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so I want to prove both sides of this identity

$$ \sin x \cos y + \cos x \sin y - \sin x = \cos x \sin y - 2\sin x \sin^2\left(\frac{1}{2} y\right) $$

I've already proved it by manipulating the right side. However, when I try to prove it by manipulating the left, I'm stuck at this point:

$$\begin{align}\sin x\cos y + \cos x\sin y - \sin x &= \cos x\sin y - 2\sin x\sin^2\left(\frac{1}{2} y\right) \\ \cos x\sin y + \sin x(1 - \cos y) &= \text{RIGHT SIDE}\end{align}$$

Am I doing something wrong from this point, or is there a better way to prove it by manipulating only the left side?

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    $\begingroup$ Well, the first thing that I notice is that you could cancel out $\cos{x} \sin{y}$ from both sides ... After that, mabe take $\sin{x}$ as a factor on both sides? Should be smooth sailing from there. $\endgroup$
    – Matti P.
    Commented Jan 23, 2020 at 11:44
  • $\begingroup$ $\sin x \cos y - \sin x=\sin x (\cos y - 1)=-2\sin x\sin^2(y/2)$. Use half angle formulas in the last step. $\endgroup$
    – cqfd
    Commented Jan 23, 2020 at 11:51

2 Answers 2

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Hint:

From here,

$$\cos \theta = 1 - 2\sin^2\left(\frac\theta2\right)$$

Also, you should change your working to

$$\sin x\cos y + \cos x\sin y - \sin x = \cos x \sin y + \sin x(\cos y - 1)$$

Now, use the identity above.

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    $\begingroup$ Thank you very much, I got the answer! $\endgroup$ Commented Jan 23, 2020 at 11:56
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$$\sin x \cos y + \cos x \sin y - \sin x = \cos x \sin y +\sin x (\cos y -1)$$

In order to get the above expression identical to the $RHS$ you need to have $$(\cos y -1) = -2\sin ^2 (y/2)$$

Note that $$\sin^2\alpha = \frac {1}{2} (1-\cos (2\alpha))$$

Let $\alpha = y/2$ and you get the desired result.

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