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.
● Conditional Trigonometric Identities
11 and 12 Grade Math
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