Multiple trigonometric functions

Question

How can you solve such a problem where multiple trigonometric functions are applied?

Find the value of $\sin(\text{arc}\cot(\tan(\arccos\frac{3}{\sqrt{13}})))$.

Answer 1

Let $\displaystyle\theta=\cos\frac3{\sqrt{13}}$

$\displaystyle\implies(i)\cos\theta=\frac3{\sqrt{13}}$ and using this, $\displaystyle(ii)0\le\theta\le\pi$

$\displaystyle\text{arccot}\left(\tan \theta\right)=\frac\pi2-\arctan\left(\tan \theta\right)=\frac\pi2-\theta$

if $\displaystyle-\frac\pi2\le\theta\le\frac\pi2$ i.e., here if $\displaystyle0\le\theta\le\frac\pi2$ ( Case $\#1$)

$\displaystyle\sin\left[\text{arccot}\{\tan\left(\theta\right)\}\right]=\sin\left(\frac\pi2-\theta\right)=\cos\theta=\cdots$

Case $\#2:$

If $\displaystyle\frac\pi2<\theta\le\pi,\arctan\left(\tan \theta\right)=\theta-\pi$

Hope you can complete it from here

Answer 2

Hint:

Let $y=\arccos\frac{3}{\sqrt{13}}$, then $\cos y=\frac{3}{\sqrt{13}}$. Hence, $\sin y=\frac{2}{\sqrt{13}}$ and $\tan y=\frac{2}{3}$. Similarly with the other trigonometric function and its inverse.

P.S. I only regard the principal value of root square.

Answer 3

One can use trig identities, like: $$\tan (\arccos (x)) = \frac{\sqrt{1-x^2}}{x}, \ \sin (\mathrm{arccot} (y)) = \frac{1}{\sqrt{1+y^2}}$$ So here you have: $$\sin(\text{arc}\cot(\tan(\arccos(x){})))=\sin\left(\text{arc}\cot\left( \frac{\sqrt{1-x^2}}{x} \right)\right)=\frac{1}{\sqrt{1+\left(\frac{\sqrt{1-x^2}}{x}\right)^2}}=x=\frac{3}{\sqrt{13}}$$

Answer 4

Let $\displaystyle\text{arc}\cot\left[\tan\left(\arccos\frac{3}{\sqrt{13}}\right)\right]=\theta$

So, we need to find $\sin\theta$

Using this, $\displaystyle(i)0<\theta<\pi\implies\sin\theta>0\ \ \ \ (1)$

$\displaystyle(ii)\cot\theta=\tan\left(\arccos\frac3{\sqrt{13}}\right)$

$\displaystyle\tan\left(\frac\pi2-\theta\right)=\tan\left(\arccos\frac3{\sqrt{13}}\right)$

$\displaystyle\implies\frac\pi2-\theta=n\pi+\arccos\frac3{\sqrt{13}}$ where $n$ is any integer

$\displaystyle\implies\theta=-n\pi+\frac\pi2-\arccos\frac3{\sqrt{13}}=-n\pi+\arcsin\frac3{\sqrt{13}}$

$\displaystyle\implies\sin\theta=\pm\frac3{\sqrt{13}}$ based on the parity of $n$ can be written as $\displaystyle\sin\theta=(-1)^n\frac3{\sqrt{13}}$

Now use $(1)$