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jimmysmith560 · Jun 21, 2022

Would the following working help?

Part (i):

If the two curves have a common tangent, then their derivatives must be equal at that point.

$\frac{d}{dx}\sin x=\cos x$

$\frac{d}{dx}\left(\sin \left(x-\alpha \right)+k\right)=\cos \left(x-\alpha \right)$

$\therefore \cos x_0=\cos \left(x_0-\alpha \right)$

Part (ii):

$\cos x_0=\cos \left(x_{0-\alpha }\right)$

So $\alpha=2x_0$

Then $\sin \left(x_{0-\alpha }\right)$

$=\sin \left(x_0-2x_0\right)$

$=\sin \left(-x_0\right)$

$=-\sin \left(x_0\right)$

$\therefore \sin \left(x_0\right)=-\sin \left(x_0-\alpha \right)$

Part (iii):

$\sin x_0=\sin \left(x_0-\alpha \right)+k$

$-\sin \left(x_0-\alpha \right)=\sin \left(x_0-\alpha \right)+k$

$-2\sin \left(x_0-\alpha \right)=k$

$k=-2\sin \left(\frac{\alpha }{2}-\alpha \right)$

$k=-2\sin \left(-\frac{\alpha }{2}\right)$

$k=2\sin \left(\frac{\alpha }{2}\right)$

Farhanthestudent005 · Jun 22, 2022

Thank you very much for helping me, this helps a lot

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