Solved on Feb 18, 2024

Find the derivative of $y = \cos^{-1}(\sin(x))$ .

STEP 1

Assumptions
1. The function $y=\cos^{-1}(\sin(x))$ is defined for values of $x$ where $\sin(x)$ falls within the range of the $\cos^{-1}$ function, which is $[-1,1]$ .
2. The range of the $\cos^{-1}$ function is $[0, \pi]$ .
3. The domain of the $\sin$ function is all real numbers.

STEP 2

Understand the relationship between the sine and cosine functions. The sine function can be expressed in terms of cosine by using the complementary angle identity:
$\sin(x) = \cos\left(\frac{\pi}{2} - x\right)$

STEP 3

Replace $\sin(x)$ with $\cos\left(\frac{\pi}{2} - x\right)$ in the original equation:
$y = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - x\right)\right)$

STEP 4

Recognize that $\cos^{-1}(\cos(\theta)) = \theta$ for $\theta$ in the range $[0, \pi]$ . Therefore, if $\frac{\pi}{2} - x$ is within this range, we can simplify the equation to:
$y = \frac{\pi}{2} - x$

STEP 5

Determine the range of $x$ for which $\frac{\pi}{2} - x$ falls within the range $[0, \pi]$ . This occurs when:
$0 \leq \frac{\pi}{2} - x \leq \pi$

STEP 6

Solve the inequality for $x$ to find the domain where the simplification is valid:
$-\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$

STEP 7

Now we need to consider the values of $x$ outside the range $[-\frac{\pi}{2}, \frac{3\pi}{2}]$ . For these values, $\cos^{-1}(\sin(x))$ does not directly simplify to $\frac{\pi}{2} - x$ because $\sin(x)$ might not be equal to $\cos(\frac{\pi}{2} - x)$ due to periodicity and the range of the $\cos^{-1}$ function.

STEP 8

For $x$ outside the range $[-\frac{\pi}{2}, \frac{3\pi}{2}]$ , we need to find an equivalent angle $\theta$ such that $\sin(x) = \sin(\theta)$ and $\theta$ is within the range $[0, \pi]$ since that is the range of the $\cos^{-1}$ function.

STEP 9

Use the fact that $\sin(x)$ has a period of $2\pi$ to find an equivalent angle $\theta$ within the desired range. We can express $\theta$ as:
$\theta = x - 2\pi k$
where $k$ is an integer chosen such that $\theta$ is in the range $[0, \pi]$ .

STEP 10

For $x$ outside the range $[-\frac{\pi}{2}, \frac{3\pi}{2}]$ , we use the angle $\theta$ to rewrite the original equation as:
$y = \cos^{-1}(\sin(\theta))$

STEP 11

Since $\theta$ is now within the range $[0, \pi]$ , we can again use the complementary angle identity:
$y = \cos^{-1}(\cos(\frac{\pi}{2} - \theta))$

STEP 12

Simplify the equation using the fact that $\cos^{-1}(\cos(\theta)) = \theta$ for $\theta$ in the range $[0, \pi]$ :
$y = \frac{\pi}{2} - \theta$

STEP 13

Replace $\theta$ with $x - 2\pi k$ to find the general solution for $y$ :
$y = \frac{\pi}{2} - (x - 2\pi k)$

STEP 14

Simplify the expression for $y$ :
$y = \frac{\pi}{2} - x + 2\pi k$

STEP 15

The final solution for $y$ depends on the value of $x$ . If $x$ is in the range $[-\frac{\pi}{2}, \frac{3\pi}{2}]$ , then $y = \frac{\pi}{2} - x$ . If $x$ is outside this range, then $y$ can be expressed as $y = \frac{\pi}{2} - x + 2\pi k$ , where $k$ is an integer chosen such that the resulting angle is in the range $[0, \pi]$ .

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Find the derivative of $y = \cos^{-1}(\sin(x))$ .

STEP 1

Assumptions
1. The function $y=\cos^{-1}(\sin(x))$ is defined for values of $x$ where $\sin(x)$ falls within the range of the $\cos^{-1}$ function, which is $[-1,1]$ .
2. The range of the $\cos^{-1}$ function is $[0, \pi]$ .
3. The domain of the $\sin$ function is all real numbers.

STEP 2

Understand the relationship between the sine and cosine functions. The sine function can be expressed in terms of cosine by using the complementary angle identity:
$\sin(x) = \cos\left(\frac{\pi}{2} - x\right)$

STEP 3

Replace $\sin(x)$ with $\cos\left(\frac{\pi}{2} - x\right)$ in the original equation:
$y = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - x\right)\right)$

STEP 4

Recognize that $\cos^{-1}(\cos(\theta)) = \theta$ for $\theta$ in the range $[0, \pi]$ . Therefore, if $\frac{\pi}{2} - x$ is within this range, we can simplify the equation to:
$y = \frac{\pi}{2} - x$

STEP 5

Determine the range of $x$ for which $\frac{\pi}{2} - x$ falls within the range $[0, \pi]$ . This occurs when:
$0 \leq \frac{\pi}{2} - x \leq \pi$

STEP 6

Solve the inequality for $x$ to find the domain where the simplification is valid:
$-\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$

STEP 7

STEP 8

For $x$ outside the range $[-\frac{\pi}{2}, \frac{3\pi}{2}]$ , we need to find an equivalent angle $\theta$ such that $\sin(x) = \sin(\theta)$ and $\theta$ is within the range $[0, \pi]$ since that is the range of the $\cos^{-1}$ function.

STEP 9

Use the fact that $\sin(x)$ has a period of $2\pi$ to find an equivalent angle $\theta$ within the desired range. We can express $\theta$ as:
$\theta = x - 2\pi k$
where $k$ is an integer chosen such that $\theta$ is in the range $[0, \pi]$ .

STEP 10

For $x$ outside the range $[-\frac{\pi}{2}, \frac{3\pi}{2}]$ , we use the angle $\theta$ to rewrite the original equation as:
$y = \cos^{-1}(\sin(\theta))$

STEP 11

Since $\theta$ is now within the range $[0, \pi]$ , we can again use the complementary angle identity:
$y = \cos^{-1}(\cos(\frac{\pi}{2} - \theta))$

STEP 12

Simplify the equation using the fact that $\cos^{-1}(\cos(\theta)) = \theta$ for $\theta$ in the range $[0, \pi]$ :
$y = \frac{\pi}{2} - \theta$

STEP 13

Replace $\theta$ with $x - 2\pi k$ to find the general solution for $y$ :
$y = \frac{\pi}{2} - (x - 2\pi k)$

STEP 14

Simplify the expression for $y$ :
$y = \frac{\pi}{2} - x + 2\pi k$

STEP 15

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Find the derivative of y=cos⁡−1(sin⁡(x))y = \cos^{-1}(\sin(x))y=cos−1(sin(x)).

STEP 1

STEP 2

Understand the relationship between the sine and cosine functions. The sine function can be expressed in terms of cosine by using the complementary angle identity:sin⁡(x)=cos⁡(π2−x)\sin(x) = \cos\left(\frac{\pi}{2} - x\right)sin(x)=cos(2π​−x)

STEP 3

Replace sin⁡(x)\sin(x)sin(x) with cos⁡(π2−x)\cos\left(\frac{\pi}{2} - x\right)cos(2π​−x) in the original equation:y=cos⁡−1(cos⁡(π2−x))y = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - x\right)\right)y=cos−1(cos(2π​−x))

STEP 4

Recognize that cos⁡−1(cos⁡(θ))=θ\cos^{-1}(\cos(\theta)) = \thetacos−1(cos(θ))=θ for θ\thetaθ in the range [0,π][0, \pi][0,π]. Therefore, if π2−x\frac{\pi}{2} - x2π​−x is within this range, we can simplify the equation to:y=π2−xy = \frac{\pi}{2} - xy=2π​−x

STEP 5

Determine the range of xxx for which π2−x\frac{\pi}{2} - x2π​−x falls within the range [0,π][0, \pi][0,π]. This occurs when:0≤π2−x≤π0 \leq \frac{\pi}{2} - x \leq \pi0≤2π​−x≤π

STEP 6

Solve the inequality for xxx to find the domain where the simplification is valid:−π2≤x≤3π2-\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}−2π​≤x≤23π​

STEP 7

STEP 8

STEP 9

Use the fact that sin⁡(x)\sin(x)sin(x) has a period of 2π2\pi2π to find an equivalent angle θ\thetaθ within the desired range. We can express θ\thetaθ as:θ=x−2πk\theta = x - 2\pi kθ=x−2πkwhere kkk is an integer chosen such that θ\thetaθ is in the range [0,π][0, \pi][0,π].

STEP 10

For xxx outside the range [−π2,3π2][-\frac{\pi}{2}, \frac{3\pi}{2}][−2π​,23π​], we use the angle θ\thetaθ to rewrite the original equation as:y=cos⁡−1(sin⁡(θ))y = \cos^{-1}(\sin(\theta))y=cos−1(sin(θ))

STEP 11

Since θ\thetaθ is now within the range [0,π][0, \pi][0,π], we can again use the complementary angle identity:y=cos⁡−1(cos⁡(π2−θ))y = \cos^{-1}(\cos(\frac{\pi}{2} - \theta))y=cos−1(cos(2π​−θ))

STEP 12

Simplify the equation using the fact that cos⁡−1(cos⁡(θ))=θ\cos^{-1}(\cos(\theta)) = \thetacos−1(cos(θ))=θ for θ\thetaθ in the range [0,π][0, \pi][0,π]:y=π2−θy = \frac{\pi}{2} - \thetay=2π​−θ

STEP 13

Replace θ\thetaθ with x−2πkx - 2\pi kx−2πk to find the general solution for yyy:y=π2−(x−2πk)y = \frac{\pi}{2} - (x - 2\pi k)y=2π​−(x−2πk)

STEP 14

Simplify the expression for yyy:y=π2−x+2πky = \frac{\pi}{2} - x + 2\pi ky=2π​−x+2πk

STEP 15

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Find the derivative of $y = \cos^{-1}(\sin(x))$ .

Understand the relationship between the sine and cosine functions. The sine function can be expressed in terms of cosine by using the complementary angle identity:
$\sin(x) = \cos\left(\frac{\pi}{2} - x\right)$

Replace $\sin(x)$ with $\cos\left(\frac{\pi}{2} - x\right)$ in the original equation:
$y = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - x\right)\right)$

Recognize that $\cos^{-1}(\cos(\theta)) = \theta$ for $\theta$ in the range $[0, \pi]$ . Therefore, if $\frac{\pi}{2} - x$ is within this range, we can simplify the equation to:
$y = \frac{\pi}{2} - x$

Determine the range of $x$ for which $\frac{\pi}{2} - x$ falls within the range $[0, \pi]$ . This occurs when:
$0 \leq \frac{\pi}{2} - x \leq \pi$

Solve the inequality for $x$ to find the domain where the simplification is valid:
$-\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$

Use the fact that $\sin(x)$ has a period of $2\pi$ to find an equivalent angle $\theta$ within the desired range. We can express $\theta$ as:
$\theta = x - 2\pi k$
where $k$ is an integer chosen such that $\theta$ is in the range $[0, \pi]$ .

For $x$ outside the range $[-\frac{\pi}{2}, \frac{3\pi}{2}]$ , we use the angle $\theta$ to rewrite the original equation as:
$y = \cos^{-1}(\sin(\theta))$

Since $\theta$ is now within the range $[0, \pi]$ , we can again use the complementary angle identity:
$y = \cos^{-1}(\cos(\frac{\pi}{2} - \theta))$

Simplify the equation using the fact that $\cos^{-1}(\cos(\theta)) = \theta$ for $\theta$ in the range $[0, \pi]$ :
$y = \frac{\pi}{2} - \theta$

Replace $\theta$ with $x - 2\pi k$ to find the general solution for $y$ :
$y = \frac{\pi}{2} - (x - 2\pi k)$

Simplify the expression for $y$ :
$y = \frac{\pi}{2} - x + 2\pi k$