Precalculus Trigonometry Review – Step-by-Step Study Guidance

Q1. Find the exact value of $\sin\left(-\frac{2\pi}{3}\right)$.

Background

Topic: Unit Circle & Exact Trigonometric Values

This question tests your understanding of the unit circle, reference angles, and how to find exact values for sine at standard angles, including negative angles.

Key Terms and Formulas

• Reference Angle: The acute angle formed by the terminal side of the given angle and the x-axis.

• Unit Circle: A circle of radius 1 centered at the origin, used to define trigonometric functions for all angles.

• Odd Function Property: $\sin(-\theta) = -\sin(\theta)$

Step-by-Step Guidance

1. Recognize that $-\frac{2\pi}{3}$ is a negative angle, so it is measured clockwise from the positive x-axis.

2. Recall that $\sin(-\theta) = -\sin(\theta)$. So, $\sin\left(-\frac{2\pi}{3}\right) = -\sin\left(\frac{2\pi}{3}\right)$.

3. Find the reference angle for $\frac{2\pi}{3}$. Since $\frac{2\pi}{3}$ is in the second quadrant, its reference angle is $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$.

4. Recall the exact value: $\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)$ (since sine is positive in the second quadrant).

Try solving on your own before revealing the answer!

Final Answer: $-\frac{\sqrt{3}}{2}$

Since $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$, and $\sin(-\theta) = -\sin(\theta)$, the answer is $-\frac{\sqrt{3}}{2}$.

Q2. Find the exact value of $\tan\left(-\frac{5\pi}{6}\right)$.

Background

Topic: Unit Circle & Tangent Values

This question tests your ability to evaluate the tangent function at negative angles and use reference angles to find exact values.

Key Terms and Formulas

• Odd Function Property: $\tan(-\theta) = -\tan(\theta)$

• Reference Angle: The smallest positive angle between the terminal side and the x-axis.

• Exact Value: $\tan\left(\frac{5\pi}{6}\right)$ can be found using the unit circle.

Step-by-Step Guidance

1. Use the odd function property: $\tan\left(-\frac{5\pi}{6}\right) = -\tan\left(\frac{5\pi}{6}\right)$.

2. Find the reference angle for $\frac{5\pi}{6}$: $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$.

3. Recall that $\tan\left(\frac{5\pi}{6}\right)$ is negative in the second quadrant, so $\tan\left(\frac{5\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right)$.

4. Recall the exact value: $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ or $\frac{\sqrt{3}}{3}$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{\sqrt{3}}$ or $\frac{\sqrt{3}}{3}$

Since $\tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}}$, then $\tan\left(-\frac{5\pi}{6}\right) = \frac{1}{\sqrt{3}}$.

Q3. Find the value of $\cos\left(-\frac{4\pi}{7}\right)$ using a calculator in radian mode. Round to four decimal places.

Background

Topic: Calculator Evaluation of Trigonometric Functions

This question tests your ability to use a calculator in radian mode to evaluate cosine at a non-standard angle and round to the required decimal places.

Key Terms and Formulas

• Calculator in Radian Mode: Make sure your calculator is set to radians, not degrees.

• Cosine Function: $\cos(\theta)$

Step-by-Step Guidance

1. Ensure your calculator is in radian mode (not degree mode).

2. Enter $-4\pi/7$ into your calculator as the argument for the cosine function.

3. Evaluate $\cos\left(-\frac{4\pi}{7}\right)$ and round your answer to four decimal places.

Try solving on your own before revealing the answer!

Final Answer: $-0.2225$

Using a calculator in radian mode, $\cos\left(-\frac{4\pi}{7}\right) \approx -0.2225$.

Q4. In which quadrants is the tangent function positive?

Background

Topic: Signs of Trigonometric Functions by Quadrant

This question tests your knowledge of the ASTC (All Students Take Calculus) rule, which tells you the sign of each trig function in each quadrant.

Key Terms and Formulas

• Quadrants: I, II, III, IV (counterclockwise from positive x-axis)

• ASTC Rule: All (I), Sine (II), Tangent (III), Cosine (IV) are positive in their respective quadrants.

Step-by-Step Guidance

1. Recall that tangent is positive where both sine and cosine have the same sign.

2. In Quadrant I, both sine and cosine are positive, so tangent is positive.

3. In Quadrant III, both sine and cosine are negative, so their ratio (tangent) is positive.

Try solving on your own before revealing the answer!

Final Answer: Quadrants I and III

Tangent is positive in Quadrants I and III because sine and cosine have the same sign in these quadrants.

Q5. The graph of $y = \csc(x)$ has which of the following properties?

Background

Topic: Graphs of Reciprocal Trigonometric Functions

This question tests your understanding of the properties of the cosecant function, including its asymptotes and zeros.

Key Terms and Formulas

• Cosecant Function: $\csc(x) = \frac{1}{\sin(x)}$

• Vertical Asymptotes: Occur where $\sin(x) = 0$

• Zeros: $\csc(x)$ has no zeros because $\frac{1}{0}$ is undefined.

Step-by-Step Guidance

1. Recall that $\csc(x)$ is undefined wherever $\sin(x) = 0$.

2. Find where $\sin(x) = 0$: at $x = n\pi$, where $n$ is an integer.

3. Therefore, $y = \csc(x)$ has vertical asymptotes at $x = n\pi$ and no zeros.

Try solving on your own before revealing the answer!

Final Answer: No zeros; asymptotes at $x = n\pi$

The cosecant function is undefined wherever sine is zero, so it has vertical asymptotes at $x = n\pi$ and no zeros.