MHB How can trigonometric functions be simplified using specific values?

[Guest2]

$$\arcsin\sin\left(\frac{11\pi}{5}\right)$$

How do you simplify this?

 

[kaliprasad]

$$\arcsin\sin\left(\frac{11\pi}{5}\right)$$

How do you simplify this?

For us to help you better you should specify what you have tried.

now for the solution

$\arcsin\sin\left(t\right)$ shall lie between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$

so convert $sin(\frac{11\pi}{5})$ to $sin(t)$ with t in $-\frac{\pi}{2}$ and $\frac{\pi}{2}$
we get
$\sin(\frac{11\pi}{5})=sin(\frac{\pi}{5}) $
hence
$\arcsin\sin\left(\frac{11\pi}{5}\right)= \frac{\pi}{5}$

 

[Guest2]

Thank you. So the idea is that $\arcsin(\sin{t}) = t$ when $-\frac{\pi}{2} \le t \le \frac{\pi}{2}$? If so, is the same true for $\arccos(\cos{t})$?

 

[kaliprasad]

Thank you. So the idea is that $\arcsin(\sin{t}) = t$ when $-\frac{\pi}{2} \le t \le \frac{\pi}{2}$? If so, is the same true for $\arccos(\cos{t})$?

it is true for $\arctan(\tan{t})$ but not for $\arccos(\cos{t})$ because $\cos -t= \cos t$ and hence range for $\arccos(t)$ is $0$ to $\pi$

 

[Guest2]

it is true for $\arctan(\tan{t})$ but not for $\arccos(\cos{t})$ because $\cos -t= \cos t$ and hence range for $\arccos(t)$ is $0$ to $\pi$

Thank you, makes perfect sense!