OBJECTIVES

When you finish this section, you should be able to:

1. Define the inverse trigonometric functions (p. 58)
2. Use the inverse trigonometric functions (p. 61)
3. Solve trigonometric equations (p. 61)

IN WORDS

We read \(y=\sin ^{-1}{x}\) as "\({y}\) is the number (or angle) whose sine equals \(x\)".

spanDEFINITIONspan Inverse Sine Function

The inverse sine function, denoted by \(y= \sin ^{-1}x,\) or \(y=\arcsin x,\) is defined as \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\begin{array}{l} y = \sin ^{ - 1} \,x\,\quad{\rm{if}}\,{\rm{and}}\,{\rm{only}}\,{\rm{if}}\,\quad\,x = \sin \,y \\ \,\,\,\,\,\,{\rm{where}}\, - {\rm{1}} \le x \le 1\,\quad{\rm{and}}\,\quad - \frac{\pi }{2} \le y \le \frac{\pi }{2} \\ \end{array}}} \]

CAUTION

The superscript \(^{-1}\) that appears in \(y=\sin ^{-1}{x}\) is not an exponent but the symbol used to denote the inverse function. (To avoid confusion, \(y=\sin ^{-1}{x}\) is sometimes written \(y= {\rm arcsin}\ x\).)

Finding the Values of an Inverse Sine Function

Find the exact value of:

1. \(\sin ^{-1}\dfrac{\sqrt{3}}{2}\)
2. \(\sin ^{-1}\left( - \dfrac{\sqrt{2}}{2}\right)\)
3. \(\sin ^{-1}\left( \sin \dfrac{ \pi }{8} \right)\)
4. \(\sin ^{-1}\left( \sin \dfrac{5\pi }{8 } \right)\)

Solution (a) We seek an angle whose sine is \(\dfrac{\sqrt{3}}{2} .\) Since \(-1{<}\dfrac{\sqrt{3}}{2}{<}1,\) let \(y=\sin ^{-1}\dfrac{\sqrt{3}}{2}\). Then by definition, \(\sin y=\dfrac{\sqrt{3}}{2}\), where \(-\dfrac{\pi }{2} \leq y\leq \dfrac{\pi }{2}\). Although \(\sin y=\dfrac{\sqrt{3}}{2}\) has infinitely many solutions, the only number in the interval \(\left[ -\dfrac{ \pi }{2},\dfrac{\pi }{2}\right] \), for which \(\sin y=\dfrac{\sqrt{3}}{2}\), is \(\dfrac{\pi }{3}.\) So \(\sin ^{-1}\dfrac{\sqrt{3}}{2}=\dfrac{\pi }{3}\).

(b) We seek an angle whose sine is \(-\dfrac{\sqrt{2}}{2}.\) Since \( -1<-\dfrac{\sqrt{2}}{2}<1\), let \(y=\sin ^{-1}\left( -\dfrac{\sqrt{2}}{2}\right)\). Then by definition, \(\sin y=-\dfrac{\sqrt{2}}{2}\), where \(-\dfrac{\pi }{2}\leq y\leq \dfrac{\pi }{2}\). The only number in the interval \( \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right] ,\) whose sine is \(-\dfrac{ \sqrt{2}}{2},\) is \(-\dfrac{\pi }{4}\). So \(\sin ^{-1}\left( -\dfrac{\sqrt{2}}{ 2}\right) =-\dfrac{\pi }{4}\).

(c) Since the number \(\dfrac{\pi }{8}\) is in the interval \(\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right]\), we use (1). \[ \sin ^{-1}\left( \sin \dfrac{\pi }{8}\right) =\dfrac{\pi }{8} \]

Figure 84

(d) Since the number \(\dfrac{5\pi }{8}\) is not in the closed interval \(\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right] ,\) we cannot use (1). Instead, we find a number \(x\) in the interval \(\left[ -\dfrac{\pi }{2} ,\dfrac{\pi }{2}\right] \) for which \(\sin x=\sin \dfrac{5\pi }{8}.\) Using Figure 84, we see that \(\sin \dfrac{5\pi }{8}=y=\sin \dfrac{3\pi }{8}.\) The number \(\dfrac{3\pi }{8}\) is in the interval \( \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right]\), so \[ \sin ^{-1}\left( \sin \dfrac{5\pi }{8}\right) =\sin ^{-1}\left( \sin \dfrac{ 3\pi }{8}\right) =\dfrac{3\pi }{8} \]

NOW WORK

Problems 9 and 17.

spanDEFINITIONspan Inverse Tangent Function

The inverse tangent function, symbolized by \( y= \tan ^{-1}x,\) or \(y=\arctan x,\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\begin{array}{l} \,\,\,\,\,\,\,\,\,y = \tan ^{ - 1} \,x\,\quad{\rm{if}}\,{\rm{and}}\,{\rm{only}}\,{\rm{if}}\,\quad\,x = \tan \,y \\ {\rm{where}}\, - \infty < x < \infty \,\quad{\rm{and}}\,\quad - \frac{\pi }{2} < y < \frac{\pi }{2} \\ \end{array}}} \]

NOW WORK

Problem 13.

spanDEFINITIONspan Inverse Secant Function

The inverse secant function, symbolized by \(y= \sec ^{-1}x\), or \(y\) = arcsec \(x\), is \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\begin{array}{c} y= \sec ^{-1}x\quad \hbox{ if and only if }\quad x=\sec y \\[6pt] \hbox{where } \vert x\vert \geq 1\quad \hbox{and}\quad 0\leq y< \dfrac{\pi }{2}\quad \hbox{or}\quad \pi \leq y<\dfrac{3\pi }{2} \end{array}}} \]

spanDEFINITIONspan

• Inverse Cosine Function \(y= \cos ^{-1}x\) if and only if \(x=\cos y, {\rm{where}}\, -1\leq x\leq 1 \quad {\rm{and}}\, \quad 0\leq y\leq \pi \)
• Inverse Cotangent Function \(y=\cot ^{-1}x\) if and only if \(x=\cot y, {\rm{where}}\, -\infty <x<\infty \quad {\rm{and}}\, \quad 0<y<\pi \)
• Inverse Cosecant Function \(y=\csc ^{-1}x\) if and only if \(x=\csc y, {\rm{where}}\, \vert x\vert \geq 1\, {\rm{and}}\, \\ -\pi <y\leq - \dfrac{\pi }{2} \quad {\rm{or}} \quad 0<y\leq \dfrac{\pi }{2}\)

THEOREM

• \(\cos ^{-1}x=\dfrac{\pi }{2}-\sin ^{-1}x,\) \(-1\leq x\leq 1\)
• \(\cot ^{-1}x=\dfrac{\pi }{2}-\tan ^{-1}x\), \(-\infty <x<\infty \)
• \(\csc ^{-1}x= \dfrac{\pi }{2}-\sec ^{-1}x,\quad \vert x\vert \geq 1\)

Writing a Trigonometric Expression as an Algebraic Expression

Write \(\sin ( \tan ^{-1}u) \) as an algebraic expression containing \(u.\)

Figure 88 \(\tan \theta =u\); \(-\dfrac{\pi }{2}<\theta < \dfrac{\pi }{2}\).

Solution Let \(\theta =\tan ^{-1}u\) so that \(\tan \theta =u\), where \(-\dfrac{\pi }{2}<\theta <\dfrac{\pi }{2}\). We note that in the interval \(\left( -\dfrac{\pi }{2},\dfrac{\pi }{2}\right) ,\) \(\sec \theta >0.\) Then \[ \begin{array}{l} \sin (\tan ^{ - 1} u) = \sin \theta = \sin \theta \cdot \frac{{\cos \theta }}{{\cos \theta }} = \tan \theta \cos \theta \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\tan \theta }}{{\sec \theta }} = \frac{{\tan \theta }}{{\sqrt {1 + \tan ^2 \theta } }} = \frac{u}{{\sqrt {1 + u^2 } }} \\ {\color{#0066A7}{ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec ^2 \theta = 1 + \mathop {\tan ^2 }\limits^ \uparrow \theta ;\sec \theta > 0}} \\ \end{array} \]

An alternate method of obtaining the solution to Example 2 uses right triangles. Let \(\theta =\tan ^{-1}u\) so that \(\tan \theta =u\), \(- \dfrac{\pi }{2}<\theta <\dfrac{\pi }{2}\), and label the right triangles drawn in Figure 88. Using the Pythagorean Theorem, the hypotenuse of each triangle is \(\sqrt{1+u^{2}}.\) Then \(\sin (\tan ^{-1}u) =\sin \theta =\dfrac{u}{\sqrt{1+u^{2}}}.\)

NOW WORK

Problem 23.

Solving Trigonometric Equations

Solve the equations:

1. \(\sin \theta =\dfrac{1}{2}\)
2. \(\cos \theta =0.4\)

Give a general formula for all the solutions and list all the solutions in the interval \([ -2\pi ,2\pi ] .\)

Figure 89

Solution (a) Use the inverse sine function \(y =\sin ^{-1} x\), \(-\dfrac{\pi}{2}\leq y\leq \dfrac{\pi}{2}\). \[ \begin{array}{l} \sin \theta = \frac{1}{2} \\ \,\,\,\theta = \sin ^{ - 1} \frac{1}{2}\, \quad {\color{#0066A7}{- \frac{\pi }{2} \le \theta \le \frac{\pi }{2}}} \\ \,\,\theta = \frac{\pi }{6} \\ \end{array} \]

Over the interval \([ 0,2\pi ] ,\) there are two angles \(\theta \) for which \(\sin \theta =\dfrac{1}{2}\). See Figure 89.

All the solutions of \(\sin \theta =\dfrac{1}{2}\) are given by the general formula \[ \theta =\dfrac{\pi }{6}+2k\pi\quad {\rm \ or \ }\quad \theta =\dfrac{5\pi }{6}+2k\pi,\qquad {\rm where }\ k \ \hbox{is any integer} \]

The solutions in the interval \([-2\pi, 2\pi] \) are \[ \left\{ -\dfrac{11\pi }{6},-\dfrac{7\pi }{6},\dfrac{\pi }{6 },\dfrac{5\pi }{6}\right\} \]

(b) A calculator must be used to solve \(\cos \theta =0.4\). Then \[ \theta =\cos ^{-1}\left( 0.4\right) \approx 1.159279 \quad 0\leq \theta \leq \pi \]

Rounded to three decimal places, \(\theta \,{=}\,\cos ^{-1}0.4=1.159\) radians. But there is another angle \(\theta \) in the interval \([ 0,2\pi ] \) for which \(\cos \theta =0.4\), namely, \(\theta \approx 2\pi -1.159\) \(\approx 5.124\) radians.

Because the cosine function has period \(2\pi \), all the solutions of \(\cos \theta =0.4\) are given by the general formulas \[ \theta \approx 1.159+2k\pi \quad \hbox{or} \quad \theta \approx 5.124+2k\pi,\qquad \hbox{ where } k \hbox{ is any integer} \]

The solutions in the interval \([ -2\pi ,2\pi ] \) are \(\left\{ -5.124,-1.159, 1.159, 5.124\right\}\).

NOW WORK

Problem 45.

Solving a Trigonometric Equation

Solve the equation \(\sin ( 2\theta ) =\dfrac{1}{2}\), where \(0\leq \theta <2\pi\).

Solution In the interval \(\left[ 0,2\pi \right) \), the sine function has the value \(\dfrac{1}{2}\) at \(\theta =\dfrac{\pi }{6}\) and at \( \theta = \dfrac{5\pi }{6}\) as shown in Figure 90. Since the period of the sine function is \({2\pi }\) and the argument in the equation \(\sin ( 2\theta ) =\dfrac{1}{2} \) is \(2\theta\), we write the general formula for all the solutions. \[ \begin{array}{l@{\quad}l@{\quad}l@{\quad}l@{\quad}l} 2\theta =\dfrac{\pi }{6}+2k\pi & {\rm or} & 2\theta =\dfrac{5\pi }{6}+2k\pi, & & \hbox{ where }k \hbox{ is any integer} \\ \theta =\dfrac{\pi }{12}+k\pi & & \theta =\dfrac{5\pi }{12}+k\pi & & \end{array} \]

The solutions of \(\sin ( 2\theta ) =\dfrac{1}{2},\) \(0\leq \theta <2\pi \), are \(\left\{ \dfrac{\pi }{12}, \dfrac{5\pi }{12}, \dfrac{13\pi }{12} , \dfrac{17\pi }{12}\right\} .\)

NOW WORK

Problem 31.

NEED TO REVIEW?

Trigonometric identities are discussed in Appendix Section A.4, pp. A-32 to A-35.

Solving a Trigonometric Equation

Solve the equation \(3\sin \theta -\cos ^{2}\theta =3\), where \(0\leq \theta <2\pi \).

Solution The equation involves both sine and cosine functions. We use the Pythagorean identity \(\sin ^{2}\theta +\cos ^{2}\theta =1\) to rewrite the equation in terms of \(\sin \theta .\) \[ \begin{eqnarray*} 3\sin \theta -\cos ^{2}\theta &=&3 \\ 3\sin \theta -( 1-\sin ^{2}\theta)~&=&3\qquad {\color{#0066A7}{\cos ^{2}{\theta =1-}\sin ^{2}{\theta }}} \\ \sin ^{2}\theta +3\sin \theta -4 &=&0 \end{eqnarray*} \]

This is a quadratic equation in \(\sin \theta\). Factor the left side and solve for \(\sin \theta .\) \[ \begin{array}{rcl@{\quad}c@{\quad}rcl} ( \sin \theta +4) (\sin \theta -1)~&=&0& \\ \sin \theta +4 &=&0&\hbox{ or }& \sin \theta -1&=&0 \\ \sin \theta &=&-4&\hbox{ or }& \sin \theta &=&1 \end{array} \]

The range of the sine function is \(-1\leq y\leq 1,\) so \(\sin \theta =-4\) has no solution. Solving \(\sin \theta =1,\) we obtain \[ \theta =\sin ^{-1}1=\dfrac{\pi }{2} \]

The only solution in the interval \(\left[ 0,2\pi \right) \) is \(\dfrac{\pi }{2 }.\)

NOW WORK

Problem 35.