Trigonometric Functions and Equations Questions and Answers

Answer: D

\(\tan 2x\) is 50.

Write \(2x = (x + y) + (x - y)\) and use the addition formula:

\[ \begin{aligned} \tan 2x &= \frac{\frac{6}{5} + \frac{4}{5}}{1 - \frac{6}{5} \cdot \frac{4}{5}} \\ &= \frac{2}{1/25} = 50 \end{aligned} \]

The condition \(0 \lt x, y \lt \frac{\pi}{4}\) keeps both angles in range so the given tangents apply directly. Dividing by \(1 + \frac{24}{25}\) instead (the formula for a difference) gives about 1.02; the listed distractors 54, 55, 60 and 62 come from arithmetic slips with the fractions.

Answer: A

The value is 1.

Since \(\frac{\pi}{12} + \frac{\pi}{6} = \frac{\pi}{4}\), take tangents of the sum:

\[ \begin{aligned} \frac{\tan A + \tan B}{1 - \tan A \tan B} &= \tan\frac{\pi}{4} = 1 \\ \tan A + \tan B &= 1 - \tan A \tan B \end{aligned} \]

So \(\tan A + \tan B + \tan A \tan B = 1\), with \(A = \frac{\pi}{12}\) and \(B = \frac{\pi}{6}\). This works for any pair of angles adding to \(\frac{\pi}{4}\). \(\sqrt{3}\) is \(\tan\frac{\pi}{3}\), the trap if the angles are added wrongly.

Answer: B

The period is \(\pi\).

\(\sin 4x\) has period \(\frac{2\pi}{4} = \frac{\pi}{2}\), and \(\cos 2x\) has period \(\frac{2\pi}{2} = \pi\). The sum repeats only when both parts repeat together, which happens first at the least common multiple of the two periods:

\[ \text{LCM}\left(\frac{\pi}{2}, \pi\right) = \pi \]

The coefficients 2 and 3 change the amplitudes, not the periods. \(\frac{\pi}{2}\) is the period of the first term alone, after which the cosine term has not yet repeated, and \(2\pi\) is a period but not the smallest one.

Answer: E

The expression equals -1.

Use \(1 - \cos 2x = 2\sin^{2}x\) and \(1 + \cos 2x = 2\cos^{2}x\):

\[ \begin{aligned} \frac{1 - \cos 2x}{1 + \cos 2x} &= \frac{2\sin^{2}x}{2\cos^{2}x} = \tan^{2}x \\ \tan^{2}x - \sec^{2}x &= -1 \end{aligned} \]

The last step is the identity \(\sec^{2}x = 1 + \tan^{2}x\). The result is a constant, so tan2x and sec2x cannot be right; +1 comes from subtracting the other way round, \(\sec^{2}x - \tan^{2}x\).

Answer: D

\(\sin^{-1}(\sin\theta)\) equals \(\pi - \theta\).

The principal value of \(\sin^{-1}\) lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). For \(\theta\) in \(\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]\), \(\theta\) itself is outside that range, so the answer is not simply \(\theta\).

Use \(\sin(\pi - \theta) = \sin\theta\). As \(\theta\) runs from \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\), \(\pi - \theta\) runs from \(\frac{\pi}{2}\) down to \(-\frac{\pi}{2}\), exactly the principal range. So \(\sin^{-1}(\sin\theta) = \pi - \theta\).

\(\theta - \pi\) has the right size but the wrong sign: its sine is \(-\sin\theta\).

Answer: D

\(x + y + z\) is -3.

The principal value of \(\cos^{-1}\) lies in \([0, \pi]\), so each term is at most \(\pi\). Three such terms can add to \(3\pi\) only if each equals \(\pi\):

\[ \begin{aligned} \cos^{-1}x = \pi &\Rightarrow x = \cos\pi = -1 \\ x + y + z &= -1 - 1 - 1 = -3 \end{aligned} \]

3 is the answer to the companion question with \(\sin^{-1}\) adding to \(\frac{3\pi}{2}\) (where each variable is 1); here cosines of \(\pi\) are -1, so the sum is negative.

Answer: E

\(\cos^{-1}x + \cos^{-1}y\) equals \(\frac{\pi}{3}\).

For each variable, \(\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}\). Adding this for x and for y:

\[ \begin{aligned} (\sin^{-1}x + \sin^{-1}y) + (\cos^{-1}x + \cos^{-1}y) &= \pi \\ \cos^{-1}x + \cos^{-1}y &= \pi - \frac{2\pi}{3} = \frac{\pi}{3} \end{aligned} \]

\(\frac{2\pi}{3}\) repeats the given sum, and \(\frac{\pi}{2}\) uses the identity for one variable only.

Answer: C

\(\cos\frac{x}{2} = \frac{3}{\sqrt{10}}\).

Use the half-angle identity \(\cos x = 2\cos^{2}\frac{x}{2} - 1\):

\[ \begin{aligned} \cos^{2}\frac{x}{2} &= \frac{1 + \cos x}{2} \\ &= \frac{1 + \frac{4}{5}}{2} \\ &= \frac{9}{10} \\ \cos\frac{x}{2} &= \frac{3}{\sqrt{10}} \end{aligned} \]

Since \(x\) is in \(\left[0, \frac{\pi}{2}\right]\), \(\frac{x}{2}\) is in \(\left[0, \frac{\pi}{4}\right]\), where cosine is positive; that rules out \(-\frac{3}{\sqrt{10}}\). \(\frac{1}{\sqrt{10}}\) is \(\sin\frac{x}{2}\), from \(\frac{1 - \cos x}{2} = \frac{1}{10}\).

Answer: B

The value is \(\frac{1}{4}\).

Use the product formula \(2\sin A \sin B = \cos(A - B) - \cos(A + B)\) with \(A = 75^{\circ}\) and \(B = 15^{\circ}\):

\[ \begin{aligned} \sin\frac{5\pi}{12}\sin\frac{\pi}{12} \\ &= \tfrac{1}{2}(\cos 60^{\circ} - \cos 90^{\circ}) \\ &= \tfrac{1}{2}\left(\tfrac{1}{2} - 0\right) \\ &= \tfrac{1}{4} \end{aligned} \]

Another way: \(\sin 75^{\circ} = \cos 15^{\circ}\), so the product is \(\sin 15^{\circ}\cos 15^{\circ} = \frac{1}{2}\sin 30^{\circ} = \frac{1}{4}\).

\(\frac{1}{2}\) forgets the factor \(\frac{1}{2}\) in the product formula. 1 and \(\frac{\sqrt{3}}{2}\) are values of single sines (\(\sin 90^{\circ}\), \(\sin 60^{\circ}\)), not of this product.

Answer: D

The expression equals \(\frac{3}{4}\tan^{2}2\theta\).

Write \(s = \sin^{2}\theta\), \(c = \cos^{2}\theta\), with \(s + c = 1\). Then:

\[ \begin{aligned} s^{3} + c^{3} &= (s + c)^{3} - 3sc(s + c) \\ &= 1 - 3\sin^{2}\theta\cos^{2}\theta \\ 1 - s^{3} - c^{3} &= 3\sin^{2}\theta\cos^{2}\theta \\ &= \tfrac{3}{4}\sin^{2}2\theta \end{aligned} \]

Dividing by \(\cos^{2}2\theta\) gives \(\frac{3}{4}\tan^{2}2\theta\).

\(\frac{3}{2}\) comes from writing \(\sin\theta\cos\theta = \frac{1}{2}\sin 2\theta\) but forgetting to square the \(\frac{1}{2}\); \(\frac{1}{4}\) drops the factor 3 from the cube identity.