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EXERCISE 1.5 · Q188

Q.x=sin⁡θx=\sin\theta, y=sin⁡3θy=\sin^3\theta, when θ=π/2\theta=\pi/2

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Given x=sin⁡θx=\sin\theta, y=sin⁡3θy=\sin^3\theta.

Step 1: dxdθ=cos⁡θ\dfrac{dx}{d\theta}=\cos\theta, dydθ=3sin⁡2θcos⁡θ\dfrac{dy}{d\theta}=3\sin^2\theta\cos\theta, so dydx=3sin⁡2θcos⁡θcos⁡θ=3sin⁡2θ\dfrac{dy}{dx}=\dfrac{3\sin^2\theta\cos\theta}{\cos\theta}=3\sin^2\theta (cancelling the common factor cos⁡θ≠0\cos\theta\ne0 near θ=π/2\theta=\pi/2).

Step 2 — differentiate this simplified form w.r.t. θ\theta: ddθ(3sin⁡2θ)=6sin⁡θcos⁡θ\dfrac{d}{d\theta}(3\sin^2\theta)=6\sin\theta\cos\theta. …

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