Chapter 7
Algebra and Trigonometry with Analytic Geometry · 348 exercises
Problem 60
Exer. 53-64: Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \(6 \sin 2 x-8 \cos x+9 \sin x-6=0 ; \quad\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
5 step solution
Problem 60
Exer. 39-62: Find the solutions of the equation that are in the interval \([0,2 \pi\) ). $$ \sec ^{5} \theta=4 \sec \theta $$
8 step solution
Problem 60
Exer. 51-60: Show that the equation is not an identity. (Hint: Find one number for which the equation is false.) $$ \cot (\tan \theta)=1 $$
4 step solution
Problem 60
Exer. 57-62: Use an addition or subtraction formula to find the solutions of the equation that are in the interval \([0, \pi\) ). $$ \sin 3 t \cos t+\cos 3 t \sin t=-\frac{1}{2} $$
5 step solution
Problem 61
Exer. 53-64: Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \((\cos x)(15 \cos x+4)=3\) \([0,2 \pi)\)
6 step solution
Problem 61
Exer. 39-62: Find the solutions of the equation that are in the interval \([0,2 \pi\) ). $$ 2 \tan t \csc t+2 \csc t+\tan t+1=0 $$
6 step solution
Problem 61
Exer. 61-64: Either show that the equation is an identity or show that the equation is not an identity. $$ (\sec x+\tan x)^{2}=2 \tan x(\tan x+\sec x) $$
5 step solution
Problem 61
Exer. 57-62: Use an addition or subtraction formula to find the solutions of the equation that are in the interval \([0, \pi\) ). $$ \tan 2 t+\tan t=1-\tan 2 t \tan t $$
3 step solution
Problem 62
Exer. 53-64: Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \(6 \sin ^{2} x=\sin x+2\) \([0,2 \pi)\)
9 step solution
Problem 62
Exer. 39-62: Find the solutions of the equation that are in the interval \([0,2 \pi\) ). $$ 2 \sin v \csc v-\csc v=4 \sin v-2 $$
5 step solution
Problem 62
Exer. 61-64: Either show that the equation is an identity or show that the equation is not an identity. $$ \frac{\tan ^{2} x}{\sec x-1}=\sec x $$
5 step solution
Problem 62
Exer. 57-62: Use an addition or subtraction formula to find the solutions of the equation that are in the interval \([0, \pi\) ). $$ \tan t-\tan 4 t=1+\tan 4 t \tan t $$
3 step solution
Problem 63
Exer. 53-64: Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \(3 \cos 2 x-7 \cos x+5=0\) \([0,2 \pi)\)
5 step solution
Problem 63
Exer. 63-68: Approximate, to the nearest 10', the solutions of the equation in the interval \(\left[0^{\circ}, 360^{\circ}\right)\). $$ \sin ^{2} t-4 \sin t+1=0 $$
6 step solution
Problem 63
Exer. 61-64: Either show that the equation is an identity or show that the equation is not an identity. $$ \cos x(\tan x+\cot x)=\csc x $$
5 step solution
Problem 64
Exer. 53-64: Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \(\sin 2 x=-1.5 \cos x\) \([0,2 \pi)\)
5 step solution
Problem 64
Exer. 63-68: Approximate, to the nearest 10', the solutions of the equation in the interval \(\left[0^{\circ}, 360^{\circ}\right)\). $$ \cos ^{2} t-4 \cos t+2=0 $$
7 step solution
Problem 64
Exer. 61-64: Either show that the equation is an identity or show that the equation is not an identity. $$ \csc ^{2} x+\sec ^{2} x=\csc ^{2} x \sec ^{2} x $$
5 step solution
Problem 65
Exer. 65-66: If an earthquake has a total horizontal displacement of \(S\) meters along its fault line, then the horizontal movement \(M\) of a point on the surface of Earth \(d\) kilometers from the fault line can be estimated using the formula $$ M=\frac{S}{2}\left(1-\frac{2}{\pi} \tan ^{-1} \frac{d}{D}\right) $$ where \(D\) is the depth (in kilometers) below the surface of the focal point of the earthquake. For the San Francisco earthquake of \(1906, S\) was 4 meters and \(D\) was \(3.5\) kilometers. Approximate \(M\) for the stated values of \(d\). (a) 1 kilometer (b) 4 kilometers (c) 10 kilometers
6 step solution
Problem 65
Exer. 63-68: Approximate, to the nearest 10', the solutions of the equation in the interval \(\left[0^{\circ}, 360^{\circ}\right)\). $$ \tan ^{2} \theta+3 \tan \theta+2=0 $$
5 step solution
Problem 66
Exer. 65-66: If an earthquake has a total horizontal displacement of \(S\) meters along its fault line, then the horizontal movement \(M\) of a point on the surface of Earth \(d\) kilometers from the fault line can be estimated using the formula $$ M=\frac{S}{2}\left(1-\frac{2}{\pi} \tan ^{-1} \frac{d}{D}\right) $$ where \(D\) is the depth (in kilometers) below the surface of the focal point of the earthquake. Approximate the depth \(D\) of the focal point of an earthquake with \(S=3 \mathrm{~m}\) if a point on the surface of Earth 5 kilometers from the fault line moved \(0.6\) meter horizontally.
8 step solution
Problem 66
Exer. 63-68: Approximate, to the nearest 10', the solutions of the equation in the interval \(\left[0^{\circ}, 360^{\circ}\right)\). $$ 2 \tan ^{2} x-3 \tan x-1=0 $$
5 step solution
Problem 66
Exer. 63-66: (a) Use the formula from Example 6 to express \(f\) in terms of the cosine function. (b) Determine the amplitude, period, and phase shift of \(f\). (c) Sketch the graph of \(f\). $$ f(x)=5 \cos 10 x-5 \sin 10 x $$
5 step solution
Problem 67
A golfer, centered in a 30-yard-wide straight fairway, hits a ball 280 yards. Approximate the largest angle the drive can have from the center of the fairway if the ball is to stay in the fairway (see the figure).
5 step solution
Problem 67
Exer. 63-68: Approximate, to the nearest 10', the solutions of the equation in the interval \(\left[0^{\circ}, 360^{\circ}\right)\). $$ 12 \sin ^{2} u-5 \sin u-2=0 $$
4 step solution
Problem 67
Exer. 67-68: For certain applications in electrical engineering, the sum of several voltage signals or radio waves of the same frequency is expressed in the compact form \(y=A \cos (B t-C)\). Express the given signal in this form. $$ y=50 \sin 60 \pi t+40 \cos 60 \pi t $$
6 step solution
Problem 68
Exer. 63-68: Approximate, to the nearest 10', the solutions of the equation in the interval \(\left[0^{\circ}, 360^{\circ}\right)\). $$ 5 \cos ^{2} \alpha+3 \cos \alpha-2=0 $$
5 step solution
Problem 68
Exer. 67-68: For certain applications in electrical engineering, the sum of several voltage signals or radio waves of the same frequency is expressed in the compact form \(y=A \cos (B t-C)\). Express the given signal in this form. $$ y=10 \sin \left(120 \pi t-\frac{\pi}{2}\right)+5 \sin 120 \pi t $$
4 step solution
Problem 69
A tidal wave of height 50 feet and period 30 minutes is approaching a sea wall that is \(12.5\) feet above sea level (see the figure). From a particular point on shore, the distance \(y\) from sea level to the top of the wave is given by $$ y=25 \cos \frac{\pi}{15} t $$ with \(t\) in minutes. For approximately how many minutes of each 30 -minute period is the top of the wave above the level of the top of the sea wall?
7 step solution
Problem 69
Exer. 69-72: Make the trigonometric substitution $$ x=a \tan \theta \text { for }-\pi / 2<\theta<\pi / 2 \text { and } a>0 \text {. } $$ Simplify the resulting expression. $$ \sqrt{a^{2}+x^{2}} $$
5 step solution
Problem 69
If a mass that is attached to a spring is raised \(y_{0}\) feet and released with an initial vertical velocity of \(v_{0} \mathrm{ft} / \mathrm{sec}\), then the subsequent position \(y\) of the mass is given by $$ y=y_{0} \cos \omega t+\frac{v_{0}}{\omega} \sin \omega t, $$ where \(t\) is time in seconds and \(\omega\) is a positive constant. (a) If \(\omega=1, y_{0}=2 \mathrm{ft}\), and \(v_{0}=3 \mathrm{ft} / \mathrm{sec}\), express \(y\) in the form \(A \cos (B t-C)\), and find the amplitude and period of the resulting motion. (b) Determine the times when \(y=0\)-that is, the times when the mass passes through the equilibrium position.
6 step solution
Problem 70
The expected low temperature \(T\) (in \({ }^{\circ} \mathrm{F}\) ) in Fairbanks, Alaska, may be approximated by $$ T=36 \sin \left[\frac{2 \pi}{365}(t-101)\right]+14 $$ where \(t\) is in days, with \(t=0\) corresponding to January 1 . For how many days during the year is the low temperature expected to be below \(-4^{\circ} \mathrm{F}\) ?
4 step solution
Problem 70
Exer. 69-72: Make the trigonometric substitution $$ x=a \tan \theta \text { for }-\pi / 2<\theta<\pi / 2 \text { and } a>0 \text {. } $$ Simplify the resulting expression. $$ \frac{1}{\sqrt{a^{2}+x^{2}}} $$
6 step solution
Problem 71
Exer. 71-76: Verify the identity. $$ \sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}} $$
5 step solution
Problem 71
On a clear day with \(D\) hours of daylight, the intensity of sunlight \(I\) (in calories \(/ \mathrm{cm}^{2}\) ) may be approximated by $$ I=I_{\mathrm{M}} \sin ^{3} \frac{\pi t}{D} \text { for } 0 \leq t \leq D, $$ where \(t=0\) corresponds to sunrise and \(I_{\mathrm{M}}\) is the maximum intensity. If \(D=12\), approximately how many hours after sunrise is \(I=\frac{1}{2} I_{\mathrm{M}}\) ?
3 step solution
Problem 71
Exer. 69-72: Make the trigonometric substitution $$ x=a \tan \theta \text { for }-\pi / 2<\theta<\pi / 2 \text { and } a>0 \text {. } $$ Simplify the resulting expression. $$ \frac{1}{x^{2}+a^{2}} $$
7 step solution
Problem 72
Exer. 71-76: Verify the identity. $$ \arccos x+\arccos \sqrt{1-x^{2}}=\frac{\pi}{2}, 0 \leq x \leq 1 $$
4 step solution
Problem 72
Exer. 69-72: Make the trigonometric substitution $$ x=a \tan \theta \text { for }-\pi / 2<\theta<\pi / 2 \text { and } a>0 \text {. } $$ Simplify the resulting expression. $$ \left(x^{2}+a^{2}\right)^{3 / 2} $$
6 step solution
Problem 73
Exer. 71-76: Verify the identity. $$ \arcsin (-x)=-\arcsin x $$
5 step solution
Problem 73
Exer. 73-76: Make the trigonometric substitution $$ x=a \sec \theta \text { for } 0<\theta<\pi / 2 \text { and } a>0 \text {. } $$ Simplify the resulting expression. $$ \sqrt{x^{2}-a^{2}} $$
5 step solution
Problem 74
Exer. 71-76: Verify the identity. $$ \arccos (-x)=\pi-\arccos x $$
4 step solution
Problem 74
In the study of frost penetration problems in highway engineering, the temperature \(T\) at time \(t\) hours and depth \(x\) feet is given by $$ T=T_{0} e^{-\lambda x} \sin (\omega t-\lambda x), $$ where \(T_{0}, \omega\), and \(\lambda\) are constants and the period of \(T\) is 24 hours. (a) Find a formula for the temperature at the surface. (b) At what times is the surface temperature a minimum? (c) If \(\lambda=2.5\), find the times when the temperature is a minimum at a depth of 1 foot.
4 step solution
Problem 74
Exer. 73-76: Make the trigonometric substitution $$ x=a \sec \theta \text { for } 0<\theta<\pi / 2 \text { and } a>0 \text {. } $$ Simplify the resulting expression. $$ \frac{1}{x^{2} \sqrt{x^{2}-a^{2}}} $$
5 step solution
Problem 75
Many animal populations, such as that of rabbits, fluctuate over ten-year cycles. Suppose that the number of rabbits at time \(t\) (in years) is given by $$ N(t)=1000 \cos \frac{\pi}{5} t+4000 . $$ (a) Sketch the graph of \(N\) for \(0 \leq t \leq 10\). (b) For what values of \(t\) in part (a) does the rabbit population exceed 4500 ?
5 step solution
Problem 75
Exer. 73-76: Make the trigonometric substitution $$ x=a \sec \theta \text { for } 0<\theta<\pi / 2 \text { and } a>0 \text {. } $$ Simplify the resulting expression. $$ x^{3} \sqrt{x^{2}-a^{2}} $$
4 step solution
Problem 76
Exer. 71-76: Verify the identity. $$ 2 \cos ^{-1} x=\cos ^{-1}\left(2 x^{2}-1\right), 0 \leq x \leq 1 $$
4 step solution
Problem 76
The flow rate (or water discharge rate) at the mouth of the Orinoco River in South America may be approximated by $$ F(t)=26,000 \sin \left[\frac{\pi}{6}(t-5.5)\right]+34,000, $$ where \(t\) is the time in months and \(F(t)\) is the flow rate in \(\mathrm{m}^{3} / \mathrm{sec}\). For approximately how many months each year does the flow rate exceed \(55,000 \mathrm{~m}^{3} / \mathrm{sec}\) ?
4 step solution
Problem 76
Exer. 73-76: Make the trigonometric substitution $$ x=a \sec \theta \text { for } 0<\theta<\pi / 2 \text { and } a>0 \text {. } $$ Simplify the resulting expression. $$ \frac{\sqrt{x^{2}-a^{2}}}{x^{2}} $$
8 step solution