0
$\begingroup$

I am sorry if the question looks very simple, but I am asking if maybe there is an identity that I am unaware of. I have this equation

$$ \sin^2 x=a $$ where $-\infty<x<\infty$ and $a>0$ are real parameters. By the given data, what is the value of $\sin(2x)$ in terms of $a$? Is there a way to express $\sin(2x)$ in terms of $\sin^2 x$ and $\cos^2 x$?

$\endgroup$
0

2 Answers 2

Reset to default
2
$\begingroup$

Since $\sin(2x)=2\sin(x)\cos(x)$ and $\cos^2(x)+\sin^2(x)=1$, you can write it as follows: $$\sin(2x)=2\sqrt{\sin^2(x)}\sqrt{1-\sin^2(x)}=2\sqrt{a(1-a)}.$$

$\endgroup$
4
  • 1
    $\begingroup$ Thanks; actually, I had doubts about considering $\cos x=\sqrt{1-\sin^2(x)}$; I thought we should use both signs $\cos x=\pm\sqrt{1-\sin^2(x)}$. @csch2 $\endgroup$
    – MsMath
    Commented Apr 5, 2022 at 16:12
  • 1
    $\begingroup$ For that you need to take into account in what region $x$ lies. For example, if $x$ is between $\pi/2$ and $\pi$, then $\sin(2x)$ will be negative and so the formula should take the negative sign. $\endgroup$
    – csch2
    Commented Apr 5, 2022 at 16:14
  • $\begingroup$ A factor of $(-1)^{\left\lfloor\frac{2x}{\pi}\right\rfloor}$ would be needed. $\endgroup$
    – John Wayland Bales
    Commented Apr 5, 2022 at 16:43
  • $\begingroup$ Your formula always gives a nonnegative answer, which cannot be true for $\sin(2x)$. See the answer by Jose Carlos Santos. $\endgroup$
    – N. F. Taussig
    Commented Apr 6, 2022 at 11:53
2
$\begingroup$

It cannot be done. Note that $\sin^2\left(\frac\pi4\right)=\sin^2\left(-\frac\pi4\right)=\frac12$. But $\sin\left(2\times\frac\pi4\right)=1$, whereas $\sin\left(2\times\left(-\frac\pi4\right)\right)=-1$. Therefore, in general, $\sin(2x)$ cannot be computed from $\sin^2(x)$ alone.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .