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Chapter 9: Problem 18

Solve $$ \begin{gathered} \sin \theta \cos 41^{\circ}+\sin 41^{\circ} \cos \theta=0.6100 \\ 0^{\circ} \leq \theta \leq 360^{\circ} \end{gathered} $$

Short Answer

Expert verified
Answer: To find the possible values of θ, first rewrite the equation using the angle addition formula for sine: sin(θ + 41°) = 0.6100. Then, find the principal solutions for θ by finding the inverse sine of 0.6100 and subtracting 41°. Next, find the general solutions for θ by considering the periodicity of the sine function and adding multiples of 360°. Finally, list all possible values of θ within the given domain by plugging these values into the general solution. The possible values of θ within the domain 0° ≤ θ ≤ 360° are: θ = arcsin(0.6100) - 41°, 180° - arcsin(0.6100) - 41°, and 360° - (arcsin(0.6100) + 41°).

Step by step solution

01

Rewrite the given equation using the angle addition formula

The given equation is: $$ \sin \theta \cos 41^{\circ}+\sin 41^{\circ} \cos \theta=0.6100 $$ Using the angle addition formula for sine function: $$ \sin (A+B) = \sin A \cos B + \cos A \sin B $$ We rewrite the given equation with A = θ and B = 41°: $$ \sin (\theta + 41^{\circ}) = 0.6100 $$
02

Solve the equation for θ

Now that we have the equation in terms of the sine function, we can find the principal solutions for θ by solving the following equation: $$ \sin (\theta + 41^{\circ}) = 0.6100 $$ To find the principal values of θ, we will first find the inverse sine of both sides: $$ \theta + 41^{\circ} = \arcsin(0.6100), \hspace{10mm} (*) $$ Once we find the values for these angles, we can subtract 41° from each value to find the principal solutions for θ.
03

Find the general solutions for θ

Now that we have the principal solutions for θ, let's consider more general solutions for θ by taking into account the periodicity of the sine function. The sine function has a period of 360°, meaning that: $$ \sin x = \sin y \hspace{5mm} \text{if} \hspace{5mm} x = y ± k \cdot 360^{\circ}, k \in \mathbb{Z} $$ Thus, the general solutions for θ will have the following form: $$ \theta = (\arcsin(0.6100)-41^{\circ} \pm k_1 \cdot 360^{\circ}) \ \text{or} \ (180^{\circ} - \arcsin(0.6100)-41^{\circ} \pm k_2 \cdot 360^{\circ}) \hspace{5mm} k_1, k_2 \in \mathbb{Z} $$
04

List all possible values of θ within the given domain

Now that we have the general solution, we need to find values of θ within the given domain (0° ≤ θ ≤ 360°). Plug the values into the general solution and list the values of θ that satisfy the domain: $$ θ = \arcsin(0.6100) - 41^{\circ}, \ \ 180^{\circ} - \arcsin(0.6100) - 41^{\circ}, \ 360^{\circ} - ( \arcsin(0.6100) + 41^{\circ}) $$ Calculate these values to obtain possible values of θ within the given domain. Keep in mind that while calculating, the arcsin(0.6100) should be calculated in degrees since the entire equation uses degree measures.

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Most popular questions from this chapter

Convert the following angles in degrees to radians: (a) \(12^{\circ}\) (b) \(65^{\circ}\) (c) \(200^{\circ}\) (d) \(340^{\circ}\) (e) \(1000^{\circ}\)

Show \(\sin \left(180^{\circ}+\theta\right)=-\sin \theta\).

Show \(\cos \left(180^{\circ}+\theta\right)=-\cos \theta\)

Show \(\cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta\).

Verify the identity $$ 2 \sin A \sin B=\cos (A-B)-\cos (A+B) $$ with \(A=50^{\circ}\) and \(B=15^{\circ}\).
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