Trigonometric Functions and Equations Questions and Answers

Answer: C

\(\sec^{2}\theta = \) \(\frac{2}{1-x}\).

\(\theta = \cot^{-1}\sqrt{\frac{1 - x}{1 + x}}\) means \(\cot\theta = \sqrt{\frac{1 - x}{1 + x}}\), so \(\tan^{2}\theta = \frac{1 + x}{1 - x}\).

\[ \begin{aligned} \sec^{2}\theta &= 1 + \tan^{2}\theta \\ &= \frac{(1 - x) + (1 + x)}{1 - x} = \frac{2}{1 - x} \end{aligned} \]

\(\frac{1 - x}{2}\) is \(\cos^{2}\theta\), the reciprocal of the answer, and \(\frac{1 + x}{2}\) is \(\sin^{2}\theta\); \(x\) and \(2x\) do not come from the identity at all.

Answer: E

The value is 1.

The expression is the triple-angle formula for tangent:

\[ \begin{aligned} \tan 3\theta \\ &= \frac{3\tan\theta - \tan^{3}\theta}{1 - 3\tan^{2}\theta} \end{aligned} \]

With \(\theta = 15^{\circ}\) it equals \(\tan 45^{\circ} = 1\). No numerical value of \(\tan 15^{\circ}\) is needed; recognising the pattern is the whole question. The other options do not correspond to \(\tan 45^{\circ}\).

Answer: E

The value is 0.

Use \(\sin(180^{\circ} - \theta) = \sin\theta\):

\[ \begin{aligned} \sin 100^{\circ} &= \sin 80^{\circ} \\ \sin 120^{\circ} \\ &= \sin 60^{\circ} \end{aligned} \]

So the expression is \(\sin 60^{\circ} - \sin 80^{\circ} + \sin 80^{\circ} - \sin 60^{\circ} = 0\): the terms cancel in pairs. Options such as \(\sqrt{3}\) or \(2\sqrt{3}\) come from adding the \(60^{\circ}\) and \(120^{\circ}\) terms instead of subtracting them.

Answer: E

\(x =\) \(\frac{1}{2}\).

Use \(\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}\). Adding this to the given equation:

\[ \begin{aligned} 2\cos^{-1}x &= \frac{\pi}{2} + \frac{\pi}{6} \\ &= \frac{2\pi}{3} \\ \cos^{-1}x \\ &= \frac{\pi}{3} \end{aligned} \]

so \(x = \cos\frac{\pi}{3} = \frac{1}{2}\). Check: \(\frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}\). \(\frac{\sqrt{3}}{2}\) would make the difference \(-\frac{\pi}{6}\), and \(\pm\frac{1}{\sqrt{2}}\) would make it 0 or \(\pi\).

Answer: B

The sum is \(\frac{\pi}{2}\).

For positive arguments \(\cot^{-1}a = \tan^{-1}\frac{1}{a}\). Then

\[ \begin{aligned} \tan^{-1}\tfrac{1}{2} + \tan^{-1}\tfrac{1}{3} &= \tan^{-1}\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \tan^{-1}1 = \tfrac{\pi}{4} \\ \text{sum} &= \tfrac{\pi}{4} + \tfrac{\pi}{4} = \tfrac{\pi}{2} \end{aligned} \]

\(\frac{\pi}{4}\) is only the first term or the last two terms; \(\pi\) and \(\frac{3\pi}{2}\) are too large for three angles each less than \(\frac{\pi}{4}\) or equal to it.

Answer: A

\(\alpha =\) \(\frac{5}{12}\).

Let \(\theta = \sin^{-1}\alpha\). Then \(\cos 2\theta = 1 - 2\sin^{2}\theta = 1 - 2\alpha^{2}\):

\[ \begin{aligned} 1 - 2\alpha^{2} &= \frac{47}{72} \\ \alpha^{2} &= \frac{25}{144} \;\Rightarrow\; \alpha = \frac{5}{12} \end{aligned} \]

taking the positive root since \(0 \lt \alpha \lt 1\). The options with denominator 13 come from mixing this up with the 5-12-13 triangle, and \(\frac{7}{12}\) does not satisfy the equation.

Answer: C

\(\alpha =\) \(\frac{\pi}{4}\).

\(\tan\phi = \frac{1}{\sqrt{3}}\) gives \(\phi = \frac{\pi}{6}\) in the relevant range, so

\[ \begin{aligned} \alpha &= \frac{\pi}{6} + \frac{\pi}{12} \\ &= \frac{3\pi}{12} \\ &= \frac{\pi}{4} \end{aligned} \]

which lies in \(\left(0, \frac{\pi}{2}\right)\). \(\frac{\pi}{6}\) forgets to add \(\frac{\pi}{12}\), \(\frac{\pi}{3}\) uses \(\tan^{-1}\sqrt{3}\), and the other angles do not satisfy the equation.

Answer: D

\(\frac{y}{x} =\) \(-\frac{1}{2}\).

Divide and use \(1 + \cos 2\theta = 2\cos^{2}\theta\) and \(\tan^{2}\theta - \sec^{2}\theta = -1\):

\[ \begin{aligned} \frac{y}{x} &= \frac{(\tan\theta + \sec\theta)(\tan\theta - \sec\theta)}{\sec^{2}\theta\,(1 + \cos 2\theta)} \\ &= \frac{-1}{\sec^{2}\theta \cdot 2\cos^{2}\theta} = -\frac{1}{2} \end{aligned} \]

\(\frac{1}{2}\) drops the minus sign from \(\tan^{2}\theta - \sec^{2}\theta\), and \(\pm 2\) inverts the ratio (\(x/y\)).

Answer: E

The value is 0.

Add the first two terms over a common denominator.

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

So the sum of the squares equals their product and the expression is 0 for every x where it is defined. The answers 1 or a single trigonometric function come from stopping before using \(\sin^{2}x + \cos^{2}x = 1\).

Answer: B

The product is \(\frac{1}{4\sqrt{2}}\).

Since \(\sin 75^{\circ} = \cos 15^{\circ}\), the outer factors form a double angle.

\[ \begin{aligned} \sin 15^{\circ}\cos 15^{\circ} &= \tfrac{1}{2}\sin 30^{\circ} = \tfrac{1}{4} \\ \tfrac{1}{4}\sin 45^{\circ} &= \tfrac{1}{4} \cdot \tfrac{1}{\sqrt{2}} = \tfrac{1}{4\sqrt{2}} \end{aligned} \]

\(\frac{1}{2\sqrt{2}}\) drops the factor \(\frac{1}{2}\) from the double-angle formula. The options with \(\sqrt{3}\) would need a 30 or 60 degree factor that is not in the product.