Trig functions: Algebra 2 Regents Practice

Convert $\dfrac{5\pi}{6}$ radians to degrees.

• (A) $216^\circ$
• (B) $75^\circ$
• (C) $300^\circ$
• (D) $150^\circ$ ✓

Radian-to-degree conversion multiplies by $\frac{180^\circ}{\pi}$, since $\pi$ radians equals $180^\circ$.

Express $135^\circ$ in radians.

• (A) $\dfrac{2\pi}{3}$
• (B) $\dfrac{3\pi}{2}$
• (C) $\dfrac{3\pi}{4}$ ✓
• (D) $\dfrac{4\pi}{3}$

Degree-to-radian conversion multiplies by $\frac{\pi}{180^\circ}$, then the fraction is reduced to lowest terms.

Expressed in radian measure, $135^\circ$ is equivalent to

• (A) $\dfrac{3}{4}$
• (B) $\dfrac{4\pi}{3}$
• (C) $\dfrac{3\pi}{8}$
• (D) $\dfrac{3\pi}{4}$ ✓

To convert from degrees to radians, multiply the degree measure by $\frac{\pi}{180}$.

Which angle, measured in radians, is coterminal with $\dfrac{13\pi}{4}$ and lies in the interval $0 \le \theta < 2\pi$?

• (A) $\dfrac{\pi}{4}$
• (B) $-\dfrac{3\pi}{4}$
• (C) $\dfrac{9\pi}{4}$
• (D) $\dfrac{5\pi}{4}$ ✓

Coterminal angles differ by full rotations of $2\pi$; subtracting $2\pi$ once gives the unique value in $[0, 2\pi)$.

What is the period of the function $y = 3\sin(2x) + 1$?

• (A) $2$
• (B) $\pi$ ✓
• (C) $2\pi$
• (D) $3$

The period of a sine function $y = a\sin(bx) + c$ is $\frac{2\pi}{|b|}$. With $b = 2$, the period is $\pi$. The value 3 is the amplitude and 2 is the frequency factor, not the period.

What is the period of the graph of $y = 4\sin(3x)$?

• (A) $3$
• (B) $\dfrac{2\pi}{3}$ ✓
• (C) $\dfrac{\pi}{3}$
• (D) $2\pi$

The period of $y = a\sin(bx)$ is $2\pi/b$, and here $b = 3$, so the period is $2\pi/3$. The amplitude 4 does not affect the period, and confusing $b$ itself or halving the period gives the distractors.

What is the period of $y = -4\sin\!\left(\dfrac{2}{3}x\right) + 1$?

• (A) $2\pi$
• (B) $3\pi$ ✓
• (C) $\dfrac{4\pi}{3}$
• (D) $4$

For $y=a\sin(bx)+k$, the period is $\frac{2\pi}{|b|}$; the amplitude and vertical shift do not affect it.

What is the range of $y = 3\sin x - 2$?

• (A) $[-2,\ 3]$
• (B) $[-5,\ 1]$ ✓
• (C) $[-3,\ 3]$
• (D) $[-1,\ 5]$

A sinusoid ranges from midline minus amplitude to midline plus amplitude; here the midline is $-2$ and the amplitude is $3$.

Find the exact value of $\cos\!\left(\dfrac{4\pi}{3}\right)$.

• (A) $-\dfrac{\sqrt{3}}{2}$
• (B) $-\dfrac{1}{2}$ ✓
• (C) $\dfrac{\sqrt{3}}{2}$
• (D) $\dfrac{1}{2}$

Evaluate using the reference angle, then attach the sign based on the quadrant; cosine is negative in Quadrant III.

A central angle of $\dfrac{\pi}{3}$ radians is drawn in a circle with radius $9$ cm. Find the length of the intercepted arc.

• (A) $27\pi$ cm
• (B) $3\pi$ cm ✓
• (C) $\dfrac{\pi}{3}$ cm
• (D) $\dfrac{27\pi}{2}$ cm

Arc length equals radius times the central angle in radians: $s=r\theta$.

What is the reference angle for $\dfrac{4\pi}{3}$?

• (A) $\dfrac{\pi}{3}$ ✓
• (B) $\dfrac{2\pi}{3}$
• (C) $\dfrac{4\pi}{3}$
• (D) $\dfrac{\pi}{6}$

The reference angle is the acute angle between the terminal side and the x-axis; for a Quadrant III angle it equals $\theta-\pi$.

If $\sin\theta = \dfrac{3}{5}$ and $\theta$ is an angle in Quadrant II, what is the value of $\cos\theta$?

• (A) $-\dfrac{16}{25}$
• (B) $-\dfrac{4}{5}$ ✓
• (C) $-\dfrac{3}{4}$
• (D) $\dfrac{4}{5}$

Use the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$, then choose the sign of cosine from the quadrant (cosine is negative in Quadrant II).

What is the range of the function $y = 2\sin(3x) - 5$?

• (A) $-7 \le y \le -3$ ✓
• (B) $-8 \le y \le -2$
• (C) $-3 \le y \le 7$
• (D) $-2 \le y \le 2$

For $y=a\sin(bx)+d$, the outputs run from $d-|a|$ to $d+|a|$; the coefficient $b$ changes the period, not the amplitude.

The terminal side of an angle $\theta$ in standard position passes through the point $(-8, 6)$. What is the value of $\sin\theta$?

• (A) $\dfrac{3}{5}$ ✓
• (B) $\dfrac{3}{4}$
• (C) $-\dfrac{4}{5}$
• (D) $-\dfrac{3}{4}$

For a point $(x,y)$ on the terminal side, $\sin\theta=\frac{y}{r}$ where $r=\sqrt{x^2+y^2}$ is always positive.

If $\cos\theta = -\dfrac{3}{5}$ and $\theta$ terminates in Quadrant II, what is the value of $\tan\theta$?

• (A) $\dfrac{4}{3}$
• (B) $-\dfrac{4}{3}$ ✓
• (C) $\dfrac{3}{4}$
• (D) $-\dfrac{3}{4}$

Use the Pythagorean identity to find $\sin\theta$, choosing the positive root because sine is positive in Quadrant II, then divide sine by cosine. In Quadrant II tangent is negative, which eliminates the positive choices.

If $\sin\theta = \dfrac{3}{5}$ and $\theta$ terminates in Quadrant II, what is the value of $\cos\theta$?

• (A) $\dfrac{4}{5}$
• (B) $-\dfrac{4}{5}$ ✓
• (C) $-\dfrac{3}{5}$
• (D) $-\dfrac{4}{3}$

The Pythagorean identity gives $\cos^2\theta = \frac{16}{25}$, so $\cos\theta = \pm\frac{4}{5}$. Because $\theta$ lies in Quadrant II, where $x$-coordinates are negative, cosine must be negative, giving $-\frac{4}{5}$.

What are all solutions of the equation $2\cos\theta + 1 = 0$ on the interval $0 \leq \theta < 2\pi$?

• (A) $\dfrac{\pi}{3}$ and $\dfrac{5\pi}{3}$
• (B) $\dfrac{5\pi}{6}$ and $\dfrac{7\pi}{6}$
• (C) $\dfrac{\pi}{3}$ and $\dfrac{2\pi}{3}$
• (D) $\dfrac{2\pi}{3}$ and $\dfrac{4\pi}{3}$ ✓

Isolating cosine gives $\cos\theta = -\frac{1}{2}$, whose reference angle is $\frac{\pi}{3}$. Cosine is negative in Quadrants II and III, so the solutions on $[0, 2\pi)$ are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$. The distractor $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ solves $\cos\theta = +\frac{1}{2}$.

Solve the equation $2\cos^2\theta + \cos\theta - 1 = 0$ algebraically for all values of $\theta$ in the interval $0^\circ \le \theta < 360^\circ$.

The equation is quadratic in $\cos\theta$, so it factors just like $2u^2+u-1=(2u-1)(u+1)$. Solving each factor gives the reference angle solutions $60^\circ$ and $300^\circ$ from $\cos\theta = \tfrac{1}{2}$, and the single solution $180^\circ$ from $\cos\theta = -1$.

Solve the equation $2\cos^2 x + 3\cos x - 2 = 0$ algebraically for all values of $x$ on the interval $0 \le x < 2\pi$. Express your answers in exact radian form.

The equation is quadratic in $\cos x$, so it factors like $2u^2 + 3u - 2 = (2u - 1)(u + 2)$. The root $\cos x = -2$ is impossible because cosine is bounded between $-1$ and $1$, leaving $\cos x = \tfrac{1}{2}$, which occurs at the reference angle $\tfrac{\pi}{3}$ in Quadrants I and IV on the given interval.

If $\sin\theta = \dfrac{3}{5}$ and $\theta$ terminates in Quadrant II, what is the value of $\tan\theta$?

• (A) $\dfrac{4}{3}$
• (B) $-\dfrac{3}{4}$ ✓
• (C) $-\dfrac{4}{3}$
• (D) $\dfrac{3}{4}$

Using the Pythagorean identity, $\cos\theta = \pm 4/5$, and the Quadrant II condition forces cosine to be negative. Then $\tan\theta = \sin\theta / \cos\theta = -3/4$. Choosing the wrong sign or inverting the ratio gives the distractors.

Solve the equation $2\cos^2\theta-\cos\theta-1=0$ for all values of $\theta$ on the interval $0^\circ \le \theta < 360^\circ$.

The equation is quadratic in $\cos\theta$, so it factors just like $2u^2-u-1=(2u+1)(u-1)$. Each cosine value is then converted to angles in the given interval, giving $0^\circ$, $120^\circ$, and $240^\circ$.

Given $\cos\theta=-\frac{3}{5}$, where $\frac{\pi}{2}<\theta<\pi$, the value of $\sin 2\theta$ is

• (A) $-\frac{12}{25}$
• (B) $-\frac{24}{25}$ ✓
• (C) $-\frac{7}{25}$
• (D) $\frac{24}{25}$

Since $\theta$ is in Quadrant II, sine is positive, so the Pythagorean identity gives $\sin\theta=\frac{4}{5}$. The double angle formula $\sin 2\theta=2\sin\theta\cos\theta$ then gives $2\cdot\frac{4}{5}\cdot(-\frac{3}{5})=-\frac{24}{25}$. Forgetting the factor of 2 gives $-\frac{12}{25}$, and $-\frac{7}{25}$ is $\cos 2\theta$.

What is the period of the function $y=2\cos\left(\dfrac{x}{3}\right)+1$?

• (A) $\frac{2\pi}{3}$
• (B) $2\pi$
• (C) $3$
• (D) $6\pi$ ✓

The period of a cosine function $y=a\cos(bx)+c$ is $\frac{2\pi}{|b|}$. Here $b=\frac{1}{3}$, so the period is $2\pi\div\frac{1}{3}=6\pi$. Dividing incorrectly gives $\frac{2\pi}{3}$, and the amplitude and vertical shift do not affect the period.

If $\sin\theta = \frac{3}{5}$ and $\theta$ terminates in Quadrant II, what is the value of $\cos\theta$?

• (A) $\frac{3}{4}$
• (B) $-\frac{4}{5}$ ✓
• (C) $\frac{4}{5}$
• (D) $-\frac{3}{4}$

Using the Pythagorean identity, $\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$, so $\cos\theta = \pm\frac{4}{5}$. Because $\theta$ is in Quadrant II, where cosine values are negative, $\cos\theta = -\frac{4}{5}$. Choosing $-\frac{3}{4}$ confuses cosine with tangent.