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If x and y are connected parametrically by the equations given in Exercise x=\cos \theta-\cos 2 \theta, y=\sin \theta-\sin 2 \theta without eliminating the parameter, find dy/dx.
CBSE | Class 12 | Continuity and Differentiability | Exercise 5.6 | Maths | NCERT
If x and y are connected parametrically by the equations given in Exercise x=\cos \theta-\cos 2 \theta, y=\sin \theta-\sin 2 \theta without eliminating the parameter, find dy/dx.

Solution:

The provided functions are x=\cos \theta-\cos 2 \theta and y=\sin \theta-\sin 2 \theta

\frac{d x}{d \theta}=\frac{d}{d \theta} \cos \theta-\frac{d}{d \theta} \cos 2 \theta

\frac{d x}{d \theta}=-\sin \theta-(-\sin 2 \theta) \frac{d}{d \theta} 2 \theta

\frac{d x}{d \theta}=-\sin \theta+(-\sin 2 \theta) 2

\frac{d x}{d \theta}=2 \sin 2 \theta-\sin \theta

And \frac{d y}{d \theta}=\frac{d}{d \theta} \sin \theta-\frac{d}{d \theta} \sin 2 \theta

\frac{d y}{d \theta}=\cos \theta-\cos 2 \theta \frac{d}{d \theta} 2 \theta

\frac{d y}{d \theta}=\cos \theta-\cos 2 \theta \times 2

\frac{d y}{d \theta}=\cos \theta-2 \cos 2 \theta

So now,

\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}

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