If a sin θ-b cos θ=0 and a sin 3 θ+b cos 3 θ= sin 2 θ, then find the value of a2+b2.

Question

If a sin θ-b cos θ=0 and a sin 3 θ+b cos 3 θ= sin 2 θ, then find the value of a2+b2. (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

Given, a sin θ-b cos θ=0 a sin θ=b cos θ a sin 3 θ+b cos 3 θ= sin 2 θ a sin θ sin 2 θ+b cos θ cos 2 θ=2 sin θ cos θ b cos θ sin 2 θ+b cos θ cos 2 θ=2 sin θ cos θ b cos θ( sin 2 θ+ cos 2 θ)=2 sin θ cos θ b cos θ=2 sin θ cos θ b=2 sin θ text (1) ⇒ a sin θ=2 sin θ cos θ a=2 cos θ a2+b2 =(2 cos θ)2+(2 sin θ)2 =4 cos 2 θ+4 sin 2 θ =4( cos 2 θ+ sin 2 θ)=4

Q. If asinθ−bcosθ=0 and asin3θ+bcos3θ=sin2θ, then find the value of a2+b2.

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Given, asinθ−bcosθ=0 asinθ=bcosθ asin3θ+bcos3θ=sin2θ asinθsin2θ+bcosθcos2θ=2sinθcosθ bcosθsin2θ+bcosθcos2θ=2sinθcosθ bcosθ(sin2θ+cos2θ)=2sinθcosθ bcosθ=2sinθcosθ b=2sinθ (1) ⇒asinθ=2sinθcosθ a=2cosθ a2+b2=(2cosθ)2+(2sinθ)2 =4cos2θ+4sin2θ =4(cos2θ+sin2θ)=4

Q. If $a sin θ - b cos θ = 0$ and $a sin^{3} θ + b cos^{3} θ = sin 2 θ$ , then find the value of $a^{2} + b^{2}$ .