Answer :
Let's tackle each part of the problem step-by-step:
### Part (a)
To find the value of \(\arcsin\left(\sin\left(\frac{13\pi}{12}\right)\right)\), we need to consider the properties of the sine and arcsine functions.
The sine function, \(\sin(\theta)\), is periodic with a period of \(2\pi\). The arcsine function, \(\arcsin(x)\), is the inverse of the sine function restricted to the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
Given \(\theta = \frac{13\pi}{12}\), we observe that this angle is outside the principal range of \([- \frac{\pi}{2}, \frac{\pi}{2}]\). To bring it within the principal range, we use the property that \(\sin(\theta) = \sin(\pi - \theta)\).
For \(\theta = \frac{13\pi}{12}\):
[tex]\[ \theta = \frac{13\pi}{12} = \pi + \left(\frac{\pi}{12} - \pi\right) \][/tex]
Thus, we get:
[tex]\[ \sin\left(\frac{13\pi}{12}\right) = \sin\left(\frac{13\pi}{12} - \pi\right) = \sin\left(-\frac{\pi}{12}\right) \][/tex]
Since \(\sin(-x) = -\sin(x)\), we have:
[tex]\[ \sin\left(-\frac{\pi}{12}\right) = -\sin\left(\frac{\pi}{12}\right) \][/tex]
Now, applying the arcsin function:
[tex]\[ \arcsin\left(\sin\left(\frac{13\pi}{12}\right)\right) = \arcsin\left(-\sin\left(\frac{\pi}{12}\right)\right) = -\frac{\pi}{12} \][/tex]
Converting \(-\frac{\pi}{12}\) to a numerical value, we get:
[tex]\[ \boxed{-0.26179938779914946} \][/tex]
### Part (b)
To find \(\arccos\left(\cos\left(\frac{8\pi}{5}\right)\right)\), again consider the properties of the cosine and arccosine functions.
The cosine function, \(\cos(\theta)\), is periodic with a period of \(2\pi\). The arccosine function, \(\arccos(x)\), is the inverse of the cosine function restricted to the interval \([0, \pi]\).
Given \(\theta = \frac{8\pi}{5}\), this angle is outside the principal range of \([0, \pi]\). We use the property that \(\cos(\theta) = \cos(2\pi - \theta)\).
For \(\theta = \frac{8\pi}{5}\):
[tex]\[ \theta = \frac{8\pi}{5} = 2\pi - \left(2\pi - \frac{8\pi}{5}\right) = 2\pi - \frac{8\pi}{5} \][/tex]
Thus:
[tex]\[ \cos\left(\frac{8\pi}{5}\right) = \cos\left(2\pi - \frac{8\pi}{5}\right) = \cos\left(\frac{2\pi}{5}\right) \][/tex]
So the value is:
[tex]\[ \arccos\left(\cos\left(\frac{8\pi}{5}\right)\right) = \arccos\left(\cos\left(\frac{2\pi}{5}\right)\right) = \frac{2\pi}{5} \][/tex]
Converting \(\frac{2\pi}{5}\) to a numerical value, we get:
[tex]\[ \boxed{1.2566370614359175} \][/tex]
### Part (c)
To find \(\arctan\left(\tan\left(\frac{5\pi}{4}\right)\right)\), we consider the properties of the tangent and arctangent functions.
The tangent function, \(\tan(\theta)\), is periodic with a period of \(\pi\). The arctangent function, \(\arctan(x)\), is the inverse of the tangent function restricted to the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\).
For \(\theta = \frac{5\pi}{4}\), we bring it within the principal range. Observe:
[tex]\[ \theta = \frac{5\pi}{4} = \pi + \left(\frac{\pi}{4}\right) = \pi + \frac{\pi}{4} \][/tex]
Thus,
[tex]\[ \tan\left(\frac{5\pi}{4}\right) = \tan\left(\frac{5\pi}{4} - \pi\right) = \tan\left(\frac{\pi}{4}\right) \][/tex]
So,
[tex]\[ \arctan\left(\tan\left(\frac{5\pi}{4}\right)\right) = \arctan\left(\tan\left(\frac{\pi}{4}\right)\right) = \frac{\pi}{4} \][/tex]
Converting \(\frac{\pi}{4}\) to a numerical value, we get:
[tex]\[ \boxed{0.7853981633974482} \][/tex]
Thus, the values are:
[tex]\[ \boxed{-0.26179938779914946}, \boxed{1.2566370614359175}, \boxed{0.7853981633974482} \][/tex]
### Part (a)
To find the value of \(\arcsin\left(\sin\left(\frac{13\pi}{12}\right)\right)\), we need to consider the properties of the sine and arcsine functions.
The sine function, \(\sin(\theta)\), is periodic with a period of \(2\pi\). The arcsine function, \(\arcsin(x)\), is the inverse of the sine function restricted to the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
Given \(\theta = \frac{13\pi}{12}\), we observe that this angle is outside the principal range of \([- \frac{\pi}{2}, \frac{\pi}{2}]\). To bring it within the principal range, we use the property that \(\sin(\theta) = \sin(\pi - \theta)\).
For \(\theta = \frac{13\pi}{12}\):
[tex]\[ \theta = \frac{13\pi}{12} = \pi + \left(\frac{\pi}{12} - \pi\right) \][/tex]
Thus, we get:
[tex]\[ \sin\left(\frac{13\pi}{12}\right) = \sin\left(\frac{13\pi}{12} - \pi\right) = \sin\left(-\frac{\pi}{12}\right) \][/tex]
Since \(\sin(-x) = -\sin(x)\), we have:
[tex]\[ \sin\left(-\frac{\pi}{12}\right) = -\sin\left(\frac{\pi}{12}\right) \][/tex]
Now, applying the arcsin function:
[tex]\[ \arcsin\left(\sin\left(\frac{13\pi}{12}\right)\right) = \arcsin\left(-\sin\left(\frac{\pi}{12}\right)\right) = -\frac{\pi}{12} \][/tex]
Converting \(-\frac{\pi}{12}\) to a numerical value, we get:
[tex]\[ \boxed{-0.26179938779914946} \][/tex]
### Part (b)
To find \(\arccos\left(\cos\left(\frac{8\pi}{5}\right)\right)\), again consider the properties of the cosine and arccosine functions.
The cosine function, \(\cos(\theta)\), is periodic with a period of \(2\pi\). The arccosine function, \(\arccos(x)\), is the inverse of the cosine function restricted to the interval \([0, \pi]\).
Given \(\theta = \frac{8\pi}{5}\), this angle is outside the principal range of \([0, \pi]\). We use the property that \(\cos(\theta) = \cos(2\pi - \theta)\).
For \(\theta = \frac{8\pi}{5}\):
[tex]\[ \theta = \frac{8\pi}{5} = 2\pi - \left(2\pi - \frac{8\pi}{5}\right) = 2\pi - \frac{8\pi}{5} \][/tex]
Thus:
[tex]\[ \cos\left(\frac{8\pi}{5}\right) = \cos\left(2\pi - \frac{8\pi}{5}\right) = \cos\left(\frac{2\pi}{5}\right) \][/tex]
So the value is:
[tex]\[ \arccos\left(\cos\left(\frac{8\pi}{5}\right)\right) = \arccos\left(\cos\left(\frac{2\pi}{5}\right)\right) = \frac{2\pi}{5} \][/tex]
Converting \(\frac{2\pi}{5}\) to a numerical value, we get:
[tex]\[ \boxed{1.2566370614359175} \][/tex]
### Part (c)
To find \(\arctan\left(\tan\left(\frac{5\pi}{4}\right)\right)\), we consider the properties of the tangent and arctangent functions.
The tangent function, \(\tan(\theta)\), is periodic with a period of \(\pi\). The arctangent function, \(\arctan(x)\), is the inverse of the tangent function restricted to the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\).
For \(\theta = \frac{5\pi}{4}\), we bring it within the principal range. Observe:
[tex]\[ \theta = \frac{5\pi}{4} = \pi + \left(\frac{\pi}{4}\right) = \pi + \frac{\pi}{4} \][/tex]
Thus,
[tex]\[ \tan\left(\frac{5\pi}{4}\right) = \tan\left(\frac{5\pi}{4} - \pi\right) = \tan\left(\frac{\pi}{4}\right) \][/tex]
So,
[tex]\[ \arctan\left(\tan\left(\frac{5\pi}{4}\right)\right) = \arctan\left(\tan\left(\frac{\pi}{4}\right)\right) = \frac{\pi}{4} \][/tex]
Converting \(\frac{\pi}{4}\) to a numerical value, we get:
[tex]\[ \boxed{0.7853981633974482} \][/tex]
Thus, the values are:
[tex]\[ \boxed{-0.26179938779914946}, \boxed{1.2566370614359175}, \boxed{0.7853981633974482} \][/tex]
