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Question

Mathematics Question on applications of integrals

If sin(πcosθ)=cos(πsinθ),\sin(\pi \cos \theta) = \cos(\pi \sin \theta), then sin2θ\sin 2 \theta equals

A

±34\pm \frac{3}{4}

B

±2\pm \sqrt{2}

C

±13\pm \frac{1}{\sqrt{3}}

D

±12\pm \frac{1}{2}

Answer

±34\pm \frac{3}{4}

Explanation

Solution

sin(πcosθ)=cos(πsinθ)\sin\left(\pi \cos \theta\right) =\cos\left(\pi \sin\theta\right)
cos(π2πcosθ)=cos(πsinθ)\Rightarrow\cos \left(\frac{\pi}{2} -\pi \cos \theta\right) =\cos \left(\pi \sin \theta\right)
π2πcosθ±πsinθ\Rightarrow \frac{\pi}{2} - \pi \cos \theta -\pm \pi \sin \theta
π2=πcosθ±πsinθ\Rightarrow \frac{\pi }{2} =\pi \cos \theta \pm \pi \sin \theta
12=cosθ±sinθ\frac{1 }{2} = \cos \theta \pm \sin \theta
Squaring both sides, we get
14=cos2θ+sin2θ±2sinθcosθ\frac{1}{4}= \cos^{2} \theta + \sin^{2} \theta \pm 2 \sin \theta \cos \theta
14=1±2sinθcosθ\Rightarrow \frac{1}{4} = 1 \pm 2 \sin \theta \cos \theta
141±sin2θ\Rightarrow \frac{1}{4} - 1 \pm \sin 2\theta
34=±sin2θsin2θ±34\Rightarrow - \frac{3}{4} = \pm \sin 2\theta \Rightarrow \sin 2\theta \pm \frac{3}{4}