Solved on Sep 06, 2023

Find the value of $f(x) = 5 \sin^{-1}(\sin(x)) + 3 \cos^{-1}(\sin(4x))$ at $x = \pi/3$ without a calculator.

STEP 1

Assumptions1. The function is given by $f(x)=5 \sin ^{-1}(\sin (x))+3 \cos ^{-1}(\sin (4 x))$ . We need to find the value of $f(x)$ at $x=\frac{\pi}{3}$

STEP 2

Let's first evaluate the first part of the function, $5 \sin ^{-1}(\sin (x))$ , at $x=\frac{\pi}{}$ .
$5 \sin ^{-1}(\sin (\frac{\pi}{}))$

STEP 3

We know that $\sin ^{-1}(\sin (x))$ is simply $x$ for $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ . But since $\frac{\pi}{3}$ is within this interval, we can simplify the above expression to $5 \times \frac{\pi}{3}$

STEP 4

Calculate the value of the first part of the function at $x=\frac{\pi}{3}$ .
$\times \frac{\pi}{3} = \frac{\pi}{3}$

STEP 5

Now, let's evaluate the second part of the function, $3 \cos ^{-1}(\sin (4 x))$ , at $x=\frac{\pi}{3}$ .
$3 \cos ^{-1}(\sin (4 \times \frac{\pi}{3}))$

STEP 6

We know that $\sin(4x)$ is periodic with period $\frac{\pi}{2}$ , so $\sin(4x)$ and $\sin(4x +2\pi)$ are the same. We can use this property to simplify the argument of the $\sin$ function to a value within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ .
$3 \cos ^{-1}(\sin (4 \times \frac{\pi}{3} -2\pi))$

STEP 7

implify the argument of the $\sin$ function.
$3 \cos ^{-1}(\sin (\frac{2\pi}{3}))$

STEP 8

We know that $\cos ^{-1}(\sin (x))$ is equal to $\frac{\pi}{2} - x$ for $0 \leq x \leq \pi$ . Since $\frac{2\pi}{3}$ is within this interval, we can simplify the above expression to $3 \times \left(\frac{\pi}{2} - \frac{2\pi}{3}\right)$

STEP 9

Calculate the value of the second part of the function at $x=\frac{\pi}{3}$ .
$3 \times \left(\frac{\pi}{2} - \frac{2\pi}{3}\right) = \frac{\pi}{2}$

STEP 10

Finally, add the values of the two parts of the function to get the value of $f(x)$ at $x=\frac{\pi}{3}$ .
$f\left(\frac{\pi}{3}\right) = \frac{5\pi}{3} + \frac{\pi}{2}$

STEP 11

Calculate the value of $f(x)$ at $x=\frac{\pi}{3}$ .
$f\left(\frac{\pi}{3}\right) = \frac{5\pi}{3} + \frac{\pi}{} = \frac{11\pi}{6}$ The value of the function $f(x)$ at $x=\frac{\pi}{3}$ is $\frac{11\pi}{6}$ .

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Find the value of $f(x) = 5 \sin^{-1}(\sin(x)) + 3 \cos^{-1}(\sin(4x))$ at $x = \pi/3$ without a calculator.

STEP 1

Assumptions1. The function is given by $f(x)=5 \sin ^{-1}(\sin (x))+3 \cos ^{-1}(\sin (4 x))$ . We need to find the value of $f(x)$ at $x=\frac{\pi}{3}$

STEP 2

Let's first evaluate the first part of the function, $5 \sin ^{-1}(\sin (x))$ , at $x=\frac{\pi}{}$ .
$5 \sin ^{-1}(\sin (\frac{\pi}{}))$

STEP 3

We know that $\sin ^{-1}(\sin (x))$ is simply $x$ for $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ . But since $\frac{\pi}{3}$ is within this interval, we can simplify the above expression to $5 \times \frac{\pi}{3}$

STEP 4

Calculate the value of the first part of the function at $x=\frac{\pi}{3}$ .
$\times \frac{\pi}{3} = \frac{\pi}{3}$

STEP 5

Now, let's evaluate the second part of the function, $3 \cos ^{-1}(\sin (4 x))$ , at $x=\frac{\pi}{3}$ .
$3 \cos ^{-1}(\sin (4 \times \frac{\pi}{3}))$

STEP 6

STEP 7

implify the argument of the $\sin$ function.
$3 \cos ^{-1}(\sin (\frac{2\pi}{3}))$

STEP 8

We know that $\cos ^{-1}(\sin (x))$ is equal to $\frac{\pi}{2} - x$ for $0 \leq x \leq \pi$ . Since $\frac{2\pi}{3}$ is within this interval, we can simplify the above expression to $3 \times \left(\frac{\pi}{2} - \frac{2\pi}{3}\right)$

STEP 9

Calculate the value of the second part of the function at $x=\frac{\pi}{3}$ .
$3 \times \left(\frac{\pi}{2} - \frac{2\pi}{3}\right) = \frac{\pi}{2}$

STEP 10

Finally, add the values of the two parts of the function to get the value of $f(x)$ at $x=\frac{\pi}{3}$ .
$f\left(\frac{\pi}{3}\right) = \frac{5\pi}{3} + \frac{\pi}{2}$

STEP 11

Calculate the value of $f(x)$ at $x=\frac{\pi}{3}$ .
$f\left(\frac{\pi}{3}\right) = \frac{5\pi}{3} + \frac{\pi}{} = \frac{11\pi}{6}$ The value of the function $f(x)$ at $x=\frac{\pi}{3}$ is $\frac{11\pi}{6}$ .

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Find the value of f(x)=5sin⁡−1(sin⁡(x))+3cos⁡−1(sin⁡(4x))f(x) = 5 \sin^{-1}(\sin(x)) + 3 \cos^{-1}(\sin(4x))f(x)=5sin−1(sin(x))+3cos−1(sin(4x)) at x=π/3x = \pi/3x=π/3 without a calculator.

STEP 1

Assumptions1. The function is given by f(x)=5sin⁡−1(sin⁡(x))+3cos⁡−1(sin⁡(4x))f(x)=5 \sin ^{-1}(\sin (x))+3 \cos ^{-1}(\sin (4 x))f(x)=5sin−1(sin(x))+3cos−1(sin(4x)) . We need to find the value of f(x)f(x)f(x) at x=π3x=\frac{\pi}{3}x=3π​

STEP 2

Let's first evaluate the first part of the function, 5sin⁡−1(sin⁡(x))5 \sin ^{-1}(\sin (x))5sin−1(sin(x)), at x=πx=\frac{\pi}{}x=π​.5sin⁡−1(sin⁡(π))5 \sin ^{-1}(\sin (\frac{\pi}{}))5sin−1(sin(π​))

STEP 3

STEP 4

Calculate the value of the first part of the function at x=π3x=\frac{\pi}{3}x=3π​.×π3=π3 \times \frac{\pi}{3} = \frac{\pi}{3}×3π​=3π​

STEP 5

Now, let's evaluate the second part of the function, 3cos⁡−1(sin⁡(4x))3 \cos ^{-1}(\sin (4 x))3cos−1(sin(4x)), at x=π3x=\frac{\pi}{3}x=3π​.3cos⁡−1(sin⁡(4×π3))3 \cos ^{-1}(\sin (4 \times \frac{\pi}{3}))3cos−1(sin(4×3π​))

STEP 6

STEP 7

implify the argument of the sin⁡\sinsin function.3cos⁡−1(sin⁡(2π3))3 \cos ^{-1}(\sin (\frac{2\pi}{3}))3cos−1(sin(32π​))

STEP 8

STEP 9

Calculate the value of the second part of the function at x=π3x=\frac{\pi}{3}x=3π​.3×(π2−2π3)=π23 \times \left(\frac{\pi}{2} - \frac{2\pi}{3}\right) = \frac{\pi}{2}3×(2π​−32π​)=2π​

STEP 10

Finally, add the values of the two parts of the function to get the value of f(x)f(x)f(x) at x=π3x=\frac{\pi}{3}x=3π​.f(π3)=5π3+π2f\left(\frac{\pi}{3}\right) = \frac{5\pi}{3} + \frac{\pi}{2}f(3π​)=35π​+2π​

STEP 11

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Find the value of $f(x) = 5 \sin^{-1}(\sin(x)) + 3 \cos^{-1}(\sin(4x))$ at $x = \pi/3$ without a calculator.

Assumptions1. The function is given by $f(x)=5 \sin ^{-1}(\sin (x))+3 \cos ^{-1}(\sin (4 x))$ . We need to find the value of $f(x)$ at $x=\frac{\pi}{3}$

Let's first evaluate the first part of the function, $5 \sin ^{-1}(\sin (x))$ , at $x=\frac{\pi}{}$ .
$5 \sin ^{-1}(\sin (\frac{\pi}{}))$

Calculate the value of the first part of the function at $x=\frac{\pi}{3}$ .
$\times \frac{\pi}{3} = \frac{\pi}{3}$

Now, let's evaluate the second part of the function, $3 \cos ^{-1}(\sin (4 x))$ , at $x=\frac{\pi}{3}$ .
$3 \cos ^{-1}(\sin (4 \times \frac{\pi}{3}))$

implify the argument of the $\sin$ function.
$3 \cos ^{-1}(\sin (\frac{2\pi}{3}))$

Calculate the value of the second part of the function at $x=\frac{\pi}{3}$ .
$3 \times \left(\frac{\pi}{2} - \frac{2\pi}{3}\right) = \frac{\pi}{2}$

Finally, add the values of the two parts of the function to get the value of $f(x)$ at $x=\frac{\pi}{3}$ .
$f\left(\frac{\pi}{3}\right) = \frac{5\pi}{3} + \frac{\pi}{2}$