Trigonometric Functions and Equations Questions and Answers

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.

Answer: D

\(\frac{\alpha + 1}{\alpha - 1}\) equals \(-\cot\left(\frac{A+B}{2}\right)\cot\left(\frac{A-B}{2}\right)\).

Substitute \(\alpha = \frac{\cos A}{\cos B}\) and multiply through by \(\cos B\), then use the sum-to-product formulas:

\[ \begin{aligned} \frac{\alpha + 1}{\alpha - 1} &= \frac{\cos A + \cos B}{\cos A - \cos B} \\ &= \frac{2\cos\frac{A+B}{2}\cos\frac{A-B}{2}}{-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}} \end{aligned} \]

Each cosine over the matching sine is a cotangent, and the minus sign from \(\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\) stays in front.

The first option loses that minus sign. The options with a tangent pair a cosine with the wrong sine, and the last drops the second factor entirely.

Answer: A

\(x = \frac{1}{6}\).

Use \(\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a + b}{1 - ab}\) and \(\tan\frac{\pi}{4} = 1\):

\[ \begin{aligned} \frac{2x + 3x}{1 - 6x^{2}} &= 1 \\ 6x^{2} + 5x - 1 &= 0 \\ (6x - 1)(x + 1) &= 0 \end{aligned} \]

The roots are \(x = \frac{1}{6}\) and \(x = -1\). The root \(-1\) is rejected: both inverse tangents are then negative, so their sum is negative (in fact \(-\frac{3\pi}{4}\)), not \(\frac{\pi}{4}\).

Check \(x = \frac{1}{6}\): \(\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{2}\) has tangent \(\frac{5/6}{5/6} = 1\). The other options do not satisfy the quadratic.

Answer: B

The domain is \([-1, 2)\).

\(\cos^{-1}\) accepts arguments from \(-1\) to 1, so we need \(-1 \le [x] \le 1\). Since \([x]\) is an integer, it must be \(-1\), 0 or 1:

• \([x] = -1\) for \(-1 \le x \lt 0\)
• \([x] = 0\) for \(0 \le x \lt 1\)
• \([x] = 1\) for \(1 \le x \lt 2\)

Together these give \(-1 \le x \lt 2\). The left end is included, because \([-1] = -1\); the right end is excluded, because \([2] = 2\) is too large.

\([-1, 2]\) wrongly includes 2, and \((-1, 1)\) confuses the condition on \([x]\) with a condition on \(x\) itself.

Answer: E

\(3\sin 2\alpha - 4\cos 2\alpha = 5\).

\(\sin^{-1}y = \frac{\pi}{2}\) only when \(y = 1\), so:

\[ \begin{aligned} \frac{3\sin 2\alpha}{5 + 4\cos 2\alpha} &= 1 \\ 3\sin 2\alpha &= 5 + 4\cos 2\alpha \\ 3\sin 2\alpha - 4\cos 2\alpha &= 5 \end{aligned} \]

As a check, \(5\) is the largest value \(3\sin t - 4\cos t\) can take, since \(\sqrt{3^{2} + 4^{2}} = 5\), so such an \(\alpha\) exists.

3 and 4 are the coefficients themselves, and 1 is the value of the argument of \(\sin^{-1}\), not of the asked expression.

Answer: B

The range is \([-\sqrt{2} + 3, \sqrt{2} + 3]\).

Write \(\sin x + \cos x\) as a single sine:

\[ \sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) \]

This takes every value from \(-\sqrt{2}\) to \(\sqrt{2}\), so adding 3 gives every value from \(3 - \sqrt{2}\) to \(3 + \sqrt{2}\).

In general \(a\sin x + b\cos x\) ranges over \([-\sqrt{a^{2} + b^{2}}, \sqrt{a^{2} + b^{2}}]\); the options with \(\sqrt{3}\) use the wrong amplitude, and \([-\sqrt{2} - 3, 2 + \sqrt{3}]\) mixes the shift and the amplitude.

Answer: E

The product equals 1.

Pair complementary angles, using \(\cot(90^{\circ} - \theta) = \tan\theta\), so \(\cot\theta\cot(90^{\circ} - \theta) = 1\):

• \(\cot 10^{\circ}\cot 80^{\circ} = 1\)
• \(\cot 20^{\circ}\cot 70^{\circ} = 1\)
• \(\cot 30^{\circ}\cot 60^{\circ} = \sqrt{3} \times \frac{1}{\sqrt{3}} = 1\)
• \(\cot 45^{\circ} = 1\)

The whole product is therefore 1. The options \(\sqrt{3}\) and \(3\sqrt{3}\) come from evaluating \(\cot 30^{\circ}\) alone or multiplying only some of the factors.

Answer: D

The value is \(\frac{400}{81}\).

\[ \begin{aligned} \tan 2\alpha &= \frac{2\tan\alpha}{1 - \tan^{2}\alpha} = \frac{1}{1 - \frac{1}{4}} = \frac{4}{3} \\ \sec^{2}2\alpha &= 1 + \tan^{2}2\alpha = \frac{25}{9} \end{aligned} \]

\[ \begin{aligned} \tan^{2}2\alpha\sec^{2}2\alpha &= \frac{16}{9} \times \frac{25}{9} \\ &= \frac{400}{81} \end{aligned} \]

The option \(\frac{400}{9}\) multiplies by 25 instead of \(\frac{25}{9}\), and the \(\frac{200}{...}\) options halve the product somewhere along the way.

Answer: E

\(k = -8\).

Expand the left side and rewrite \(\sec^{2}x\) as \(1 + \tan^{2}x\):

\[ \begin{aligned} (3\cos x - 2\sec x)^{2} &= 9\cos^{2}x - 12 + 4\sec^{2}x \\ &= 9\cos^{2}x - 12 + 4 + 4\tan^{2}x \\ &= 9\cos^{2}x + 4\tan^{2}x - 8 \end{aligned} \]

The cross term is \(-2 \times 3\cos x \times 2\sec x = -12\), since \(\cos x\sec x = 1\). Comparing gives \(k = -8\). The option \(-12\) stops before converting \(\sec^{2}x\) to \(1 + \tan^{2}x\).

Answer: D

The value is \(2 + \sqrt{2}\).

Expand and group the squares:

\[ \begin{aligned} &(\sin\alpha + \sin\beta)^{2} + (\cos\alpha + \cos\beta)^{2} \\ &= 2 + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) \\ &= 2 + 2\cos(\alpha - \beta) \\ &= 2 + 2\cos\tfrac{\pi}{4} = 2 + \sqrt{2} \end{aligned} \]

Each pair \(\sin^{2} + \cos^{2}\) gives 1, which supplies the 2. Using \(\cos(\alpha + \beta)\) by mistake or dropping the factor 2 leads to the other options.