Trigonometric Functions and Equations Questions and Answers

Answer: A

\(\frac{\beta}{\alpha} = 10\).

Let \(t = \sin^{-1}x\), which ranges over \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). The expression \(\frac{\pi^{2}}{8} + 2\left(t - \frac{\pi}{4}\right)^{2}\) is smallest when \(t = \frac{\pi}{4}\) and largest at the end of the interval farthest from \(\frac{\pi}{4}\), which is \(t = -\frac{\pi}{2}\).

\[ \begin{aligned} \alpha &= \frac{\pi^{2}}{8} \\ \beta &= \frac{\pi^{2}}{8} + 2\left(\frac{3\pi}{4}\right)^{2} = \frac{\pi^{2}}{8} + \frac{9\pi^{2}}{8} \\ &= \frac{10\pi^{2}}{8} \end{aligned} \]

So \(\frac{\beta}{\alpha} = 10\). Using \(t = \frac{\pi}{2}\) for the maximum gives \(\beta = \frac{2\pi^{2}}{8}\) and a ratio of 2, the trap of choosing the wrong endpoint.

Answer: A

The value is 1.

Use the sum-to-product formulas on the numerator and denominator.

\[ \begin{aligned} \sin 40^\circ + \sin 20^\circ &= 2\sin 30^\circ\cos 10^\circ \\ \cos 40^\circ + \cos 20^\circ &= 2\cos 30^\circ\cos 10^\circ \\ \text{value} &= \sqrt{3}\tan 30^\circ = \sqrt{3} \times \tfrac{1}{\sqrt{3}} = 1 \end{aligned} \]

Forgetting the factor \(\sqrt{3}\) in front gives \(\frac{1}{\sqrt{3}}\), and \(\sqrt{3}\) itself is \(\tan 60^\circ\), the wrong half-sum.

Answer: D

\(\tan 2\theta\) is \(-\sqrt{3}\).

Replace \(\sec^{2}\theta\) by \(1 + \tan^{2}\theta\).

\[ \begin{aligned} 1 + 2\tan^{2}\theta &= 7 \Rightarrow \tan^{2}\theta = 3 \\ \tan\theta &= \sqrt{3},\ \theta = \tfrac{\pi}{3} \\ \tan 2\theta &= \tan\tfrac{2\pi}{3} = -\sqrt{3} \end{aligned} \]

The positive root is taken because \(\theta\) is in the first quadrant. The double angle \(\frac{2\pi}{3}\) is in the second quadrant, where tangent is negative, so \(\sqrt{3}\) is the sign trap. The formula \(\frac{2\tan\theta}{1 - \tan^{2}\theta} = \frac{2\sqrt{3}}{-2}\) gives the same result.

Answer: E

The value is \(\frac{2}{\sqrt{3}}\).

Use \(\tan 75^\circ = 2 + \sqrt{3}\) and \(\tan 15^\circ = 2 - \sqrt{3}\).

\[ \begin{aligned} &\frac{(2 + \sqrt{3}) + (2 - \sqrt{3})}{(2 + \sqrt{3}) - (2 - \sqrt{3})} \\ &= \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}} \end{aligned} \]

Equivalently, the expression is \(\frac{\sin 90^\circ}{\sin 60^\circ}\), since \(\tan A \pm \tan B = \frac{\sin(A \pm B)}{\cos A\cos B}\). \(\sqrt{3}\) and \(\frac{1}{\sqrt{3}}\) are tangents of 60° and 30°, not this ratio.

Answer: A

The value is 1.

Use the known values \(\cos 36^\circ = \frac{\sqrt{5} + 1}{4}\) and \(\cos 72^\circ = \frac{\sqrt{5} - 1}{4}\).

\[ \begin{aligned} 4\cos 36^\circ\cos 72^\circ &= 4 \times \frac{(\sqrt{5} + 1)(\sqrt{5} - 1)}{16} \\ &= \frac{5 - 1}{4} = 1 \end{aligned} \]

Two options both read 2 in the paper; neither is correct. 0 would need one of the cosines to vanish, and \(\sqrt{2}\) has no connection with these angles.

Answer: C

The maximum value is 385.

\(8\sin x + 15\cos x\) ranges between \(\pm\sqrt{8^{2} + 15^{2}} = \pm 17\). Adding 3 gives the range [-14, 20] for the expression inside the bracket.

\[ \begin{aligned} \max (8\sin x + 15\cos x + 3)^{2} &= 20^{2} = 400 \\ \max f &= 400 - 15 = 385 \end{aligned} \]

The square is largest at the end of the range farther from zero, which is 20, not -14 (whose square, 196, is smaller). Taking 17 without the 3 gives 274, and \(17^{2} + 3^{2}\) or similar slips give the other options.

Answer: E

The value is \(\pi\).

The principal ranges are \([0, \pi]\) for \(\cos^{-1}\) and \([-\frac{\pi}{2}, \frac{\pi}{2}]\) for \(\sin^{-1}\).

\[ \begin{aligned} \cos^{-1}\left(\cos\tfrac{2\pi}{3}\right) &= \tfrac{2\pi}{3} \\ \sin^{-1}\left(\sin\tfrac{2\pi}{3}\right) &= \sin^{-1}\left(\sin\tfrac{\pi}{3}\right) = \tfrac{\pi}{3} \\ \text{sum} &= \pi \end{aligned} \]

\(\frac{2\pi}{3}\) lies outside the range of \(\sin^{-1}\), so the second term is \(\pi - \frac{2\pi}{3}\). Taking it as \(\frac{2\pi}{3}\) would give \(\frac{4\pi}{3}\), which is not offered.

Answer: A

The value is 8.

Let \(\theta = \sec^{-1}3\), so \(\sec\theta = 3\). Then use \(\tan^{2}\theta = \sec^{2}\theta - 1\).

\[ \tan^{2}\theta = 3^{2} - 1 = 8 \]

Using \(\sec\theta - 1\) without squaring gives 2, and adding instead of subtracting gives 10, the trap in the options. Drawing a right triangle with hypotenuse 3 and adjacent side 1 gives opposite side \(\sqrt{8}\), so \(\tan\theta = \sqrt{8}\) and its square is 8.

Answer: A

The value is \(\frac{\pi}{6} - x\).

Divide the numerator and denominator by 2 and recognise the compound-angle forms.

\[ \begin{aligned} \frac{\tfrac{1}{2}\cos x - \tfrac{\sqrt{3}}{2}\sin x}{\tfrac{\sqrt{3}}{2}\cos x + \tfrac{1}{2}\sin x} &= \frac{\cos(x + \frac{\pi}{3})}{\sin(x + \frac{\pi}{3})} \\ &= \tan\left(\tfrac{\pi}{6} - x\right) \end{aligned} \]

For \(0 \lt x \lt \frac{\pi}{2}\), \(\frac{\pi}{6} - x\) lies in \((-\frac{\pi}{3}, \frac{\pi}{6})\), inside the principal range of \(\tan^{-1}\), so \(\tan^{-1}\) returns it unchanged. \(\frac{\pi}{3} - x\) uses the angle \(\frac{\pi}{3}\) without converting the cotangent into a tangent.