Advanced Trigonometry: MCQ test, Set 3

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

If $\displaystyle \sin\theta = \frac{15}{17}$ and $\theta$ is acute, then $\cot\theta =$
A $\displaystyle \frac{8}{17}$ B $\displaystyle \frac{17}{8}$ C $\displaystyle \frac{8}{15}$ D $\displaystyle \frac{15}{8}$

Answer and solution

Answer: (C) $\displaystyle \frac{8}{15}$

$\displaystyle \sin\theta = \frac{\text{opp}}{\text{hyp}} = \frac{15}{17}$ : take opp $= 15$ , hyp $= 17$

adj $= \sqrt{17^2 - 15^2} = \sqrt{64} = 8$

$\displaystyle \cot\theta = \frac{\text{adj}}{\text{opp}} = \frac{8}{15}$
$270^\circ$ in grades is
A $330^g$ B $243^g$ C $270^g$ D $300^g$

Answer and solution

Answer: (D) $300^g$

$\displaystyle 1^\circ = \frac{10}{9}^g$

$\displaystyle 270^\circ = 270 \times \frac{10}{9} = 300^g$
The sign of $\sin 100^\circ\cos 200^\circ\tan 300^\circ$ is
A positive B not defined C negative D zero

Answer and solution

Answer: (A) positive

$100^\circ \in \text{II}_Q \Rightarrow \sin 100^\circ$ is +ve

$200^\circ \in \text{III}_Q \Rightarrow \cos 200^\circ$ is -ve

$300^\circ \in \text{IV}_Q \Rightarrow \tan 300^\circ$ is -ve

So the product is positive
$\csc 30^\circ + \sec 60^\circ =$
A $3$ B $4$ C $\displaystyle \frac{4 \sqrt{3}}{3}$ D $1$

Answer and solution

Answer: (B) $4$

$\displaystyle \csc 30^\circ = \frac{1}{\sin 30^\circ} = 2$ , $\displaystyle \sec 60^\circ = \frac{1}{\cos 60^\circ} = 2$

$= 4$
$\operatorname{csc}(180^\circ + \theta) =$
A $\operatorname{csc}\theta$ B $\operatorname{sec}\theta$ C $-\operatorname{sec}\theta$ D $-\operatorname{csc}\theta$

Answer and solution

Answer: (D) $-\operatorname{csc}\theta$

$180^\circ + \theta$ is of the form $180^\circ \pm \theta$ or $360^\circ \pm \theta$ , so $\csc$ does not change

For acute $\theta$ , $180^\circ + \theta \in \text{III}_Q$ , where $\csc$ is -ve

$\therefore \csc(180^\circ + \theta) = -\csc\theta$
$\sin 810^\circ + \cos 900^\circ =$
A $2$ B $0$ C $-2$ D $1$

Answer and solution

Answer: (B) $0$

$\sin 810^\circ = \sin(720^\circ + 90^\circ) = \sin 90^\circ = 1$

$\cos 900^\circ = \cos(720^\circ + 180^\circ) = \cos 180^\circ = -1$

$1 + (-1)$

$= 0$
The domain of $\tan x$ is
A $\displaystyle R - \left\{(2n + 1)\frac{\pi}{2} : n \in Z\right\}$ B $R - \{n\pi : n \in Z\}$ C $R$ D $[-1, 1]$

Answer and solution

Answer: (A) $\displaystyle R - \left\{(2n + 1)\frac{\pi}{2} : n \in Z\right\}$

$\displaystyle \tan x = \frac{\sin x}{\cos x}$ needs $\cos x \ne 0$

$\displaystyle \cos x = 0 \Rightarrow x = (2n + 1)\frac{\pi}{2}$
The values $\cos 1$ , $\cos 2$ , $\cos 3$ (angles in radians) satisfy
A $\operatorname{cos} 3 > \operatorname{cos} 2 > \operatorname{cos} 1$ B $\operatorname{cos} 2 > \operatorname{cos} 1 > \operatorname{cos} 3$ C $\operatorname{cos} 1 > \operatorname{cos} 3 > \operatorname{cos} 2$ D $\operatorname{cos} 1 > \operatorname{cos} 2 > \operatorname{cos} 3$

Answer and solution

Answer: (D) $\operatorname{cos} 1 > \operatorname{cos} 2 > \operatorname{cos} 3$

$1 \approx 57^\circ$ , $2 \approx 114^\circ$ , $3 \approx 171^\circ$

From the graph, $\cos x$ decreases from $0^\circ$ to $180^\circ$

$\therefore \cos 1 > \cos 2 > \cos 3$
If $\sin 3A = \cos(A - 26^\circ)$ , where $3A$ is acute, then $A =$
A $32^\circ$ B $58^\circ$ C $16^\circ$ D $29^\circ$

Answer and solution

Answer: (D) $29^\circ$

$\sin 3A = \cos(A - 26^\circ) \Rightarrow 3A + (A - 26^\circ) = 90^\circ$

$4A = 116^\circ \Rightarrow A = 29^\circ$
$\displaystyle \frac{\sin^4\theta + \cos^4\theta - 1}{\sin^6\theta + \cos^6\theta - 1} =$
A $1$ B $\displaystyle \frac{1}{3}$ C $\displaystyle \frac{2}{3}$ D $\displaystyle \frac{3}{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{2}{3}$

Numerator $= -2\sin^2\theta\cos^2\theta$ , denominator $= -3\sin^2\theta\cos^2\theta$

Ratio $\displaystyle = \frac{2}{3}$
The least value of $16\tan^2\theta + \cot^2\theta$ is
A $2$ B $17$ C $8$ D $4$

Answer and solution

Answer: (C) $8$

Consider $16\tan^2\theta$ and $\cot^2\theta$ ; both are +ve

A.M $\ge$ G.M: $\displaystyle \frac{16\tan^2\theta + \cot^2\theta}{2} \ge \sqrt{16\tan^2\theta \cdot \cot^2\theta} = \sqrt{16} = 4$

$\Rightarrow 16\tan^2\theta + \cot^2\theta \ge 8$ ; equality when $\displaystyle \tan^2\theta = \frac{1}{4}$
$\cos 105^\circ =$
A $\displaystyle \frac{\sqrt{3}-1}{2\sqrt{2}}$ B $\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}$ C $\displaystyle -\frac{\sqrt{3}+1}{2\sqrt{2}}$ D $\displaystyle \frac{1-\sqrt{3}}{2\sqrt{2}}$

Answer and solution

Answer: (D) $\displaystyle \frac{1-\sqrt{3}}{2\sqrt{2}}$

$\cos(60^\circ + 45^\circ) = \cos 60^\circ\cos 45^\circ - \sin 60^\circ\sin 45^\circ$

$\displaystyle = \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} = \frac{1-\sqrt{3}}{2\sqrt{2}}$
$\displaystyle \frac{\cos 12^\circ + \sin 12^\circ}{\cos 12^\circ - \sin 12^\circ} =$
A $\operatorname{tan} 57^\circ$ B $\operatorname{tan} 24^\circ$ C $\operatorname{tan} 12^\circ$ D $\operatorname{tan} 33^\circ$

Answer and solution

Answer: (A) $\operatorname{tan} 57^\circ$

$\displaystyle \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} = \frac{1 + \tan\theta}{1 - \tan\theta} = \tan(45^\circ + \theta)$

$= \tan(45^\circ + 12^\circ) = \tan 57^\circ$
$\displaystyle \cos^2 37\tfrac{1}{2}^\circ - \sin^2 7\tfrac{1}{2}^\circ =$
A $\displaystyle \frac{1}{2\sqrt{2}}$ B $\displaystyle \frac{\sqrt{3}}{2\sqrt{2}}$ C $\displaystyle \frac{\sqrt{3}}{2}$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (B) $\displaystyle \frac{\sqrt{3}}{2\sqrt{2}}$

$\displaystyle = \cos 45^\circ \cos 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}$
If $A + B + C = 90^\circ$ , then $\tan A\tan B + \tan B\tan C + \tan C\tan A =$
A $-1$ B $\operatorname{tan} A\operatorname{tan} B\operatorname{tan} C$ C $1$ D $0$

Answer and solution

Answer: (C) $1$

$A + B = 90^\circ - C \Rightarrow \cot(A + B) = \tan C$

$\displaystyle \frac{1 - \tan A\tan B}{\tan A + \tan B} = \tan C \Rightarrow \Sigma\tan A\tan B = 1$
$(1 - \cot 29^\circ)(1 - \cot 16^\circ) =$
A $-2$ B $2$ C $1$ D $0$

Answer and solution

Answer: (B) $2$

$29^\circ + 16^\circ = 45^\circ \Rightarrow (1 - \cot A)(1 - \cot B) = 2$
$\cos^2 15^\circ - \sin^2 15^\circ =$
A $\displaystyle \frac{\sqrt{3}}{2}$ B $\displaystyle \frac{1}{\sqrt{2}}$ C $\displaystyle \frac{1}{2}$ D $1$

Answer and solution

Answer: (A) $\displaystyle \frac{\sqrt{3}}{2}$

$\cos^2 A - \sin^2 A = \cos 2A$

$\displaystyle = \cos 30^\circ = \frac{\sqrt{3}}{2}$
$3\sin 20^\circ - 4\sin^3 20^\circ =$
A $\sqrt{3}$ B $\operatorname{sin} 20^\circ$ C $\displaystyle \frac{\sqrt{3}}{2}$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{\sqrt{3}}{2}$

$3\sin A - 4\sin^3 A = \sin 3A$

$\displaystyle = \sin 60^\circ = \frac{\sqrt{3}}{2}$
$\cos 144^\circ =$
A $\displaystyle \frac{\sqrt{5}+1}{4}$ B $\displaystyle -\frac{\sqrt{5}-1}{4}$ C $\displaystyle -\frac{\sqrt{5}+1}{4}$ D $\displaystyle \frac{\sqrt{5}-1}{4}$

Answer and solution

Answer: (C) $\displaystyle -\frac{\sqrt{5}+1}{4}$

$\displaystyle \cos 144^\circ = \cos(180^\circ - 36^\circ) = -\cos 36^\circ = -\frac{\sqrt{5}+1}{4}$
$\tan 10^\circ(1 + \sec 20^\circ) =$
A $\operatorname{tan} 20^\circ$ B $\operatorname{tan} 40^\circ$ C $\operatorname{tan} 10^\circ$ D $\operatorname{cot} 20^\circ$

Answer and solution

Answer: (A) $\operatorname{tan} 20^\circ$

$\displaystyle \tan\frac{\theta}{2}(1 + \sec\theta) = \tan\theta$ with $\theta = 20^\circ$
$\displaystyle \cos\frac{2\pi}{13} + \cos\frac{4\pi}{13} + \cos\frac{6\pi}{13} + \cdots + \cos\frac{12\pi}{13} =$
A $\displaystyle -\frac{1}{2}$ B $-1$ C $0$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (A) $\displaystyle -\frac{1}{2}$

$n = 6$ , $\displaystyle d = \frac{2\pi}{13}$ : sum $\displaystyle = \frac{\cos\frac{7\pi}{13}\sin\frac{6\pi}{13}}{\sin\frac{\pi}{13}} = -\frac{\sin\frac{12\pi}{13}}{2\sin\frac{\pi}{13}} = -\frac{1}{2}$
$\sin^2 20^\circ + \sin^2 40^\circ + \sin^2 80^\circ =$
A $\displaystyle \frac{3}{4}$ B $2$ C $1$ D $\displaystyle \frac{3}{2}$

Answer and solution

Answer: (D) $\displaystyle \frac{3}{2}$

$\displaystyle \sin^2\theta + \sin^2(60^\circ + \theta) + \sin^2(60^\circ - \theta) = \frac{3}{2}$ with $\theta = 20^\circ$
If $\displaystyle \frac{x}{y} = \frac{\sin A}{\sin B}$ , then $\displaystyle \frac{x + y}{x - y} =$
A $\displaystyle \frac{\operatorname{cos}\frac{A + B}{2}}{\operatorname{cos}\frac{A - B}{2}}$ B $\displaystyle \frac{\operatorname{tan}\frac{A + B}{2}}{\operatorname{tan}\frac{A - B}{2}}$ C $\displaystyle \frac{\operatorname{tan}\frac{A - B}{2}}{\operatorname{tan}\frac{A + B}{2}}$ D $\displaystyle \frac{\operatorname{sin}(A + B)}{\operatorname{sin}(A - B)}$

Answer and solution

Answer: (B) $\displaystyle \frac{\operatorname{tan}\frac{A + B}{2}}{\operatorname{tan}\frac{A - B}{2}}$

Componendo and dividendo: $\displaystyle \frac{x + y}{x - y} = \frac{\sin A + \sin B}{\sin A - \sin B}$

$\displaystyle = \frac{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}}{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}} = \frac{\tan\frac{A + B}{2}}{\tan\frac{A - B}{2}}$
$\displaystyle \frac{\sin 7x + \sin 3x}{\cos 7x + \cos 3x} =$
A $\operatorname{cot} 5x$ B $\operatorname{tan} 10x$ C $\operatorname{tan} 5x$ D $\operatorname{tan} 2x$

Answer and solution

Answer: (C) $\operatorname{tan} 5x$

$\displaystyle = \frac{2\sin 5x\cos 2x}{2\cos 5x\cos 2x} = \tan 5x$
The period of $\cos^4 x$ is
A $4\pi$ B $\pi$ C $\displaystyle \frac{\pi}{2}$ D $2\pi$

Answer and solution

Answer: (B) $\pi$

$\cos^n x$ with $n$ even has period $\pi$

( $\cos^4(x + \pi) = (-\cos x)^4 = \cos^4 x$ )

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MCQ test, Set 3