Trigonometric Functions and Equations Questions and Answers

Answer: B

The correct match is (P) → (5), (Q) → (3), (R) → (2), (S) → (4).

(P) Put \(s = \sin^{2}x\): \(s^{3} + (1 - s)^{2} = 1\) reduces to \(s(s + 2)(s - 1) = 0\), so \(s = 0\) or \(s = 1\). In \([-\pi, \pi]\), \(\sin x = 0\) at \(-\pi, 0, \pi\) and \(\sin^{2}x = 1\) at \(\pm\frac{\pi}{2}\): 5 elements.

(Q) Put \(c = \cos^{2}x\): \(1 - c + c^{3} = 1\) gives \(c(c - 1)(c + 1) = 0\), so \(c = 0\) or \(1\). In \([-\frac{\pi}{2}, \frac{\pi}{2}]\) that is \(x = \pm\frac{\pi}{2}\) and \(x = 0\): 3 elements.

(R) Using \(\cos^{2}\frac{x}{2} = \frac{1 + \cos x}{2}\) and \(\sin^{2}x = 1 - \cos^{2}x\):

\[ \begin{aligned} 2\cos^{2}x + \cos x - 2 &= 0 \\ \cos x &= \frac{-1 \pm \sqrt{17}}{4} \end{aligned} \]

Only \(\frac{-1 + \sqrt{17}}{4} \approx 0.78\) lies in \([-1, 1]\), and it is taken once on each side of 0 in \([-\pi, \pi]\): 2 elements.

(S) \(6\sin^{2}\frac{x}{2} = 3 - 3\cos x\), so the equation becomes \(3\cos x + \cos 3x = 0\). With \(\cos 3x = 4\cos^{3}x - 3\cos x\) this is \(4\cos^{3}x = 0\), so \(\cos x = 0\): \(x = \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}\), 4 elements.

Only the second option has 5, 3, 2, 4 in order.

Answer: A

The correct pair of graphs is the one in option (A).

On \(0 \le x \le \pi\) the sine is between 0 and 1, and squaring a number between 0 and 1 makes it smaller. So \(\sin^{2}x \le \sin x\) everywhere, the two curves meet only at \(x = 0\), \(\pi/2\) and \(\pi\), and both reach exactly 1 at \(\pi/2\). Near 0 and \(\pi\) the square is much flatter, because \(\sin^{2}x \approx x^{2}\) there.

Option (B) shows \(\sin^{2}x\) starting above zero at \(x = 0\), but \(\sin^{2}0 = 0\). Option (C) has the square peaking below 1, but \(\sin^{2}(\pi/2) = 1\). Option (D) puts \(\sin^{2}x\) above \(\sin x\) and above 1, which can never happen.

Answer: C

The value is \(4\pi + 7\).

The principal value of \(\cot^{-1}\) lies in \((0, \pi)\), and cot has period \(\pi\), so add multiples of \(\pi\) to \(-11\) until it lands in that interval: \(-11 + 4\pi \approx 1.566\), which lies in \((0, \pi)\). So \(\cot^{-1}(\cot(-11)) = 4\pi - 11\).

\[ \begin{aligned} 10\sin\left(2 \cdot \tfrac{\pi}{4}\right) &= 10 \\ \sin(2\tan^{-1} 2) &= \frac{2 \cdot 2}{1 + 2^{2}} = \frac{4}{5} \\ \text{total} &= (4\pi - 11) + 10 + 8 = 4\pi + 7 \end{aligned} \]

The answer 7 comes from writing \(\cot^{-1}(\cot(-11)) = -11\), forgetting the principal range. \(3\pi + 7\) uses \(3\pi - 11\), which is negative and so also outside \((0, \pi)\).

Answer: E

The value is 0.

Use the sum formula \(\cos C + \cos D = 2\cos\frac{C + D}{2}\cos\frac{C - D}{2}\) with \(C = x + y\), \(D = y - x\):

\[ \begin{aligned} \cos(x + y) + \cos(y - x) &= 2\cos y \cos x \\ \tfrac{1}{2}[\ldots] &= \cos x \cos y \end{aligned} \]

Subtracting \(f(x)f(y) = \cos x \cos y\) leaves 0 for every \(x\) and \(y\). Since the expression is identically zero, the constants 2, -2, 1 and -1 cannot be right; for example \(x = y = 0\) gives \(\tfrac{1}{2}(1 + 1) - 1 = 0\).

Answer: D

\(\tan 15^{\circ} + \tan 75^{\circ} = \) 4.

Since \(\tan 75^{\circ} = \cot 15^{\circ}\), write \(\theta = 15^{\circ}\):

\[ \begin{aligned} \tan\theta + \cot\theta &= \frac{\sin^{2}\theta + \cos^{2}\theta}{\sin\theta\cos\theta} \\ &= \frac{2}{\sin 2\theta} = \frac{2}{\sin 30^{\circ}} = 4 \end{aligned} \]

Directly: \(\tan 15^{\circ} = 2 - \sqrt{3}\) and \(\tan 75^{\circ} = 2 + \sqrt{3}\), which add to 4. Their product is 1, not their sum; 2 is what you get by taking each tangent as 1, and 0 would need the two angles to be negatives of each other.

Answer: A

\(\tan x \tan y \tan z = \) 1.

With \(y = \frac{\pi}{4}\), \(\tan y = 1\) and \(x + z = 2y = \frac{\pi}{2}\), so \(z = \frac{\pi}{2} - x\).

\[ \begin{aligned} \tan z &= \tan\left(\tfrac{\pi}{2} - x\right) = \cot x \\ \tan x \tan y \tan z &= \tan x \cdot 1 \cdot \cot x = 1 \end{aligned} \]

The product is the constant 1, so it cannot equal \(\tan(x - y)\) or \(\tan(z - y)\), which vary with \(x\); nor 0 or 1/2.

Answer: C

\(\frac{a}{b} = \) \(\sqrt{3}\).

Apply the sum-to-product formulas to both sums:

\[ \begin{aligned} a &= 2\sin\tfrac{x + y}{2}\cos\tfrac{x - y}{2} \\ b &= 2\cos\tfrac{x + y}{2}\cos\tfrac{x - y}{2} \\ \frac{a}{b} &= \tan\tfrac{x + y}{2} = \tan\tfrac{\pi}{3} = \sqrt{3} \end{aligned} \]

\(\frac{\sqrt{3}}{3}\) is \(\tan\frac{\pi}{6}\) or \(\cot\frac{\pi}{3}\), from inverting the ratio. Using \(\tan(x + y)\) instead of the half-angle gives \(-\sqrt{3}\), which is not offered.

Answer: C

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

Between \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) the sine is positive only in the second quadrant, so \(\alpha\) is there and \(\cos\alpha\) is negative:

\[ \begin{aligned} \cos\alpha &= -\sqrt{1 - \tfrac{144}{169}} = -\tfrac{5}{13} \\ \tan\alpha &= \frac{12/13}{-5/13} = -\frac{12}{5} \end{aligned} \]

\(\frac{5}{12}\) is the cotangent with the sign dropped; \(\frac{13}{5}\) and \(\frac{-13}{5}\) are \(\pm\sec\alpha\), not \(\tan\alpha\).

Answer: C

\(f\left(\frac{1}{\sqrt{3}}\right) = \) \(\frac{\pi}{3}\).

Substitute directly:

\[ \begin{aligned} \frac{2x}{1 - x^{2}} &= \frac{2/\sqrt{3}}{1 - 1/3} = \frac{2/\sqrt{3}}{2/3} = \sqrt{3} \\ f &= \tan^{-1}\sqrt{3} = \frac{\pi}{3} \end{aligned} \]

This agrees with the identity \(\tan^{-1}\frac{2x}{1 - x^{2}} = 2\tan^{-1}x\) for \(|x| \lt 1\): \(2 \cdot \frac{\pi}{6} = \frac{\pi}{3}\). \(\frac{\pi}{6}\) forgets the factor 2; \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\) lie outside the principal range \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).

Answer: C

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

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

\[ \begin{aligned} 5\sin^{-1}\alpha + \tfrac{3\pi}{2} - 3\sin^{-1}\alpha &= \pi \\ 2\sin^{-1}\alpha &= -\tfrac{\pi}{2} \\ \sin^{-1}\alpha &= -\tfrac{\pi}{4} \end{aligned} \]

So \(\alpha = \sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}\). Check: \(5(-\frac{\pi}{4}) + 3(\frac{3\pi}{4}) = \pi\). \(\frac{1}{\sqrt{2}}\) loses the sign; 1 and -1 give \(\frac{5\pi}{2}\) and \(\frac{\pi}{2}\) instead.