Answer :
Sure, let's solve each part of the question step-by-step.
### (a) \(\arcsin \left(\sin \frac{13 \pi}{12}\right)\)
To solve \(\arcsin (\sin x)\), we need to consider the principal range of the \(\arcsin\) function, which is \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
1. \(\frac{13 \pi}{12}\) is outside the principal range of \(\arcsin\), so we have to bring it within this range by adjusting the angle.
2. It's helpful to use angle equivalence modulo \(2\pi\). Notice that \(\frac{13 \pi}{12} - 2\pi = \frac{13 \pi}{12} - \frac{24 \pi}{12} = - \frac{11 \pi}{12}\).
3. \(-\frac{11 \pi}{12}\) is still outside \([- \frac{\pi}{2}, \frac{\pi}{2}]\), so simplify it further to find the equivalent angle within the principal range of \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
Finally, \(\arcsin(\sin(\frac{13\pi}{12})) = \boxed{0.26179938779914913}\).
### (b) \(\arccos \left(\cos \frac{8 \pi}{5}\right)\)
To solve \(\arccos (\cos x)\), we need to consider the principal range of the \(\arccos\) function, which is \([0, \pi]\).
1. \(\frac{8 \pi}{5}\) is outside the principal range of \(\arccos\), so we equivalently reduce it modulo \(2\pi\).
2. Observe \(\frac{8 \pi}{5} - 2\pi = \frac{8 \pi}{5} - \frac{10 \pi}{5} = -\frac{2 \pi}{5}\).
3. Adding \(2\pi\) to \(-\frac{2 \pi}{5}\) gives us a positive angle within the maximum range due to periodicity: \(-\frac{2 \pi}{5} + 2\pi = \frac{8 \pi}{5}\).
In this case, the simplified form falls within the principal range for \(\arccos\).
Thus, \(\arccos(\cos(\frac{8\pi}{5})) = \boxed{5.026548245743669}\).
### (c) \(\arctan \left(\tan \frac{5 \pi}{4}\right)\)
To solve \(\arctan (\tan x)\), the principal range of the \(\arctan\) function is \((-\frac{\pi}{2}, \frac{\pi}{2})\).
1. \(\frac{5 \pi}{4}\) is outside the principal range of \(\arctan\), so we should adjust this angle to fall within \((-\frac{\pi}{2}, \frac{\pi}{2})\).
2. Consider the angle equivalence: \(\frac{5 \pi}{4} - \pi = \frac{\pi}{4}\). Since \(\tan(\frac{\pi}{4}) = 1\), \(\frac{\pi}{4}\) lies within \((-\frac{\pi}{2}, \frac{\pi}{2})\).
Then, \(\arctan(\tan(\frac{5\pi}{4})) = \boxed{0.7853981633974483}\).
### Summary
\(\arcsin \left(\sin \frac{13 \pi}{12}\right) = \boxed{0.26179938779914913}\)
\(\arccos \left(\cos \frac{8 \pi}{5}\right) = \boxed{5.026548245743669}\)
[tex]\(\arctan \left(\tan \frac{5 \pi}{4}\right) = \boxed{0.7853981633974483}\)[/tex]
### (a) \(\arcsin \left(\sin \frac{13 \pi}{12}\right)\)
To solve \(\arcsin (\sin x)\), we need to consider the principal range of the \(\arcsin\) function, which is \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
1. \(\frac{13 \pi}{12}\) is outside the principal range of \(\arcsin\), so we have to bring it within this range by adjusting the angle.
2. It's helpful to use angle equivalence modulo \(2\pi\). Notice that \(\frac{13 \pi}{12} - 2\pi = \frac{13 \pi}{12} - \frac{24 \pi}{12} = - \frac{11 \pi}{12}\).
3. \(-\frac{11 \pi}{12}\) is still outside \([- \frac{\pi}{2}, \frac{\pi}{2}]\), so simplify it further to find the equivalent angle within the principal range of \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
Finally, \(\arcsin(\sin(\frac{13\pi}{12})) = \boxed{0.26179938779914913}\).
### (b) \(\arccos \left(\cos \frac{8 \pi}{5}\right)\)
To solve \(\arccos (\cos x)\), we need to consider the principal range of the \(\arccos\) function, which is \([0, \pi]\).
1. \(\frac{8 \pi}{5}\) is outside the principal range of \(\arccos\), so we equivalently reduce it modulo \(2\pi\).
2. Observe \(\frac{8 \pi}{5} - 2\pi = \frac{8 \pi}{5} - \frac{10 \pi}{5} = -\frac{2 \pi}{5}\).
3. Adding \(2\pi\) to \(-\frac{2 \pi}{5}\) gives us a positive angle within the maximum range due to periodicity: \(-\frac{2 \pi}{5} + 2\pi = \frac{8 \pi}{5}\).
In this case, the simplified form falls within the principal range for \(\arccos\).
Thus, \(\arccos(\cos(\frac{8\pi}{5})) = \boxed{5.026548245743669}\).
### (c) \(\arctan \left(\tan \frac{5 \pi}{4}\right)\)
To solve \(\arctan (\tan x)\), the principal range of the \(\arctan\) function is \((-\frac{\pi}{2}, \frac{\pi}{2})\).
1. \(\frac{5 \pi}{4}\) is outside the principal range of \(\arctan\), so we should adjust this angle to fall within \((-\frac{\pi}{2}, \frac{\pi}{2})\).
2. Consider the angle equivalence: \(\frac{5 \pi}{4} - \pi = \frac{\pi}{4}\). Since \(\tan(\frac{\pi}{4}) = 1\), \(\frac{\pi}{4}\) lies within \((-\frac{\pi}{2}, \frac{\pi}{2})\).
Then, \(\arctan(\tan(\frac{5\pi}{4})) = \boxed{0.7853981633974483}\).
### Summary
\(\arcsin \left(\sin \frac{13 \pi}{12}\right) = \boxed{0.26179938779914913}\)
\(\arccos \left(\cos \frac{8 \pi}{5}\right) = \boxed{5.026548245743669}\)
[tex]\(\arctan \left(\tan \frac{5 \pi}{4}\right) = \boxed{0.7853981633974483}\)[/tex]
