TITLE: Show $\sin(x+h) \cdot \cos x - \cos(x+h) \cdot \sin x = \sin h$ (without limits please - straight trigonometry only).
QUESTION [5 upvotes]: I've tried an algebraic approach using the identity
$\sin(x) = \sin(x+h-h) = \sin(x+h)\cos(h) - \cos(x+h)\sin(h)$, leading to a complicated expression I'm having trouble simplifying:
$[\sin(x)\cos(h) + \cos(x)\sin(h)]\cos(x) - [\cos(x)\cos(h) - \sin(x)\sin(h)][\sin(x+h)\cos(h)-\cos(x+h)\sin(h)]$
(further expansion of $\sin(x+h)$ and $\cos(x+h)$ in this expression did not help me).
Use of this identity was a hint in the book I came across this in, What is Mathematics, by Courant and Robbins, p. 422.
Note: Since posting I realized how to show symbolically in the updated diagram below, which has been updated since posting (including also the correct algebraic answer given by @Andre Nicholas).

REPLY [6 votes]: Recall that $\sin(a-b)=\sin a\cos b -\cos a\sin b$. Let $a=x+h$ and $b=x$.
Remark: You used precisely the same approach, except that instead of using $\sin(x+h-x)$, as above, you used $\sin(x+h-h)$.