As many people have pointed out by now, $\sin^2 x$ is simply a "nickname" for $(\sin x)^2$. Therefore, $\sin^2\ 30 = (\sin 30)^2 = (1/2)^2 = 1/4$. As it happens, though, there is another useful thing we can say about $\sin^2 x$: $$\sin^2 x = (\sin x)^2 = \frac12 (1 - \cos (2 x)).$$ We can see this using the double-angle formula for cosines ...

"sin" seems like the meat and body of the concept whereas (x) is just the input. It seems to make more sense to put modifications closer the the "sin". Or it does to me. $\sin(x)^2$ looks like the power is trying to modify the (x) and I have to look twice to verify that it is actually the [sin(x)] that is being squared.

May 6, 2012 · Hence the answer separates $\frac{\sin^2{x}}{x^2}$ into three parts all which have singularity at zero. The upper and lower are picked in a way that controls the growth of the numerator. $\endgroup$ – Shuheng Zheng

I'm working with derivatives and need to know if $\\sin^2(x)$ the same as $\\sin(x^2)$? I almost don't want to ask because my last question was closed. It was a valid question and so is this one. I've

I need to prove that $\\sin(x) > \\frac{x}{2}$ if $0<x<\\pi/2$ I've started working with the derivative, but if it's possible, I'd rather something simpler than that.

Apr 26, 2019 · Another possible way is to differentiate $\sin^2x$ and observe that $$[\sin^2x]'=2\sin x\cos x=\sin 2x ...

Feb 22, 2024 · Now if that is the case, then shouldn't $\sin^2 (x)$ be equal to $\sin(\sin x)$, however, I have seen people use it as $\sin(x)^2$. Could someone please clarify my confusion. Could someone please clarify my confusion.

Here is a cannon to shoot a fly.. Rewrite the integral as \begin{equation} I(n):= \int_0^\pi\frac{1-\cos2nx}{1-\cos2x}\,dx\stackrel{2x\,\mapsto\, x}\Longrightarrow ...

I really don't know where I'm going wrong, I use the sum to product formula but always end up far from $\\sin(2x)\\sin(2y)$. Any help is appreciated, thanks.

May 29, 2015 · By some simple manipulations you can use what was shown in the related question: $$ \int_{-\pi}^\pi {\sin^2(nx)\over \sin^2(x)} = 2 n \pi \tag{1} $$