Solve (a) \(\sin \theta=0.3510,0^{\circ} \leq \theta \leq 360^{\circ}\) (b) \(\sin \theta=0.4161,0 \leq \theta \leq 2 \pi\) (c) \(\cos t=-0.3778,0 \leq t \leq 2 \pi\) (d) \(\cos x=0.7654,0^{\circ} \leq x \leq 360^{\circ}\) (e) \(\tan y=1.7136,0^{\circ} \leq y \leq 360^{\circ}\) (f) \(\tan y=-0.3006,0^{\circ} \leq y \leq 360^{\circ}\)

Short Answer

Expert verified
a) \(\sin\theta=0.3510\) in the range \(0^{\circ} \leq \theta \leq 360^{\circ}\) The possible angles are \(\theta_1\approx20.62^{\circ}\) and \(\theta_2\approx159.38^{\circ}\). b) \(\sin\theta=0.4161\) in the range \(0 \leq \theta \leq 2\pi\) The possible angles are \(\theta_1\approx0.4327\) and \(\theta_2\approx2.7089\) radians. c) \(\cos t=-0.3778\) in the range \(0 \leq t \leq 2\pi\) The possible angles are \(t_1\approx1.9802\) and \(t_2\approx5.1218\) radians. d) \(\cos x=0.7654\) in the range \(0^{\circ} \leq x \leq 360^{\circ}\) The possible angles are \(x_1\approx40.06^{\circ}\) and \(x_2\approx319.94^{\circ}\). e) \(\tan y=1.7136\) in the range \(0^{\circ} \leq y \leq 360^{\circ}\) The possible angles are \(y_1\approx59.96^{\circ}\) and \(y_2\approx239.96^{\circ}\). f) \(\tan y=-0.3006\) in the range \(0^{\circ} \leq y \leq 360^{\circ}\) The possible angles are \(y_2\approx343.11^{\circ}\) and \(y_3\approx163.11^{\circ}\).

Step by step solution

01

Solve for (a) \(\sin\theta=0.3510\) in the range \(0^{\circ} \leq \theta \leq 360^{\circ}\)

Use inverse sine function to find the principal angle: \(\theta_1=\sin^{-1}(0.3510)\approx20.62^{\circ}\). Since sine is positive in both first and second quadrants, find the angle in the second quadrant: \(\theta_2=180^{\circ}-\theta_1=180^{\circ}-20.62^{\circ}\approx159.38^{\circ}\). Thus, the angles are \(\theta_1\approx20.62^{\circ}\) and \(\theta_2\approx159.38^{\circ}\).
02

Solve for (b) \(\sin\theta=0.4161\) in the range \(0 \leq \theta \leq 2\pi\)

Use inverse sine function to find the principal angle: \(\theta_1=\sin^{-1}(0.4161)\approx0.4327\) radians. Since sine is positive in both first and second quadrants, find the angle in the second quadrant: \(\theta_2=\pi-\theta_1=\pi-0.4327\approx2.7089\) radians. Thus, the angles are \(\theta_1\approx0.4327\) and \(\theta_2\approx2.7089\) radians.
03

Solve for (c) \(\cos t=-0.3778\) in the range \(0 \leq t \leq 2\pi\)

Use inverse cosine function to find the principal angle: \(t_1=\cos^{-1}(-0.3778)\approx1.9802\) radians. Since cosine is negative in both second and third quadrants, find the angle in the third quadrant: \(t_2=\pi+t_1=\pi+1.9802\approx5.1218\) radians. Thus, the angles are \(t_1\approx1.9802\) and \(t_2\approx5.1218\) radians.
04

Solve for (d) \(\cos x=0.7654\) in the range \(0^{\circ} \leq x \leq 360^{\circ}\)

Use inverse cosine function to find the principal angle: \(x_1=\cos^{-1}(0.7654)\approx40.06^{\circ}\). Since cosine is positive in both first and fourth quadrants, find the angle in the fourth quadrant: \(x_2=360^{\circ}-x_1=360^{\circ}-40.06^{\circ}\approx319.94^{\circ}\). Thus, the angles are \(x_1\approx40.06^{\circ}\) and \(x_2\approx319.94^{\circ}\).
05

Solve for (e) \(\tan y=1.7136\) in the range \(0^{\circ} \leq y \leq 360^{\circ}\)

Use inverse tangent function to find the principal angle: \(y_1=\tan^{-1}(1.7136)\approx59.96^{\circ}\). Since tangent has a period of \(180^{\circ}\), find the angle in the third quadrant: \(y_2=y_1+180^{\circ}=59.96^{\circ}+180^{\circ}\approx239.96^{\circ}\). Thus, the angles are \(y_1\approx59.96^{\circ}\) and \(y_2\approx239.96^{\circ}\).
06

Solve for (f) \(\tan y=-0.3006\) in the range \(0^{\circ} \leq y \leq 360^{\circ}\)

Use inverse tangent function to find the principal angle: \(y_1=\tan^{-1}(-0.3006)\approx-16.89^{\circ}\). Since the angle is negative, we need to find the angle in the range. Calculate first angle in the specified range: \(y_2=360^{\circ}+y_1=360^{\circ}-16.89^{\circ}\approx343.11^{\circ}\). Now, add \(180^{\circ}\) to find second angle: \(y_3=343.11^{\circ}+180^{\circ}\approx163.11^{\circ}\). Thus, the angles are \(y_2\approx343.11^{\circ}\) and \(y_3\approx163.11^{\circ}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free