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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 sin2 A+ cos2 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 = π2)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,
cos2 A + cos2 B - cos2 C = 1 - 2 sin A sin B cos C.
Solution:
L.H.S. = cos2 A + cos2 B - cos2 C
= cos2 A + (1 - sin2 B) - cos2 C
= 1 + [cos2 A - sin2 B] - cos2 C
= 1 + cos (A + B) cos (A - B) - cos2 C
= 1 + cos (π - C) cos (A - B) - cos2 C, [Since A + B + C = π ⇒ A + B = π - C]
= 1 - cos C cos (A - B) - cos2 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,
sin2 A2 + sin2 A2 + sin2 A2 = 1 - 2 sin A2 - sin B2 sin C2
Solution:
L.H.S. = sin2 A2 + sin2 B2 +
sin2 C2
= 12(1 - cos A) + 12(1 - cos B) + sin2 C2, [Since, 2 sin2 A2 = 1 - cos A
⇒ sin2 A2 = 12(1
- cos A)
Similarly, sin2 B2
= 12( 1 - cos B)]
= 1 - 12(cos A + cos B) + sin2 C2
= 1 - 12 ∙ 2 cos A+B2 ∙ cos A−B2 + sin2 C2
=1 - sin C2 cos A−B2 + sin 2 C2
[A + B + C = π ⇒ A+B2 = π2 - C2.
Therefore, cos A+B2 = cos (π2 - C2) = sin C2]
= 1 - sin C2[cos A−B2 - sin C2]
= 1 - sin C2[cos A−B2 - cos A+B2] [Since, sin C2 = cos A+B2]
= 1 - sin C2[2 sin A2 ∙ sin B2]
= 1 - 2 sin A2 sin B2 sin C2 = R.H.S. Proved.
3. If A + B + C = π, prove that,
cos2 A2 + cos2 B2 -
cos2 C2 = 2 cos A2 cos B2
sin C2
Solution:
L.H.S. = cos2 A2 + cos2 B2 - cos2 C2
= 12(1 + cos A) + 12(1 + cos B) - cos2 C2, [Since, 2 cos2 A2
= 1 + cos A ⇒ cos2
A2 = 12(1 + cos A)
Similarly, cos2 B2 = 12(1 + cos B)]
= 1 + 12(cos A + cos B) - cos2 C2
= 1 + 12 ∙ 2 cos A+B2 cos A−B2 - 1 + sin2 C2
= cos A+B2 cos A−B2 + sin2 C2
= sin C/2 cos A−B2 + sin2 C2
[Since, A + B + C = π ⇒ A+B2 = π2 - C2.
Therefore, cos (A+B2) = cos (π2 - C2) = sin C2]
= sin C2 [cos A−B2 + sin C2]
= sin C2 [cos A−B2 + cos A+B2], [Since, sin C2 = cos A−B2]
= sin C2 [2 cos A2 cos B2]
= 2 cos A2 cos B2 sin C2 = R.H.S. Proved.
● Conditional Trigonometric Identities
11 and 12 Grade Math
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