# help (1 Viewer)

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
Would the following working help?

Part (i):

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

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

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

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

Part (ii):

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

So $\bg_white \alpha=2x_0$

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

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

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

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

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

Part (iii):

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

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

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

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

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

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

#### Farhanthestudent005

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