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

  • $\begingroup$ 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

You are correct that:




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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.