Trigonometric Functions and Equations Questions and Answers

Answer: B

\(f\left(\frac{\pi}{12}\right)\) = \(2\sqrt{3}\).

Factor out 4 and use \(\sin(x + \frac{\pi}{4})\):

\[ \begin{aligned} f(x) &= 4\left(\tfrac{1}{\sqrt{2}}\sin x + \tfrac{1}{\sqrt{2}}\cos x\right) \\ &= 4\sin\left(x + \tfrac{\pi}{4}\right) \end{aligned} \]

\[ \begin{aligned} f\left(\tfrac{\pi}{12}\right) &= 4\sin\left(\tfrac{\pi}{12} + \tfrac{3\pi}{12}\right) \\ &= 4\sin\tfrac{\pi}{3} = 4 \cdot \tfrac{\sqrt{3}}{2} = 2\sqrt{3} \end{aligned} \]

\(\sqrt{3}\) and \(\frac{\sqrt{3}}{2}\) lose the amplitude 4; \(2\sqrt{2}\) is the coefficient, not a value of f.

Answer: E

\(\theta\) lies in the first quadrant or in the third quadrant.

The product \(\sin\theta\cos\theta\) is positive when the two factors have the same sign. Both are positive in the first quadrant, and both are negative in the third quadrant. Equivalently, \(\sin\theta\cos\theta = \frac{1}{2}\sin 2\theta\), and \(\tan\theta \gt 0\) in exactly these quadrants.

In the second and fourth quadrants the signs differ, so the product is negative, ruling out the options naming them. 'Only in the first quadrant' misses the third.

Answer: C

The values are 30°, 60°, 120°.

Treat the equation as a quadratic in \(s = \sin x\). It factorises:

\[ \begin{aligned} 4s^{2} - 2(1 + \sqrt{3})s + \sqrt{3} &= 0 \\ (2s - 1)(2s - \sqrt{3}) &= 0 \end{aligned} \]

So \(\sin x = \frac{1}{2}\) or \(\sin x = \frac{\sqrt{3}}{2}\). In \(15^{\circ} \lt x \lt 150^{\circ}\): \(\sin x = \frac{1}{2}\) gives \(x = 30^{\circ}\) (150° is excluded by the strict inequality), and \(\sin x = \frac{\sqrt{3}}{2}\) gives \(x = 60^{\circ}\) and \(120^{\circ}\).

45° and 90° give \(\sin x = \frac{1}{\sqrt{2}}\) and 1, which are not roots, so every other option fails.

Answer: D

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

Use the addition formula for tangent:

\[ \begin{aligned} \tan(\alpha + \beta) &= \frac{\frac{5}{6} + \frac{1}{11}}{1 - \frac{5}{6}\cdot\frac{1}{11}} \\ &= \frac{61/66}{61/66} = 1 \end{aligned} \]

Both angles are acute with small tangents, so \(0 \lt \alpha + \beta \lt \pi\) and \(\tan(\alpha + \beta) = 1\) gives \(\alpha + \beta = \frac{\pi}{4}\). \(\frac{\pi}{3}\) and \(\frac{\pi}{6}\) would need tangents \(\sqrt{3}\) and \(\frac{1}{\sqrt{3}}\).

Answer: D

The value is \(\frac{1}{4}(\sin 18^{\circ} + \cos 36^{\circ})\).

Use the identity \(\sin A \sin(60^{\circ} - A)\sin(60^{\circ} + A) = \frac{1}{4}\sin 3A\).

First term: \(\cos 36^{\circ} = \sin 54^{\circ}\), so with \(A = 6^{\circ}\) the product \(\sin 6^{\circ}\sin 54^{\circ}\sin 66^{\circ} = \frac{1}{4}\sin 18^{\circ}\).

Second term: \(\cos 12^{\circ} = \sin 78^{\circ}\), so with \(A = 18^{\circ}\) the product \(\sin 18^{\circ}\sin 42^{\circ}\sin 78^{\circ}\) equals \(\frac{1}{4}\sin 54^{\circ} = \frac{1}{4}\cos 36^{\circ}\).

Adding, the value is \(\frac{1}{4}(\sin 18^{\circ} + \cos 36^{\circ})\), about \(\frac{1}{4}(1.118) \approx 0.2795\), which matches a direct numerical check. The other coefficients use a wrong factor in place of \(\frac{1}{4}\).

Answer: B

The domain is [0,1].

\(\sin^{-1}u\) is defined only for \(-1 \leq u \leq 1\). With \(u = 2x - 1\):

\[ \begin{aligned} -1 \leq 2x - 1 &\leq 1 \\ 0 \leq 2x &\leq 2 \\ 0 \leq x &\leq 1 \end{aligned} \]

The constant \(-\frac{\pi}{4}\) and the factor 2 do not affect the domain. [-1,1] is the domain of \(\sin^{-1}x\) itself, and [0,2] forgets to divide by 2.

Answer: D

The value is \(\frac{\pi}{3}\).

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

\[ \begin{aligned} \sin\left(\tfrac{5\pi}{9} + \tfrac{\pi}{9}\right) &= \sin\tfrac{6\pi}{9} \\ &= \sin\tfrac{2\pi}{3} \end{aligned} \]

But \(\sin^{-1}\) returns a value in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), and \(\frac{2\pi}{3}\) is outside that range. Since \(\sin\frac{2\pi}{3} = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\), the principal value is \(\frac{\pi}{3}\).

\(\frac{2\pi}{3}\) is the trap: it ignores the principal range of \(\sin^{-1}\).

Answer: D

The value is \(\frac{24}{25}\).

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

\[ \begin{aligned} \sin 2\theta &= 2\sin\theta\cos\theta \\ &= 2 \cdot \tfrac{3}{5} \cdot \tfrac{4}{5} = \tfrac{24}{25} \end{aligned} \]

\(\frac{18}{25}\) comes from using \(2\sin^{2}\theta\), and \(\frac{7}{25}\) (not offered) would be \(\cos 2\theta\). The remaining options do not arise from a standard formula.

Answer: E

The value is -2.

Reduce each angle using periods of 180° for tan and cot, and their oddness:

\[ \begin{aligned} \cot(-405^{\circ}) &= -\cot 45^{\circ} = -1 \\ \tan 315^{\circ} &= -\tan 45^{\circ} = -1 \\ \cot 495^{\circ} &= \cot 135^{\circ} = -1 \\ \tan(-585^{\circ}) &= -\tan 45^{\circ} = -1 \end{aligned} \]

So the expression is \(3(-1)(-1) - 5(-1)(-1) = 3 - 5 = -2\). Sign slips in any one of the four reductions lead to 8, 2 or -8.

Answer: A

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

Pair the terms and use \(\sin(A + x) - \sin(A - x) = 2\cos A\sin x\) and \(\cos(A + x) - \cos(A - x) = -2\sin A\sin x\), with \(A = \frac{3\pi}{4}\):

\[ \begin{aligned} &2\cos\tfrac{3\pi}{4}\sin x - 2\sin\tfrac{3\pi}{4}\sin x \\ &= 2\left(-\tfrac{1}{\sqrt{2}}\right)\sin x - 2\left(\tfrac{1}{\sqrt{2}}\right)\sin x \\ &= -2\sqrt{2}\sin x \end{aligned} \]

The trap is the sign of \(\cos\frac{3\pi}{4}\), which is negative in the second quadrant; taking it as positive gives 0 instead.