# Does $\frac{\sin^2(x)}{\sin(x)} = \sin(x)$?

Is $\sin^2 (x)$ equal to $(\sin(x))^2$, and if so, does that mean that $\frac{\sin^2(x)}{\sin(x)} = \sin(x)$?

• Yes. (Assuming $\sin x \ne 0$) – Shailesh Sep 21 '15 at 3:41
• Yes, $\sin^2(x) = \left[\sin(x)\right]^2$. Thus, $\frac{\sin^2(x)}{\sin(x)} = \frac{[\sin(x)]^2}{\sin(x)} = \frac{\sin(x) \cdot \sin(x)}{\sin(x)} = \frac{\sin(x)}{\sin(x)} \cdot \sin(x) = 1\cdot \sin(x) = \sin(x)$, such that $\sin(x) \neq 0$. – Decaf-Math Sep 21 '15 at 3:42

$$\sin^2(x)=(\sin(x))^2$$
$$\frac{\sin^2(x)}{\sin(x)}=\sin(x)$$
is not quite true. You need to specify that $\sin(x)\neq 0$. The equality is true for all $x$ such that $\sin(x) \neq 0$, however.
• You might add that you can define the LHS to be equal to the RHS by definition when $\sin x=0$. Then you can say you have equality for all $x$ (although this might be slightly excessive). +$1$ by the way. – Clayton Sep 21 '15 at 3:44