Just Simple Manipulation

Geometry Level pending

If sin ⁡ θ + sin ⁡ 2 θ + sin ⁡ 3 θ = 1 \sin \theta +\sin^2 \theta +\sin^3 \theta =1 then what is the value of 2 ( cos ⁡ 6 θ − 4 cos ⁡ 4 θ + 8 cos ⁡ 2 θ 2(\cos^6 \theta -4 \cos^4 \theta +8\cos^2 \theta )?

6 4 2 8 1 0

Achal Jain by Achal Jain , 19, India

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1 solution

sin ⁡ θ + sin ⁡ 3 θ = 1 − sin ⁡ 2 θ = cos ⁡ 2 θ \sin\theta+\sin^3\theta=1-\sin^2\theta=\cos^2\theta
⟹ sin ⁡ θ ( 1 + sin ⁡ 2 θ ⏟ 1 − cos ⁡ 2 θ ) = cos ⁡ 2 θ \implies \sin \theta(1+\underbrace{\sin^2\theta}_{1-\cos^2\theta})=\cos^2\theta Squaring both sides:

sin ⁡ 2 θ ⏟ 1 − cos ⁡ 2 θ ( 2 − cos ⁡ 2 θ ) 2 = cos ⁡ 4 θ \underbrace{\sin^2\theta}_{1-\cos^2\theta}(2-\cos^2\theta)^2=\cos^4\theta

( 1 − cos ⁡ 2 θ ) ( cos ⁡ 4 θ − 4 cos ⁡ 2 θ + 4 ) = cos ⁡ 4 θ (1-\cos^2\theta)(\cos^4\theta-4\cos^2\theta +4)=\cos^4\theta

⟹ 2 ( cos ⁡ 6 θ − 4 cos ⁡ 4 θ + 8 cos ⁡ 2 θ ) = 2 × 4 = 8 \implies2(\cos^6\theta-4\cos^4\theta+8\cos^2\theta)=2×4=\large\boxed{\color{#69047E}{8}}

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Thanx for the correction. Is it legit now?

Achal Jain - 4 years, 9 months ago

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Hmm..Its ok now.

A Former Brilliant Member - 4 years, 9 months ago

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