Evaluate Expressions That Include the Inverse Tangent, Cosecant, Secant, or Cotangent Function - Pre-Calculus

Card 0 of 48

Evaluate:

$cos^$-^1$\left($\frac{\sqrt3}{2}\right)+sin^$-^1$\left($\frac{\sqrt3}{2}\right)=$

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To determine the value of $cos^$-^1$($\frac{\sqrt3}{2}$)+sin^$-^1$($\frac{\sqrt3}{2}$)$ , solve each of the terms first.

$cos^$-^1$($\frac{\sqrt3}{2}$)$

The inverse cosine has a domain and range restriction.

The domain exists from , and the range from $[0,\pi]$ . The inverse cosine asks for the angle when the x-value of the existing coordinate is $\frac{\sqrt3}{2}$ . The only possibility is $\frac{\pi}{6}$ since the coordinate can only exist in the first quadrant.

$sin^$-^1$($\frac{\sqrt3}{2}$)$

The inverse sine also has a domain and range restriction.

The domain exists from , and the range from $[-$\frac{\pi}{2}$,$\frac{\pi}{2}$]$ . The inverse sine asks for the angle when the y-value of the existing coordinate is $\frac{\sqrt3}{2}$ . The only possibility is $\frac{\pi}{3}$ since the coordinate can only exist in the first quadrant.

Therefore:

$cos^$-^1$($\frac{\sqrt3}{2}$)+sin^$-^1$($\frac{\sqrt3}{2}$)= $\frac{\pi}{6}$+\frac{\pi}{3}$=\frac{\pi}{2}$$

Approximate:

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:

There is a restriction for the range of the inverse tangent function from $(-$\frac{\pi}{2}$,$\frac{\pi}{2}$)$ .

The inverse tangent of a value asks for the angle where the coordinate lies on the unit circle under the condition that $$\frac{y}{x}$=99999$ . For this to be valid on the unit circle, the must be very close to 1, with an value also very close to zero, but cannot equal to zero since $\frac{y}{x}$ would be undefined.

The point is located on the unit circle when $\theta=\frac{\pi}{2}$$ , but $tan($\frac{\pi}{2}$)$ is invalid due to the existent asymptote at this angle.

An example of a point very close to that will yield $$\frac{y}{x}$=99999$ can be written as:

Therefore, the approximated rounded value of is $\frac{\pi}{2}$ .

Determine the value of in degrees.

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Rewrite and evaluate .

$csc^$-^1$(2)=sin^$-^1$($\frac{1}{2}$)=\frac{\pi}{6}$$

The inverse sine of one-half is $\frac{\pi}{6}$ since $\frac{1}{2}$ is the y-value of the coordinate when the angle is $\frac{\pi}{6}$ .

$2csc^$-^1$(2)=2($\frac{\pi}{6}$)=\frac{\pi}{3}$$

To convert from radians to degrees, replace $\pi$ with 180.

$$\frac{\pi}{3}$=\frac{180}{3}$=60$

Evaluate the following:

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For this particular problem we need to recall that the inverse cosine cancels out the cosine therefore,

.

So the expression just becomes

$f(x)=cos\left($\frac{\pi }{3}\right)$

From here, recall the unit circle for specific angles such as $\frac{\pi}{3}$ .

Thus,

$f(x)=cos\left($\frac{\pi }{3}\right)=\frac{1}{2}$$ .

Evaluate the following expression: $f(x)=tan(arctan(tan($\frac{\pi}{4}$)))$

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This one seems complicated, but becomes considerably easier once you implement the fact that the composite cancels out to 1 and you are left with $tan($\frac{\pi}{4}$)$ which is equal to 1

Evaluate:

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$arcsin($\frac{\sqrt{3}$}{2})=60^\circ\rightarrow $\frac{\pi}{3}\rightarrow 5\cdot $\frac{\pi}{3}$$

Approximate the following:

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This one is rather simple with knowledge of the unit circle: the value is extremely close to zero, of which always

Given that $sin\theta=\frac{3}{4}$$ and that $\theta$ is acute, find the value of $cot\theta$ without using a calculator.

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Given the value of the opposite and hypotenuse sides from the sine expression (3 and 4 respectively) we can use the Pythagorean Theorem to find the 3rd side (we’ll call it “t”): . From here we can easily deduce the value of $cot\theta=\frac{\sqrt{7}$}{3}$ (the adjacent side over the opposite side)

Evaluate the following expression: $f\left ( x \right )=\tan (\arctan (\tan ($\frac{\pi }{4}$)))$

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This one seems complicated but becomes considerably easier once you implement the fact that the composite $\tan \left ( \arctan \right )$ cancels out to and you are left with $\tan \left ( $\frac{\pi }{4}$ \right )$ which is equal to , and so the answer is .

Evaluate: $5\sin ^{-1}\left ( $\frac{\sqrt{3}$}{2} \right )$

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$\arcsin \left ( $\frac{\sqrt{3}$}2{} \right $)=60^{\circ}$\rightarrow $\frac{\pi }{3}\rightarrow 5\cdot $\frac{\pi }{3}$$ and so the credited answer is $\frac{5\pi }{3}$ .

Approximate the following: $\arcsin \left ( 0.00002 \right )$ is closest in value to which of the following?

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This problem is quite manageable with knowledge of the unit circle: the value is extremely close to zero, of which $\arcsin \left ( 0 \right )= 0$ always, so the only reasonable estimation of this value is 0.

Given that $\sin \theta =\frac{3}{4}$$ and that $\theta$ is acute, find the value of $\cot \theta$ without using a calculator.

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Given the value of the opposite and hypotenuse sides from the sine expression (3 and 4 respectively) we can use the Pythagorean Theorem to find the 3rd side (we’ll call it “t”): .

From here we can deduce the value of $\cot \theta =\frac{\sqrt{7}$}{3}$ (the adjacent side over the opposite side) and so the answer is $$\frac{\sqrt{7}$}{3}$ .

Evaluate:

$cos^$-^1$\left($\frac{\sqrt3}{2}\right)+sin^$-^1$\left($\frac{\sqrt3}{2}\right)=$

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To determine the value of $cos^$-^1$($\frac{\sqrt3}{2}$)+sin^$-^1$($\frac{\sqrt3}{2}$)$ , solve each of the terms first.

$cos^$-^1$($\frac{\sqrt3}{2}$)$

The inverse cosine has a domain and range restriction.

The domain exists from , and the range from $[0,\pi]$ . The inverse cosine asks for the angle when the x-value of the existing coordinate is $\frac{\sqrt3}{2}$ . The only possibility is $\frac{\pi}{6}$ since the coordinate can only exist in the first quadrant.

$sin^$-^1$($\frac{\sqrt3}{2}$)$

The inverse sine also has a domain and range restriction.

The domain exists from , and the range from $[-$\frac{\pi}{2}$,$\frac{\pi}{2}$]$ . The inverse sine asks for the angle when the y-value of the existing coordinate is $\frac{\sqrt3}{2}$ . The only possibility is $\frac{\pi}{3}$ since the coordinate can only exist in the first quadrant.

Therefore:

$cos^$-^1$($\frac{\sqrt3}{2}$)+sin^$-^1$($\frac{\sqrt3}{2}$)= $\frac{\pi}{6}$+\frac{\pi}{3}$=\frac{\pi}{2}$$

Approximate:

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:

There is a restriction for the range of the inverse tangent function from $(-$\frac{\pi}{2}$,$\frac{\pi}{2}$)$ .

The inverse tangent of a value asks for the angle where the coordinate lies on the unit circle under the condition that $$\frac{y}{x}$=99999$ . For this to be valid on the unit circle, the must be very close to 1, with an value also very close to zero, but cannot equal to zero since $\frac{y}{x}$ would be undefined.

The point is located on the unit circle when $\theta=\frac{\pi}{2}$$ , but $tan($\frac{\pi}{2}$)$ is invalid due to the existent asymptote at this angle.

An example of a point very close to that will yield $$\frac{y}{x}$=99999$ can be written as:

Therefore, the approximated rounded value of is $\frac{\pi}{2}$ .

Determine the value of in degrees.

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Rewrite and evaluate .

$csc^$-^1$(2)=sin^$-^1$($\frac{1}{2}$)=\frac{\pi}{6}$$

The inverse sine of one-half is $\frac{\pi}{6}$ since $\frac{1}{2}$ is the y-value of the coordinate when the angle is $\frac{\pi}{6}$ .

$2csc^$-^1$(2)=2($\frac{\pi}{6}$)=\frac{\pi}{3}$$

To convert from radians to degrees, replace $\pi$ with 180.

$$\frac{\pi}{3}$=\frac{180}{3}$=60$

Evaluate the following:

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For this particular problem we need to recall that the inverse cosine cancels out the cosine therefore,

.

So the expression just becomes

$f(x)=cos\left($\frac{\pi }{3}\right)$

From here, recall the unit circle for specific angles such as $\frac{\pi}{3}$ .

Thus,

$f(x)=cos\left($\frac{\pi }{3}\right)=\frac{1}{2}$$ .

Evaluate the following expression: $f(x)=tan(arctan(tan($\frac{\pi}{4}$)))$

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This one seems complicated, but becomes considerably easier once you implement the fact that the composite cancels out to 1 and you are left with $tan($\frac{\pi}{4}$)$ which is equal to 1

Evaluate:

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$arcsin($\frac{\sqrt{3}$}{2})=60^\circ\rightarrow $\frac{\pi}{3}\rightarrow 5\cdot $\frac{\pi}{3}$$

Approximate the following:

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This one is rather simple with knowledge of the unit circle: the value is extremely close to zero, of which always

Given that $sin\theta=\frac{3}{4}$$ and that $\theta$ is acute, find the value of $cot\theta$ without using a calculator.

Tap to see back →

Given the value of the opposite and hypotenuse sides from the sine expression (3 and 4 respectively) we can use the Pythagorean Theorem to find the 3rd side (we’ll call it “t”): . From here we can easily deduce the value of $cot\theta=\frac{\sqrt{7}$}{3}$ (the adjacent side over the opposite side)