Trigonometric Function Practice Questions

The sine values for the standard angles should be memorized. From the 30-60-90 special right triangle, the side opposite the 30° angle is half the hypotenuse, giving:

\(\sin 30° = \dfrac{1}{2}\)

The other choices are common sine values that students sometimes mix up:

• \(\frac{\sqrt{3}}{2}\) = \(\sin 60°\)
• \(\frac{\sqrt{2}}{2}\) = \(\sin 45°\)
• \(1\) = \(\sin 90°\)

First, note that \(\frac{5\pi}{6}\) radians = 150°, which is in Quadrant II. Find the reference angle:

\(\pi – \dfrac{5\pi}{6} = \dfrac{\pi}{6} = 30°\)

In Quadrant II, cosine is negative. Therefore:

\(\cos \dfrac{5\pi}{6} = -\cos 30° = -\dfrac{\sqrt{3}}{2}\)

Use the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\):

\(\left(\dfrac{3}{5}\right)^2 + \cos^2\theta = 1\)

\(\dfrac{9}{25} + \cos^2\theta = 1\)

\(\cos^2\theta = \dfrac{16}{25}\)

\(\cos\theta = \pm\dfrac{4}{5}\)

Since \(\theta\) is in Quadrant II, where cosine is negative:

\(\cos\theta = -\dfrac{4}{5}\)

Choice B forgets to apply the negative sign for Quadrant II. Always use the quadrant to determine the correct sign after solving.

Recall that \(\frac{\pi}{3}\) radians = 60°. Use the identity \(\tan\theta = \frac{\sin\theta}{\cos\theta}\):

\(\tan 60° = \dfrac{\sin 60°}{\cos 60°} = \dfrac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\)

Choice C gives \(\frac{\sqrt{3}}{3}\), which is the value of \(\tan 30°\) (not \(\tan 60°\)). Choice B is \(\tan 45°\), and choice D would apply to \(\tan 90°\).

The signs of the trig functions depend on the quadrant. Use the mnemonic “All Students Take Calculus” (ASTC) to remember which functions are positive in each quadrant:

• Quadrant I: All positive
• Quadrant II: Sine positive
• Quadrant III: Tangent positive
• Quadrant IV: Cosine positive

We need sine negative and cosine positive. The only quadrant where cosine is positive but sine is negative is Quadrant IV.

The reference angle is the acute angle formed between the terminal side and the \(x\)-axis. Since 225° is in Quadrant III (between 180° and 270°), subtract 180°:

\(225° – 180° = 45°\)

The rule depends on the quadrant:

• Quadrant I: Reference angle = \(\theta\)
• Quadrant II: Reference angle = \(180° – \theta\)
• Quadrant III: Reference angle = \(\theta – 180°\)
• Quadrant IV: Reference angle = \(360° – \theta\)

Secant is the reciprocal of cosine:

\(\sec\theta = \dfrac{1}{\cos\theta}\)

Since \(\cos 60° = \frac{1}{2}\):

\(\sec 60° = \dfrac{1}{\frac{1}{2}} = 2\)

Choice B gives \(\cos 60°\) itself, not its reciprocal. Choice D gives \(\sec 30°\).

This is the Pythagorean identity, one of the most fundamental identities in trigonometry:

\(\sin^2\theta + \cos^2\theta = 1\)

It holds true for every value of \(\theta\). It comes from the Pythagorean theorem applied to a unit circle, where the \(x\)-coordinate is \(\cos\theta\) and the \(y\)-coordinate is \(\sin\theta\), and the radius is 1.

The two related Pythagorean identities (obtained by dividing by \(\cos^2\theta\) or \(\sin^2\theta\)) are:

• \(\tan^2\theta + 1 = \sec^2\theta\)
• \(1 + \cot^2\theta = \csc^2\theta\)

Since \(\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}\), use the Pythagorean theorem to find the hypotenuse:

\(\text{hypotenuse} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16}\)\(\:= \sqrt{25} = 5\)

This gives \(\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}\). However, since \(\theta\) is in Quadrant III, where sine is negative:

\(\sin\theta = -\dfrac{3}{5}\)

Choice C gives \(-\frac{4}{5}\), which is \(\cos\theta\) in Quadrant III, not \(\sin\theta\).

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

\(150° \times \dfrac{\pi}{180°} = \dfrac{150\pi}{180} = \dfrac{5\pi}{6}\)

Simplify the fraction by dividing numerator and denominator by 30.

Choice D gives \(\frac{7\pi}{6}\), which is 210°. This is a common error of adding an extra 60° (or \(\frac{\pi}{3}\)).

First, identify the reference angle. Since \(\sin\theta = \frac{1}{2}\), the reference angle is 30°.

Sine is positive in Quadrants I and II, so there are two solutions in \([0°, 360°)\):

• Quadrant I: \(\theta = 30°\)
• Quadrant II: \(\theta = 180° – 30° = 150°\)

Choice A gives only one of the two solutions. Choice C pairs 30° with 330°, but \(\sin 330° = -\frac{1}{2}\) (negative, since 330° is in Quadrant IV).

Cosine is an even function, which means:

\(\cos(-\theta) = \cos\theta\)

Negating the angle does not change the cosine value. Graphically, this is because the cosine graph is symmetric about the \(y\)-axis.

In contrast, sine and tangent are odd functions:

• \(\sin(-\theta) = -\sin\theta\)
• \(\tan(-\theta) = -\tan\theta\)

Note that \(\frac{5\pi}{4}\) radians = 225°, which is in Quadrant III. Find the reference angle:

\(\dfrac{5\pi}{4} – \pi = \dfrac{\pi}{4} = 45°\)

In Quadrant III, sine is negative. Therefore:

\(\sin \dfrac{5\pi}{4} = -\sin 45° = -\dfrac{\sqrt{2}}{2}\)

The ladder forms the hypotenuse of a right triangle. The height up the wall is the side opposite the 60° angle. Use the sine ratio:

\(\sin 60° = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{h}{12}\)

Solve for \(h\):

\(h = 12 \sin 60° = 12 \cdot \dfrac{\sqrt{3}}{2} = 6\sqrt{3}\)\(\:\approx 10.39 \text{ feet}\)

Choice B uses \(\cos 60° = \frac{1}{2}\) instead of sine, which would give the distance from the wall to the base of the ladder (the adjacent side), not the height.

This is one of the three Pythagorean identities. Starting from the fundamental identity \(\sin^2\theta + \cos^2\theta = 1\), divide every term by \(\cos^2\theta\):

\(\dfrac{\sin^2\theta}{\cos^2\theta} + \dfrac{\cos^2\theta}{\cos^2\theta} = \dfrac{1}{\cos^2\theta}\)

\(\tan^2\theta + 1 = \sec^2\theta\)

Choice B gives \(\csc^2\theta\), which comes from a different Pythagorean identity: \(1 + \cot^2\theta = \csc^2\theta\). Choice C equals 1 (the original Pythagorean identity), not \(\tan^2\theta + 1\).