Square of Identities Involving Squares of Sines and Cosines

We will learn how to solve identities involving square of sines and cosines of multiples or submultiples of the angles involved.

We use the following ways to solve the identities involving square of sines and cosines.

(i) Express the first two squares of L.H.S. in terms of cos 2A (or cos A).

(ii) Either retain the third term unchanged or make a change using the formula sin\(^{2}\) A+ cos\(^{2}\) A = 1.

(iii) Keeping the numericais (if any) apart, express the sum of two cosines in the form of product.

(iv) Then use the condition A + B + C = π (or A + B + C = \(\frac{π}{2}\))and take one sine or cosine term common.

(v) Finally, express the sum or difference of two sines (or cosines) in the brackets as product.

1. If A + B + C = π, prove that,

cos\(^{2}\) A + cos\(^{2}\) B - cos\(^{2}\) C = 1 - 2 sin A sin B cos C.

Solution:

L.H.S. =  cos\(^{2}\) A + cos\(^{2}\) B - cos\(^{2}\) C

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

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

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

= 1 + cos (π - C) cos (A - B) - cos\(^{2}\) C, [Since A + B + C = π ⇒ A + B = π - C]

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

= 1 - cos C [cos (A - B) + cos C]

= 1 - cos C [cos (A - B) + cos {π - (A + B)}], [Since A + B + C = π ⇒ C = π - (A + B)]

= 1 - cos C [cos (A - B) - cos (A + B)]

= 1 - cos C [2 sin A sin B]

= 1 - 2 sin A sin B cos C = R.H.S.                    Proved.


2. If A + B + C = π, prove that,

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

Solution:

L.H.S. = sin\(^{2}\) \(\frac{A}{2}\) + sin\(^{2}\) \(\frac{B}{2}\) + sin\(^{2}\) \(\frac{C}{2}\)

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

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

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

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

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

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

[A + B + C = π ⇒ \(\frac{A + B}{2}\) = \(\frac{π}{2}\)  - \(\frac{C}{2}\).

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

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

= 1 - sin \(\frac{C}{2}\)[cos \(\frac{A - B}{2}\) - cos \(\frac{A + B}{2}\)]   [Since, sin \(\frac{C}{2}\) = cos \(\frac{A + B}{2}\)]

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

= 1 - 2 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) sin \(\frac{C}{2}\) = R.H.S.                    Proved.

 

2. If A + B + C = π, prove that,

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

Solution:

L.H.S. = cos\(^{2}\)  \(\frac{A}{2}\)  + cos\(^{2}\) \(\frac{B}{2}\) - cos\(^{2}\) \(\frac{C}{2}\)

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

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

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

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

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

= sin C/2 cos \(\frac{A - B}{2}\) + sin\(^{2}\)  \(\frac{C}{2}\)

[Since, A + B + C = π ⇒ \(\frac{A + B}{2}\)  = \(\frac{π}{2}\) - \(\frac{C}{2}\).

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

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

= sin \(\frac{C}{2}\) [cos \(\frac{A - B}{2}\) + cos \(\frac{A + B}{2}\)], [Since, sin \(\frac{C}{2}\) = cos \(\frac{A - B}{2}\)]

= sin \(\frac{C}{2}\) [2 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\)]

= 2 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) sin \(\frac{C}{2}\) = R.H.S.          Proved.






11 and 12 Grade Math

From Square of Identities Involving Squares of Sines and Cosines to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.