Answer :
To find the exact values of the given expressions, we need to evaluate some key properties of the trigonometric and inverse trigonometric functions involved. Let's go through each part step-by-step:
### (a) \(\arcsin(\sin(\frac{13\pi}{12}))\)
The function \(\arcsin(x)\) yields a value within the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\). We need to find the principal value of the angle equivalent to \(\frac{13\pi}{12}\) within this interval.
1. \(\frac{13\pi}{12}\) is greater than \(\pi/2\), so we need to find an equivalent angle within \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
2. Consider the sine function's property: \(\sin(\theta) = \sin(\pi - \theta)\).
3. Notice that \(\frac{13\pi}{12} = \pi + \frac{\pi}{12}\). Thus, its sine is \(\sin(\frac{13\pi}{12}) = \sin(\pi - (-\frac{\pi}{12})) = -\sin(\frac{\pi}{12})\).
Therefore, the principal value is:
[tex]\[ \arcsin(\sin(\frac{13\pi}{12})) = -\frac{\pi}{12} \][/tex]
In decimal form, this is approximately:
[tex]\[ \boxed{-0.26179938779914946} \][/tex]
### (b) \(\arccos(\cos(\frac{8\pi}{5}))\)
The function \(\arccos(x)\) yields a value within the interval \([0, \pi]\). We need to find the principal value of the angle equivalent to \(\frac{8\pi}{5}\) within this interval.
1. \(\frac{8\pi}{5}\) is greater than \(\pi\), so we need to find an equivalent angle within \([0, \pi]\).
2. Consider the cosine function's property: \(\cos(\theta) = \cos(2\pi - \theta)\).
3. \(\frac{8\pi}{5} = 2\pi - \frac{2\pi}{5}\). This angle can be converted to the equivalent angle in \([0, \pi]\) by subtracting one full period (\(2\pi\)):
[tex]\[ \frac{8\pi}{5} = 2\pi - \frac{2\pi}{5} \Rightarrow \frac{2\pi}{5} \][/tex]
Therefore, the principal value is:
[tex]\[ \arccos(\cos(\frac{8\pi}{5})) = \frac{2\pi}{5} \][/tex]
In decimal form, this is approximately:
[tex]\[ \boxed{1.2566370614359175} \][/tex]
### (c) \(\arctan(\tan(\frac{5\pi}{4}))\)
The function \(\arctan(x)\) yields a value within the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\). We need to find the principal value of the angle equivalent to \(\frac{5\pi}{4}\) within this interval.
1. \(\frac{5\pi}{4}\) is beyond \(\pi/2\), so we need to find an equivalent angle within \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
2. Consider the tangent function's property for \(\frac{5\pi}{4}\):
[tex]\[ \frac{5\pi}{4} = \pi + \frac{\pi}{4} \][/tex]
Since \(\tan(\theta) = \tan(\theta + n\pi)\) where \(n\) is an integer, the tangent value for this angle is:
[tex]\[ \tan(\frac{5\pi}{4}) = \tan(\frac{\pi}{4}) \][/tex]
But because it's actually in the third quadrant, the tangent value is negative, and:
[tex]\[ -\arctan(\tan(\frac{\pi}{4})) = -\frac{\pi}{4} \][/tex]
Therefore, the principal value is:
[tex]\[ \arctan(\tan(\frac{5\pi}{4})) = -\frac{\pi}{4} \][/tex]
In decimal form, this is approximately:
[tex]\[ \boxed{0.7853981633974482} \][/tex]
### (a) \(\arcsin(\sin(\frac{13\pi}{12}))\)
The function \(\arcsin(x)\) yields a value within the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\). We need to find the principal value of the angle equivalent to \(\frac{13\pi}{12}\) within this interval.
1. \(\frac{13\pi}{12}\) is greater than \(\pi/2\), so we need to find an equivalent angle within \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
2. Consider the sine function's property: \(\sin(\theta) = \sin(\pi - \theta)\).
3. Notice that \(\frac{13\pi}{12} = \pi + \frac{\pi}{12}\). Thus, its sine is \(\sin(\frac{13\pi}{12}) = \sin(\pi - (-\frac{\pi}{12})) = -\sin(\frac{\pi}{12})\).
Therefore, the principal value is:
[tex]\[ \arcsin(\sin(\frac{13\pi}{12})) = -\frac{\pi}{12} \][/tex]
In decimal form, this is approximately:
[tex]\[ \boxed{-0.26179938779914946} \][/tex]
### (b) \(\arccos(\cos(\frac{8\pi}{5}))\)
The function \(\arccos(x)\) yields a value within the interval \([0, \pi]\). We need to find the principal value of the angle equivalent to \(\frac{8\pi}{5}\) within this interval.
1. \(\frac{8\pi}{5}\) is greater than \(\pi\), so we need to find an equivalent angle within \([0, \pi]\).
2. Consider the cosine function's property: \(\cos(\theta) = \cos(2\pi - \theta)\).
3. \(\frac{8\pi}{5} = 2\pi - \frac{2\pi}{5}\). This angle can be converted to the equivalent angle in \([0, \pi]\) by subtracting one full period (\(2\pi\)):
[tex]\[ \frac{8\pi}{5} = 2\pi - \frac{2\pi}{5} \Rightarrow \frac{2\pi}{5} \][/tex]
Therefore, the principal value is:
[tex]\[ \arccos(\cos(\frac{8\pi}{5})) = \frac{2\pi}{5} \][/tex]
In decimal form, this is approximately:
[tex]\[ \boxed{1.2566370614359175} \][/tex]
### (c) \(\arctan(\tan(\frac{5\pi}{4}))\)
The function \(\arctan(x)\) yields a value within the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\). We need to find the principal value of the angle equivalent to \(\frac{5\pi}{4}\) within this interval.
1. \(\frac{5\pi}{4}\) is beyond \(\pi/2\), so we need to find an equivalent angle within \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
2. Consider the tangent function's property for \(\frac{5\pi}{4}\):
[tex]\[ \frac{5\pi}{4} = \pi + \frac{\pi}{4} \][/tex]
Since \(\tan(\theta) = \tan(\theta + n\pi)\) where \(n\) is an integer, the tangent value for this angle is:
[tex]\[ \tan(\frac{5\pi}{4}) = \tan(\frac{\pi}{4}) \][/tex]
But because it's actually in the third quadrant, the tangent value is negative, and:
[tex]\[ -\arctan(\tan(\frac{\pi}{4})) = -\frac{\pi}{4} \][/tex]
Therefore, the principal value is:
[tex]\[ \arctan(\tan(\frac{5\pi}{4})) = -\frac{\pi}{4} \][/tex]
In decimal form, this is approximately:
[tex]\[ \boxed{0.7853981633974482} \][/tex]
