Is $\cos^2 \theta$ the same as $(\cos \theta)^2$

Question

I am trying to find $\lim_{x\to0}\sin\dfrac{1}{x}$.

I'm assuming it's possible with trigonometric identities. Doing some research, I found that $\sin \theta = \pm\sqrt{1 - \cos^2{\theta}} $. However I'm a bit confused on what is $\cos^2$.

Is $\cos^2 \theta$ the same as $(\cos \theta)^2$?

Answer 1

Yes, $\cos ^2 x$ is a short cut for $(\cos x)^2$

Similarly we have $ \tan ^2 x $ and $\sin ^2 x $ and more.

We also have the confusing $\sin ^{-1} x$ which is the inverse of $\sin x$ not the reciprocal of $\sin x$ which is called $\csc x$

Answer 2

Yes, this is true: $\cos^2(\theta)$ is another notation for $(\cos \theta)^2$, likewise for $\sin$ and $\tan$, and $\log$, $\lg$, and $\ln$ too.

Answer 3

Yes. Some confusion could arise because the notation $f^2$ is also used to denote the composition of a mapping $f$ with itself.

However, the convention is that with trigonometric functions, $\cos^2$, $\sin^2$ denote what you said.

Now another pain in every student's bottom body part is that $\sin^{-1}x$ does keep the usual meaning of $f^{-1}$, that is the inverse map $\arcsin$, and not the reciprocal $\frac{1}{\sin x}$. This is one reason a lot of people (including myself) prefer never using this notation and use $\arcsin$ instead.

Answer 4

Yes of course it is the same and $\cos^n x$ is widely used intead of $(\cos x)^n$.

For the limit let consider

• $x_n=\frac2{n\pi}\to 0\implies \sin\dfrac{1}{x_n}=\sin \frac{n\pi}2$

which is equal to 0 for $n$ even and $\pm 1$ for $n$ odd, that suffices to prove that the limit doesn't exist.