Trigonometric Functions and Equations Questions and Answers

Answer: A

They satisfy \(x^{2} + y^{2} - 4x + 2y + 4 = 0\).

Let both sides equal \(\theta\). Since \(\cos^{-1}\) lies in \([0, \pi]\) and \(\sin^{-1}\) in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), a common value lies in \(\left[0, \frac{\pi}{2}\right]\), and

\[ x - 2 = \cos\theta, \quad y + 1 = \sin\theta \]

\[ \begin{aligned} (x - 2)^{2} + (y + 1)^{2} &= 1 \\ x^{2} - 4x + 4 + y^{2} + 2y + 1 &= 1 \\ x^{2} + y^{2} - 4x + 2y + 4 &= 0 \end{aligned} \]

The option ending in \(+5\) forgets to move the 1 across, and the option with \(+4x\) has the centre's sign wrong.

Answer: D

\(x = \frac{7}{5}\).

Let \(\theta = \cot^{-1}\frac{4}{3}\), so \(\cot\theta = \frac{4}{3}\), \(\cos\theta = \frac{4}{5}\) and \(\sin\theta = \frac{3}{5}\).

\[ \begin{aligned} \cos 2\theta &= \cos^{2}\theta - \sin^{2}\theta \\ &= \tfrac{16}{25} - \tfrac{9}{25} = \tfrac{7}{25} \end{aligned} \]

So \(2\theta = \cos^{-1}\frac{7}{25}\), and \(\frac{x}{5} = \frac{7}{25}\) gives \(x = \frac{7}{5}\). The option \(\frac{7}{25}\) is \(\cos 2\theta\) itself, the value of \(\frac{x}{5}\), not \(x\).

Answer: A

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

Substitute \(x = \frac{1}{6}\) and \(y = \alpha\):

\[ \begin{aligned} \sin^{-1}\tfrac{1}{2} + 2\sin^{-1}\alpha &= \tfrac{\pi}{2} \\ \tfrac{\pi}{6} + 2\sin^{-1}\alpha &= \tfrac{\pi}{2} \\ \sin^{-1}\alpha &= \tfrac{\pi}{6} \\ \alpha &= \tfrac{1}{2} \end{aligned} \]

The values 2 and 4 are outside \([-1, 1]\), so they cannot be values of \(y\) on this curve, and \(\frac{1}{4}\) halves the angle instead of the sine.

Answer: C

\(\tan t = \frac{20}{21}\).

Use \(\text{cosec}^{2}t - \cot^{2}t = 1\), which factors as \((\text{cosec}\,t + \cot t)(\text{cosec}\,t - \cot t) = 1\). So \(\text{cosec}\,t - \cot t = \frac{2}{5}\).

\[ \begin{aligned} 2\cot t &= \frac{5}{2} - \frac{2}{5} = \frac{21}{10} \\ \cot t &= \frac{21}{20},\ \tan t = \frac{20}{21} \end{aligned} \]

\(\frac{21}{20}\) is cot t, the answer before inverting; \(\frac{21}{10}\) is 2 cot t.

Answer: A

\(\tan\theta = -\frac{3}{4}\).

Since \(\sin\theta \gt 0\) and \(\cos\theta \lt 0\), \(\theta\) is in the second quadrant.

\[ \begin{aligned} \cos\theta &= -\sqrt{1 - \tfrac{9}{25}} = -\frac{4}{5} \\ \tan\theta &= \frac{3/5}{-4/5} = -\frac{3}{4} \end{aligned} \]

\(\frac{3}{4}\) ignores the sign of the cosine, \(-\frac{4}{3}\) is \(\cot\theta\), and \(-\frac{3}{5}\) is just \(-\sin\theta\).

Answer: C

The product is \(\frac{\sqrt{2}}{8}\).

Since \(\sin 75^{\circ} = \cos 15^{\circ}\):

\[ \begin{aligned} \sin 15^{\circ}\cos 15^{\circ} &= \tfrac{1}{2}\sin 30^{\circ} = \tfrac{1}{4} \\ \text{product} &= \tfrac{1}{4} \times \tfrac{\sqrt{2}}{2} = \frac{\sqrt{2}}{8} \end{aligned} \]

All three sines are positive, so the negative options are impossible. \(\frac{\sqrt{2}}{4}\) forgets the \(\frac{1}{2}\) in the double-angle step.

Answer: A

The sum equals \(\sqrt{2}\sin x\).

Use \(\sin(x + A) + \sin(x - A) = 2\sin x\cos A\) with \(A = \frac{7\pi}{4}\).

\[ \begin{aligned} \cos\tfrac{7\pi}{4} &= \cos\left(2\pi - \tfrac{\pi}{4}\right) = \tfrac{\sqrt{2}}{2} \\ \text{sum} &= 2\sin x \cdot \tfrac{\sqrt{2}}{2} = \sqrt{2}\sin x \end{aligned} \]

\(-\sqrt{2}\sin x\) comes from taking \(\cos\frac{7\pi}{4}\) as negative (it lies in the fourth quadrant, where cosine is positive). The cosine options would need \(\cos x\sin A\) terms, which cancel in this sum.

Answer: C

x = \(\frac{1}{4}\).

\[ \begin{aligned} \cos^{-1}\left(\sqrt{3}(2x - 1)\right) &= \frac{5\pi}{6} \\ \sqrt{3}(2x - 1) &= \cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} \\ 2x - 1 &= -\frac{1}{2},\ x = \frac{1}{4} \end{aligned} \]

\(\frac{5\pi}{6}\) lies in \([0, \pi]\), the range of \(\cos^{-1}\), so the step is valid. Taking the cosine as positive gives \(x = \frac{3}{4}\), which is not offered, and x = \(\frac{1}{2}\) makes the argument 0, giving \(\frac{\pi}{2}\) instead.

Answer: A

The value is \(-\frac{1}{2}\).

Use \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}\) with \(\tan A = 3\), \(\tan B = 7\):

\[ \begin{aligned} \tan(A + B) &= \frac{3 + 7}{1 - 21} \\ &= \frac{10}{-20} = -\frac{1}{2} \end{aligned} \]

The sum of the two angles exceeds \(\frac{\pi}{2}\) because \(3 \times 7 \gt 1\), which is why the tangent is negative. \(\frac{1}{2}\) drops that sign, and \(\pm\frac{1}{5}\) and 0 do not come out of the formula at all.

Answer: A

x = \(\frac{4}{5}\).

Let \(\theta = \sin^{-1}x = \cos^{-1}\frac{3x}{4}\). Then \(\sin\theta = x\) and \(\cos\theta = \frac{3x}{4}\), so

\[ \begin{aligned} x^{2} + \frac{9x^{2}}{16} &= 1 \\ \frac{25x^{2}}{16} &= 1,\ x = \pm\frac{4}{5} \end{aligned} \]

Both \(\sin^{-1}x\) and \(\cos^{-1}\frac{3x}{4}\) can be equal only when the angle is in \([0, \frac{\pi}{2}]\), so x is positive: \(x = \frac{4}{5}\). Check: \(\sin^{-1}0.8 = \cos^{-1}0.6\). \(\frac{3}{5}\) is \(\frac{3x}{4}\), not x.