MHB How can trigonometric functions be simplified using specific values?

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To simplify $$\arcsin\sin\left(\frac{11\pi}{5}\right)$$, convert the angle to one within the range of $$-\frac{\pi}{2}$$ to $$\frac{\pi}{2}$$. This results in $$\sin\left(\frac{11\pi}{5}\right) = \sin\left(\frac{\pi}{5}\right)$$, leading to the conclusion that $$\arcsin\sin\left(\frac{11\pi}{5}\right) = \frac{\pi}{5}$$. The discussion confirms that $$\arcsin(\sin{t}) = t$$ holds true within its defined range, while a similar relationship for $$\arccos(\cos{t})$$ does not apply due to its different range. The clarification emphasizes the importance of understanding the ranges of these trigonometric functions for accurate simplification.
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$$\arcsin\sin\left(\frac{11\pi}{5}\right)$$

How do you simplify this?
 
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Guest said:
$$\arcsin\sin\left(\frac{11\pi}{5}\right)$$

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
$\arcsin\sin\left(\frac{11\pi}{5}\right)= \frac{\pi}{5}$
 
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 $\arccos(\cos{t})$?
 
Guest said:
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 $\pi$
 
kaliprasad said:
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!
 
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