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

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  • $\begingroup$ Yes. (Assuming $\sin x \ne 0$) $\endgroup$ – Shailesh Sep 21 '15 at 3:41
  • $\begingroup$ 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 $\sin(x) \neq 0$. $\endgroup$ – Decaf-Math Sep 21 '15 at 3:42
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You are correct that:

$$\sin^2(x)=(\sin(x))^2$$

Although

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

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.

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  • $\begingroup$ 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. $\endgroup$ – Clayton Sep 21 '15 at 3:44
  • $\begingroup$ I see, because that would be 0/0, which is undefined. Thank you, I will accept the answer as soon as I can. $\endgroup$ – JohnDoe Sep 21 '15 at 3:44

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