JEE Novice - (27)

Geometry Level 2

{ sin ⁡ θ + sin ⁡ 2 θ = 1 cos ⁡ 12 θ + 3 cos ⁡ 10 θ + 3 cos ⁡ 8 θ + cos ⁡ 6 θ − 1 = ? \large\begin{cases} & \sin\theta + \sin^{2}\theta & = & 1 \\ & \cos^{12}\theta + 3\cos^{10}\theta+3\cos^{8}\theta+\cos^{6}\theta-1 & =& ? \end{cases}

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Nihar Mahajan by Nihar Mahajan , 20, India

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1 solution

Tanishq Varshney
May 27, 2015

The required expression can be written as ( c o s 4 θ + c o s 2 θ ) 3 − 1 (cos^{4} \theta +cos^{2} \theta)^{3}-1

Given that s i n θ = c o s 2 θ sin \theta=cos^{2} \theta

or s i n 2 θ = c o s 4 θ sin^{2} \theta= cos^{4} \theta

the expression is ( s i n θ + s i n 2 θ ) 3 − 1 (sin \theta + sin^{2} \theta)^{3}-1

as s i n θ + s i n 2 θ = 1 sin \theta +sin^{2} \theta =1

hence the answer is z e r o zero

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