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

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?

• 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. – Matti P. Jan 23 at 11:44
• $\sin x \cos y - \sin x=\sin x (\cos y - 1)=-2\sin x\sin^2(y/2)$. Use half angle formulas in the last step. – Shivering Soldier Jan 23 at 11:51

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.

• Thank you very much, I got the answer! – Charlie Lazoc Jan 23 at 11:56

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

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.