2 - $$\frac{1}{2}$$(cos 2A + cos 2B) - cos$$^{2}$$ C

= 2 - $$\frac{1}{2}$$ ∙ 2 cos (A + B) cos (A - B) - cos$$^{2}$$ C

= 2 + cos C cos (A - B) - cos$$^{2}$$ C, [Since, A + B + C = π ⇒ A + B = π - C.

Therefore, cos (A + B) = cos (π - C) = - cos C]

= 2 + cos C [cos (A - B) - cosC]

= 2 + cos C [cos (A - B) + cos (A + B)], [Since, cos C = cos (A + B)]

= 2 + cos C [2 cos A cos B]

= 2 + 2 cos A cos B cos C = R.H.S.                     Proved.

2. If A + B + C = $$\frac{π}{2}$$  prove that,

cos$$^{2}$$ A+ cos$$^{2}$$ B + cos$$^{2}$$ C = 2 + 2sin A  sin B sin C.

Solution:

L.H.S. = cos$$^{2}$$ A+ cos$$^{2}$$ B + cos$$^{2}$$ C

= $$\frac{1}{2}$$(1+ cos 2A) + $$\frac{1}{2}$$(1 + cos 2B)+ cos$$^{2}$$ C [Since, 2 cos$$^{2}$$ A = 1 + cos 2A

⇒ cos$$^{2}$$A = $$\frac{1}{2}$$(1 + cos2A)

Similarly, cos$$^{2}$$B =$$\frac{1}{2}$$(1 + cos 2B)]

= 1 + $$\frac{1}{2}$$(cos 2A + cos 2B) + cos$$^{2}$$ C

= 1+ $$\frac{1}{2}$$ ∙ [2 cos (A + B) cos (A - B)] + 1- sin$$^{2}$$ C

= 2 + sin C cos (A - B) - sin$$^{2}$$ C

[A + B + C = $$\frac{π}{2}$$

⇒ A + B = $$\frac{π}{2}$$  - C

Therefore, cos (A + B) = cos ($$\frac{π}{2}$$ - C)=  sin C]

= 2 + sin C [cos (A - B) - sin C]

= 2 + sin C [cos (A - B) - cos (A + B)], [Since, sin C = cos (A + B)]

= 2 + sin C [2 sin A sin B]

= 2 + 2 sin A sin B sin C = R.H.S.                   Proved.