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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 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 AB2 + sin2 C2

=1 - sin C2  cos AB2  + sin 2 C2

[A + B + C = π ⇒ A+B2 = π2  - C2.

 Therefore, cos A+B2 = cos (π2  - C2) = sin C2]

= 1 - sin C2[cos AB2 - sin C2]

= 1 - sin C2[cos AB2 - 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 AB2 - 1 + sin2  C2

= cos A+B2 cos AB2 + sin2  C2

= sin C/2 cos AB2 + sin2  C2

[Since, A + B + C = π ⇒ A+B2  = π2 - C2.

Therefore, cos (A+B2) = cos (π2 - C2) = sin C2]

= sin C2 [cos AB2 + sin C2]

= sin C2 [cos AB2 + cos A+B2], [Since, sin C2 = cos AB2]

= 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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