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6.3: Inverse Trigonometric Functions

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Section 6.3 Inverse Trig Functions

In previous sections, we have evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that for a one-to-one function, if $$f(a)=b$$, then an inverse function would satisfy $$f^{-1} (b)=a$$.

You probably are already recognizing an issue – that the sine, cosine, and tangent functions are not one-to-one functions. To define an inverse of these functions, we will need to restrict the domain of these functions to yield a new function that is one-to-one. We choose a domain for each function that includes the angle zero.

Sine, limited to $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$ Cosine, limited to $$\left[0,\pi \right]$$ Tangent, limited to $$\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)$$

On these restricted domains, we can define the inverse sine, inverse cosine, and inverse tangent functions.

Inverse Sine, Cosine, and Tangent Functions

For angles in the interval $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$, if $$\sin \left(\theta \right)=a$$, then $$\sin ^{-1} \left(a\right)=\theta$$

For angles in the interval $$\left[0,\pi \right]$$, if $$\cos \left(\theta \right)=a$$, then $$\cos ^{-1} \left(a\right)=\theta$$

For angles in the interval $$\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)$$, if $$\tan \left(\theta \right)=a$$, then $$\tan ^{-1} \left(a\right)=\theta$$

$$\sin ^{-1} \left(x\right)$$ has domain [-1, 1] and range $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$

$$\cos ^{-1} \left(x\right)$$ has domain [-1, 1] and range $$\left[0,\pi \right]$$

$$\tan ^{-1} \left(x\right)$$ has domain of all real numbers and range $$\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)$$

The $$\sin ^{-1} \left(x\right)$$ is sometimes called the arcsineArcsine, Arccosine and Arctangent function, and notated $$\arcsin \left(a\right)$$.

The $$\cos ^{-1} \left(x\right)$$ is sometimes called the arccosine function, and notated $$\arccos \left(a\right)$$.

The $$\tan ^{-1} \left(x\right)$$ is sometimes called the arctangent function, and notated $$\arctan \left(a\right)$$.

The graphs of the inverse functions are shown here:

$\sin ^{-1} \left(x\right) \cos ^{-1} \left(x\right) \tan ^{-1} \left(x\right)$

Notice that the output of each of these inverse functions is an angle.

Example $$\PageIndex{1}$$

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1

Evaluate

a) $$\sin ^{-1} \left(\frac{1}{2} \right)$$ b) $$\sin ^{-1} \left(-\frac{\sqrt{2} }{2} \right)$$ c) $$\cos ^{-1} \left(-\frac{\sqrt{3} }{2} \right)$$ d) $$\tan ^{-1} \left(1\right)$$

a) Evaluating $$\sin ^{-1} \left(\frac{1}{2} \right)$$ is the same as asking what angle would have a sine value of $$\frac{1}{2}$$. In other words, what angle $$\theta$$ would satisfy $$\sin \left(\theta \right)=\frac{1}{2}$$?

There are multiple angles that would satisfy this relationship, such as $$\frac{\pi }{6}$$ and $$\frac{5\pi }{6}$$ , but we know we need the angle in the range of $$\sin ^{-1} \left(x\right)$$, the interval $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$, so the answer will be $$\sin ^{-1} \left(\frac{1}{2} \right)=\frac{\pi }{6}$$.

Remember that the inverse is a function so for each input, we will get exactly one output.

b) Evaluating $$\sin ^{-1} \left(-\frac{\sqrt{2} }{2} \right)$$, we know that $$\frac{5\pi }{4}$$ and $$\frac{7\pi }{4}$$ both have a sine value of $$-\frac{\sqrt{2} }{2}$$, but neither is in the interval $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$. For that, we need the negative angle coterminal with $$\frac{7\pi }{4}$$. $$\sin ^{-1} \left(-\frac{\sqrt{2} }{2} \right)=-\frac{\pi }{4}$$.

c) Evaluating $$\cos ^{-1} \left(-\frac{\sqrt{3} }{2} \right)$$, we are looking for an angle in the interval $$\left[0,\pi \right]$$ with a cosine value of $$-\frac{\sqrt{3} }{2}$$. The angle that satisfies this is $$\cos ^{-1} \left(-\frac{\sqrt{3} }{2} \right)=\frac{5\pi }{6}$$.

d) Evaluating $$\tan ^{-1} \left(1\right)$$, we are looking for an angle in the interval $$\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)$$ with a tangent value of 1. The correct angle is $$\tan ^{-1} \left(1\right)=\frac{\pi }{4}$$.

Exercise $$\PageIndex{1}$$

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1. Evaluate

a) $$\sin ^{-1} \left(-1\right)$$ b) $$\tan ^{-1} \left(-1\right)$$ c) $$\cos ^{-1} \left(-1\right)$$ d) $$\cos ^{-1} \left(\frac{1}{2} \right)$$

Example $$\PageIndex{1}$$

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2

Evaluate $$\sin ^{-1} \left(0.97\right)$$ using your calculator.

Since the output of the inverse function is an angle, your calculator will give you a degree value if in degree mode, and a radian value if in radian mode.

In radian mode, $$\sin ^{-1} \eqref{GrindEQ__0_97_}\approx 1.3252$$ In degree mode, $$\sin ^{-1} \left(0.97\right)\approx 75.93{}^\circ$$

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2. Evaluate $$\cos ^{-1} \left(-0.4\right)$$ using your calculator.

In Section 5.5, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trig functions, we can solve for the angles of a right triangle given two sides.

Example $$\PageIndex{1}$$

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3

Solve the triangle for the angle $$\theta$$.

Since we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.

$$\cos \left(\theta \right)=\frac{9}{12}$$ Using the definition of the inverse,

$$\theta =\cos ^{-1} \left(\frac{9}{12} \right)$$ Evaluating

$$\theta \approx 0.7227$$, or about 41.4096$$\mathrm{{}^\circ}$$

There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can find exact values for the resulting expressions

Example $$\PageIndex{4}$$

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4

Evaluate $$\sin ^{-1} \left(\cos \left(\frac{13\pi }{6} \right)\right)$$.

a) Here, we can directly evaluate the inside of the composition. $\cos \left(\frac{13\pi }{6} \right)=\frac{\sqrt{3} }{2}$

Now, we can evaluate the inverse function as we did earlier. $\sin ^{-1} \left(\frac{\sqrt{3} }{2} \right)=\frac{\pi }{3}$

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3. Evaluate $$\cos ^{-1} \left(\sin \left(-\frac{11\pi }{4} \right)\right)$$.

Example $$\PageIndex{1}$$

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5

Find an exact value for $$\sin \left(\cos ^{-1} \left(\frac{4}{5} \right)\right)$$.

Beginning with the inside, we can say there is some angle so $$\theta =\cos ^{-1} \left(\frac{4}{5} \right)$$, which means $$\cos \left(\theta \right)=\frac{4}{5}$$, and we are looking for $$\sin \left(\theta \right)$$. We can use the Pythagorean identity to do this.

$$\sin ^{2} \left(\theta \right)+\cos ^{2} \left(\theta \right)=1$$ Using our known value for cosine

$$\sin ^{2} \left(\theta \right)+\left(\frac{4}{5} \right)^{2} =1$$ Solving for sine $\sin ^{2} \left(\theta \right)=1-\frac{16}{25}$ $\sin \left(\theta \right)=\pm \sqrt{\frac{9}{25} } =\pm \frac{3}{5}$

Since we know that the inverse cosine always gives an angle on the interval $$\left[0,\pi \right]$$, we know that the sine of that angle must be positive, so $$\sin \left(\cos ^{-1} \left(\frac{4}{5} \right)\right)=\sin (\theta )=\frac{3}{5}$$

Example $$\PageIndex{1}$$

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6

Find an exact value for $$\sin \left(\tan ^{-1} \left(\frac{7}{4} \right)\right)$$.

While we could use a similar technique as in the last example, we will demonstrate a different technique here. From the inside, we know there is an angle so $$\tan \left(\theta \right)=\frac{7}{4}$$. We can envision this as the opposite and adjacent sides on a right triangle.

Using the Pythagorean Theorem, we can find the hypotenuse of this triangle: $4^{2} +7^{2} =hypotenuse\; ^{2}$ $hypotenuse=\sqrt{65}$

Now, we can represent the sine of the angle as opposite side divided by hypotenuse. $\sin \left(\theta \right)=\frac{7}{\sqrt{65} }$

This gives us our desired composition $\sin \left(\tan ^{-1} \left(\frac{7}{4} \right)\right)=\sin (\theta )=\frac{7}{\sqrt{65} } .$

Exercise $$\PageIndex{1}$$

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4. Evaluate $$\cos \left(\sin ^{-1} \left(\frac{7}{9} \right)\right)$$.

We can also find compositions involving algebraic expressions

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7

Find a simplified expression for $$\cos \left(\sin ^{-1} \left(\frac{x}{3} \right)\right)$$, for $$-3\le x\le 3$$.

We know there is an angle $$\theta$$ so that $$\sin \left(\theta \right)=\frac{x}{3}$$. Using the Pythagorean Theorem,

$$\sin ^{2} \left(\theta \right)+\cos ^{2} \left(\theta \right)=1$$ Using our known expression for sine

$$\left(\frac{x}{3} \right)^{2} +\cos ^{2} \left(\theta \right)=1$$ Solving for cosine $\cos ^{2} \left(\theta \right)=1-\frac{x^{2} }{9}$ $\cos \left(\theta \right)=\pm \sqrt{\frac{9-x^{2} }{9} } =\pm \frac{\sqrt{9-x^{2} } }{3}$ Since we know that the inverse sine must give an angle on the interval $$\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]$$, we can deduce that the cosine of that angle must be positive. This gives us

$\cos \left(\sin ^{-1} \left(\frac{x}{3} \right)\right)=\frac{\sqrt{9-x^{2} } }{3}$

Exercise $$\PageIndex{1}$$

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5. Find a simplified expression for $$\sin \left(\tan ^{-1} \left(4x\right)\right)$$, for $$-\frac{1}{4} \le x\le \frac{1}{4}$$.

Important Topics of This Section

• Inverse trig functions: arcsine, arccosine and arctangent
• Domain restrictions
• Evaluating inverses using unit circle values and the calculator
• Simplifying numerical expressions involving the inverse trig functions
• Simplifying algebraic expressions involving the inverse trig functions

Try it Now Answers

1. a) $$-\frac{\pi }{2}$$ b) $$-\frac{\pi }{4}$$ c) $$\pi$$ d) $$\frac{\pi }{3}$$

2. 1.9823 or 113.578$$\mathrm{{}^\circ}$$

$3. \sin \left(-\frac{11\pi }{4} \right)=-\frac{\sqrt{2} }{2} . \cos ^{-1} \left(-\frac{\sqrt{2} }{2} \right)=\frac{3\pi }{4}$

4. Let $$\theta =\sin ^{-1} \left(\frac{7}{9} \right)$$ so $$\sin (\theta )=\frac{7}{9}$$. .Using Pythagorean Identity, $$\sin ^{2} \theta +\cos ^{2} \theta =1$$, so $$\left(\frac{7}{9} \right)^{2} +\cos ^{2} \theta =1$$. Solving, $$\cos \left(\sin ^{-1} \left(\frac{7}{9} \right)\right)=\cos \left(\theta \right)=\frac{4\sqrt{2} }{9}$$ .

5. Let $$\theta =\tan ^{-1} \left(4x\right)$$, so $$\tan (\theta )=4x$$. We can represent this on a triangle as $$\tan (\theta )=\frac{4x}{1}$$.The hypotenuse of the triangle would be $$\sqrt{\left(4x\right)^{2} +1}$$. $$\sin \left(\tan ^{-1} \left(4x\right)\right)=\sin (\theta )=\frac{4x}{\sqrt{16x^{2} +1} }$$