Trigonometry #107

Geometry Level 2

For θ ∈ ( 0 , π 2 ) \theta\in\left(0,\frac{\pi}{2}\right) , if x sin ⁡ 3 θ + y cos ⁡ 3 θ = sin ⁡ θ cos ⁡ θ x\sin^{3}\theta+y\cos^{3}\theta=\sin\theta\cos\theta and x sin ⁡ θ = y cos ⁡ θ x\sin\theta=y\cos\theta then which of the following holds true?

This problem is part of the set Trigonometry .

x 2 − y 2 = 1 x^{2}-y^{2}=1 x 2 + y 2 = 1 x^{2}+y^{2}=1 x 3 + y 3 = 1 x^{3}+y^{3}=1 x 4 + y 4 = 1 x^{4}+y^{4}=1

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Omkar Kulkarni
Feb 12, 2015

x sin ⁡ 3 θ + y cos ⁡ 3 θ = sin ⁡ θ cos ⁡ θ x\sin^{3}\theta+y\cos^{3}\theta=\sin\theta\cos\theta

x sin ⁡ 3 θ + ( y cos ⁡ θ ) ( cos ⁡ 2 θ ) = sin ⁡ θ cos ⁡ θ x\sin^{3}\theta+(y\cos\theta)(\cos^{2}\theta)=\sin\theta\cos\theta

x sin ⁡ 3 θ + ( x sin ⁡ θ ) ( cos ⁡ 2 θ ) = sin ⁡ θ cos ⁡ θ x\sin^{3}\theta+(x\sin\theta)(\cos^{2}\theta)=\sin\theta\cos\theta

x sin ⁡ θ ( sin ⁡ 2 θ + cos ⁡ 2 θ ) = sin ⁡ θ cos ⁡ θ x\sin\theta(\sin^{2}\theta+\cos^{2}\theta)=\sin\theta\cos\theta

x sin ⁡ θ = sin ⁡ θ cos ⁡ θ x\sin\theta=\sin\theta\cos\theta

x = cos ⁡ θ x=\cos\theta

Similarly we can obtain the result y = cos ⁡ θ y=\cos\theta

∴ x 2 + y 2 = 1 \therefore\boxed{x^{2}+y^{2}=1}

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