Question

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

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

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 \(\theta_2\approx159.38^{\circ}\).

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.

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.

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 \(x_2\approx319.94^{\circ}\).

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 \(y_2\approx239.96^{\circ}\).

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 \(y_3\approx163.11^{\circ}\).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools in solving trigonometric equations. These functions allow us to find the angles whose trigonometric values (sine, cosine, and tangent) are known. When we have a value and need to determine the angle, we use inverse functions such as \( \sin^{-1} \, \cos^{-1} \, \text{and} \, \tan^{-1} \). For instance, if \( \sin \theta = 0.3510 \), we use the inverse sine function \( \theta = \sin^{-1}(0.3510) \) to find the principal angle.

These inverse functions help find the principal value, which is the smallest positive angle. They are especially useful because there are often multiple angles that can yield the same trigonometric value due to the periodic nature of these functions. Therefore, understanding these inverse operations is critical:

\( \bullet \) Inverse Sine (\( \sin^{-1} \)): used to find angles when the sine value is known.
\( \bullet \) Inverse Cosine (\( \cos^{-1} \)): used to find angles when the cosine value is known.
\( \bullet \) Inverse Tangent (\( \tan^{-1} \)): used to find angles when the tangent value is known.

Inverse trigonometric functions play a significant role in finding all possible angles, especially when we extend the search to other quadrants.

Angle Measurement

Understanding angle measurement is crucial when solving trigonometric equations. Angles can be measured in degrees or radians, and knowing how to work with both is essential.

Degrees are commonly used due to their convenience for human interpretation. They divide a circle into 360 equal parts, facilitating easy calculations and interpretations in everyday problems.

However, radians provide a more natural measure in mathematics and physics. A radian measures the angle created when the radius is wrapped along the circle's edge. There are \( 2\pi \) radians in a circle, offering a seamless integration with other mathematical formulas.

When solving trigonometric equations, it is crucial to pay attention to the units specified in the problem. This will inform how you interpret the solutions and any necessary conversions you might have to perform.

Quadrants of Angles

The concept of quadrants is fundamental when dealing with trigonometric functions. A unit circle is divided into four quadrants, each with different characteristics for sine, cosine, and tangent values.

\( \bullet \) **First Quadrant**: Both sine and cosine values are positive, making angles between \( 0^{\circ} \) and \( 90^{\circ} \) very straightforward to work with.
\( \bullet \) **Second Quadrant**: Sine values are positive, while cosine and tangent values are negative. This includes angles between \( 90^{\circ} \) and \( 180^{\circ} \).
\( \bullet \) **Third Quadrant**: Here, sine and cosine values are negative, but tangent values are positive, covering angles between \( 180^{\circ} \) and \( 270^{\circ} \).
\( \bullet \) **Fourth Quadrant**: Sine values are negative, while cosine values are positive, spanning angles between \( 270^{\circ} \) and \( 360^{\circ} \).

When solving a trigonometric equation, it's essential to consider which quadrants the angle might be in, as this could affect whether the function yields a positive or negative value.
Knowing the quadrant allows us to determine which angles satisfy the equation, considering their sine, cosine, or tangent values accordingly.

Radian and Degree Conversion

The conversion between radians and degrees is a necessary skill when solving trigonometric problems. It ensures accurate interpretations and consistency in calculations.

The conversion factor between degrees and radians is \( \pi \) radians being equal to \( 180^{\circ} \), or equivalently:

\( \bullet \) To convert degrees to radians: multiply by \( \frac{\pi}{180} \)
\( \bullet \) To convert radians to degrees: multiply by \( \frac{180}{\pi} \)

This skill is instrumental in problems that involve both formats. For example, in a problem operating in radians with an answer stated in degrees, a quick and accurate conversion will allow a smooth transition.

Mastery of these conversions also enhances comprehension of the solutions, avoiding confusion and ensuring one achieves the correct result according to the initial conditions presented.