Trigonometric Functions and Equations Questions and Answers

Answer: B

The expression equals \(2\sin x\).

Use \(\sin\theta\cos\theta = \frac{1}{2}\sin 2\theta\) repeatedly, starting from the smallest angle:

\[ \begin{aligned} \sin\tfrac{x}{2^{10}}\cos\tfrac{x}{2^{10}} &= \tfrac{1}{2}\sin\tfrac{x}{2^{9}} \\ \tfrac{1}{2}\sin\tfrac{x}{2^{9}}\cos\tfrac{x}{2^{9}} &= \tfrac{1}{2^{2}}\sin\tfrac{x}{2^{8}} \end{aligned} \]

Each of the 10 cosines halves the product and doubles the angle, so the product of the sine and all ten cosines is \(\frac{1}{2^{10}}\sin x\). Multiplying by \(2^{11}\):

\[ 2^{11} \times \frac{\sin x}{2^{10}} = 2\sin x \]

Treating \(2^{11}\) as cancelling exactly would give \(\sin x\); it leaves one factor of 2 over.

Answer: E

The product is 0.

The angles run 5°, 10°, 15°, ..., 240° in steps of 5°, so 180° is one of them. Since \(\sin 180^{\circ} = 0\), the whole product is 0.

The other options are of the form produced by product formulas for sines of angles below 90°; they would apply only if no factor vanished. Spotting a zero factor first is the quick route here.

Answer: A

The domain is \(\left[1, \sqrt{\tfrac{3}{2}}\right] \cup \left[-\sqrt{\tfrac{3}{2}}, -1\right]\).

Both \(\sin^{-1}\) and \(\cos^{-1}\) need their argument in \([-1, 1]\):

\[ \begin{aligned} -1 \leq 5 - 4x^{2} &\leq 1 \\ 1 \leq x^{2} &\leq \tfrac{3}{2} \end{aligned} \]

So \(1 \leq |x| \leq \sqrt{\frac{3}{2}}\), which gives the two intervals in the answer. Options with \(\sqrt{\frac{5}{2}}\), 2 or -2 as an endpoint come from solving one of the two inequalities wrongly.

Answer: A

The value is \(\dfrac{4\pi}{9}\).

The bracket is the expansion of \(\cos(A - B)\) with \(A = \frac{5\pi}{9}\), \(B = \frac{\pi}{9}\):

\[ \begin{aligned} \cos\tfrac{5\pi}{9}\cos\tfrac{\pi}{9} + \sin\tfrac{5\pi}{9}\sin\tfrac{\pi}{9} \\ &= \cos\tfrac{4\pi}{9} \end{aligned} \]

Since \(\frac{4\pi}{9}\) lies in \([0, \pi]\), the principal range of \(\cos^{-1}\), \(\cos^{-1}\left(\cos\frac{4\pi}{9}\right) = \frac{4\pi}{9}\). Using \(\cos(A + B)\) by mistake gives \(\frac{6\pi}{9}\), and \(\frac{5\pi}{9}\) just repeats one of the angles.

Answer: E

\(A = 0\) and \(B = \pi\).

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

\[ 2\sin^{-1}x + \cos^{-1}x = \sin^{-1}x + \frac{\pi}{2} \]

As \(x\) runs over \([-1, 1]\), \(\sin^{-1}x\) runs over \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), so the expression runs from \(-\frac{\pi}{2} + \frac{\pi}{2} = 0\) to \(\frac{\pi}{2} + \frac{\pi}{2} = \pi\). The least and greatest values are 0 and \(\pi\); the other pairs do not match these bounds.

Answer: B

The arc length is \(\pi\) cm.

Arc length is \(l = r\theta\) with \(\theta\) in radians. 45° is \(\frac{\pi}{4}\) radians.

\[ l = 4 \times \frac{\pi}{4} = \pi \text{ cm} \]

\(\frac{\pi}{4}\) is the angle in radians, forgetting to multiply by the radius, and the other fractions use the wrong angle.

Answer: C

\(\sin 2\theta = \frac{-3}{5}\).

Let \(t = \tan\theta\). Then

\[ \begin{aligned} \tan\left(\frac{\pi}{4} + \theta\right) &= \frac{1 + t}{1 - t} = \frac{1}{2} \\ 2 + 2t &= 1 - t \\ t &= -\frac{1}{3} \end{aligned} \]

\[ \begin{aligned} \sin 2\theta &= \frac{2t}{1 + t^{2}} = \frac{-2/3}{10/9} \\ &= -\frac{3}{5} \end{aligned} \]

\(\frac{4}{5}\) is \(\cos 2\theta = \frac{1 - t^{2}}{1 + t^{2}}\), and dropping the sign of \(t\) gives \(+\frac{3}{5}\), which is not offered.

Answer: B

\(\sin\frac{x}{2} = \sqrt{\frac{2 - \alpha}{2}}\).

Use the half-angle identity \(\sin^{2}\frac{x}{2} = \frac{1 - \cos x}{2}\). Since \(\cos x = \alpha - 1\),

\[ \begin{aligned} \sin^{2}\frac{x}{2} &= \frac{1 - (\alpha - 1)}{2} \\ &= \frac{2 - \alpha}{2} \end{aligned} \]

For \(0 \le x \le \frac{\pi}{2}\), \(\frac{x}{2}\) lies in the first quadrant, so \(\sin\frac{x}{2} \ge 0\) and the positive root is taken. \(\sqrt{\frac{\alpha}{2}}\) would be \(\cos\frac{x}{2}\); the options with \(1 \pm \alpha\) forget that \(\alpha\) already includes the 1.

Answer: E

The value is \(\pi\).

The principal value of \(\cos^{-1}\) lies in \([0, \pi]\), and \(\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}\), so \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}\). Also \(\sin^{-1}\frac{1}{2} = \frac{\pi}{6}\).

\[ \frac{5\pi}{6} + \frac{\pi}{6} = \pi \]

Taking \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)\) as \(\frac{\pi}{6}\), ignoring the minus sign, gives \(\frac{\pi}{3}\); taking it as \(-\frac{\pi}{6}\) gives 0. Both use the wrong principal value.

Answer: E

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

Use \(2\tan^{-1}x = \tan^{-1}\frac{2x}{1 - x^{2}}\) for \(|x| \lt 1\):

\[ \begin{aligned} 2\tan^{-1}\frac{1}{3} &= \tan^{-1}\frac{2/3}{8/9} = \tan^{-1}\frac{3}{4} \\ \cot^{-1}\frac{3}{4} &= \tan^{-1}\frac{4}{3} \end{aligned} \]

Then \(\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{4}{3} = \frac{\pi}{2}\), because \(\tan^{-1}y + \tan^{-1}\frac{1}{y} = \frac{\pi}{2}\) for \(y \gt 0\).

The common slip is to treat \(\cot^{-1}\frac{3}{4}\) as \(\tan^{-1}\frac{3}{4}\); it is \(\tan^{-1}\frac{4}{3}\), the complementary angle, which is what makes the sum exactly a right angle.