Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
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]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
