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Asked by sejal2208 | 11th Feb, 2010, 06:30: PM

Expert Answer:

Dear Student

a3+b3=(a+b)3-3ab(a+b)

sin 6 θ + cos 6 θ = (sin 2 θ + cos 2 θ )3 -3 sin 2 θ cos 2 θ(sin 2 +θ cos 2 θ)=1- 3sin 2 θ cos 2 θ

2 (sin 6 θ + cos 6 θ )= 2- 6sin 2 θ cos 2 θ-------(1)

a2+b2=(a+b)2-2ab

 (sin 4 θ + cos 4 θ )= (sin 2 θ + cos 2 θ )2 - 2sin 2 θ cos 2 θ=1-2sin 2 θ cos 2 θ

3 (sin 4 θ + cos 4 θ )=3- 6sin 2 θ cos 2 θ---------(2)

(2)-(1) ===> 3 (sin 4 θ + cos 4 θ ) - 2 ( sin 6 θ + cos 6 θ ) = 1

Regards

Team

Topperlearning

Answered by  | 11th Feb, 2010, 07:01: PM

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