Question paper, Set 4

$\displaystyle \cos\theta = \frac{\text{adj}}{\text{hyp}} = \frac{24}{25}$: take adj $= 24$, hyp $= 25$

opp $= \sqrt{25^2 - 24^2} = \sqrt{49} = 7$

$\displaystyle \tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{7}{24}$, $\displaystyle \csc\theta = \frac{\text{hyp}}{\text{opp}} = \frac{25}{7}$

Answer: $\displaystyle \tan\theta = \frac{7}{24}$, $\displaystyle \csc\theta = \frac{25}{7}$

$200^g = 180^\circ$, so $\displaystyle 1^g = \frac{9}{10}$ degree

$\displaystyle 120^g = 120 \times \frac{9}{10} = 108^\circ$

Answer: $108^\circ$

$\displaystyle = \frac{1}{2} + \frac{2}{1}$

$\displaystyle = \frac{5}{2}$

Answer: $\displaystyle \frac{5}{2}$

Numerator: $\csc(270^\circ - \theta) = -\sec\theta$, $\tan(180^\circ - \theta) = -\tan\theta$, $\sec(90^\circ + \theta) = -\csc\theta$

Denominator: $\sec(180^\circ - \theta) = -\sec\theta$, $\cot(90^\circ - \theta) = \tan\theta$, $\csc(360^\circ - \theta) = -\csc\theta$

G.E $\displaystyle = \frac{(-\sec\theta)(-\tan\theta)(-\csc\theta)}{(-\sec\theta)(\tan\theta)(-\csc\theta)} = -1$

Answer: $-1$

$\sec 840^\circ = \sec(720^\circ + 120^\circ) = \sec 120^\circ$

$120^\circ \in \text{II}_Q$, where $\sec$ is -ve

$\sec 120^\circ = \sec(180^\circ - 60^\circ) = -\sec 60^\circ = -2$

Answer: $-2$

$-1 \le \sin x \le 1$

$\Rightarrow -2 \le 2\sin x \le 2$

$\Rightarrow 1 \le 2\sin x + 3 \le 5$

Range $= [1, 5]$

Answer: $[1, 5]$

$2 = 2^c \approx 114^\circ$

$\sin 114^\circ = \sin(180^\circ - 66^\circ) = \sin 66^\circ$

$\sin x$ increases from $0^\circ$ to $90^\circ$ and $2^\circ < 66^\circ$, so $\sin 2^\circ < \sin 2$

Answer: $\sin 2$ is greater

$\tan 1^\circ \tan 89^\circ = \tan 1^\circ \cot 1^\circ = 1$ (as $\tan(90^\circ - \theta) = \cot\theta$)

The 89 factors make 44 such pairs, and $\tan 45^\circ = 1$ is left

G.E $= 1^{44} \times 1 = 1$

Answer: $1$

Consider $4\sin^2\theta$ and $\csc^2\theta$; both are +ve

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

$\Rightarrow 4\sin^2\theta + \csc^2\theta \ge 4$; equality when $\displaystyle \sin^2\theta = \frac{1}{2}$

Least value $= 4$

Answer: $4$

$\displaystyle \sin A = \frac{12}{13} \Rightarrow \cos A = \frac{5}{13}$ (A acute)

$\displaystyle \cos B = \frac{8}{17} \Rightarrow \sin B = \frac{15}{17}$ (B acute)

$\cos(A - B) = \cos A\cos B + \sin A\sin B$

$\displaystyle = \frac{5}{13} \cdot \frac{8}{17} + \frac{12}{13} \cdot \frac{15}{17} = \frac{40}{221} + \frac{180}{221} = \frac{220}{221}$

Answer: $\displaystyle \frac{220}{221}$

$\displaystyle 0 < A < B < \frac{\pi}{4} \Rightarrow 0 < A + B < \frac{\pi}{2}$ and $A - B < 0$

$\displaystyle \sin(A + B) = \frac{3}{5} \Rightarrow \cos(A + B) = \frac{4}{5}$, $\displaystyle \tan(A + B) = \frac{3}{4}$

$\displaystyle \sin(A - B) = -\frac{7}{25} \Rightarrow \cos(A - B) = \frac{24}{25}$, $\displaystyle \tan(A - B) = -\frac{7}{24}$

$\displaystyle \tan 2B = \tan[(A + B) - (A - B)] = \frac{\tan(A + B) - \tan(A - B)}{1 + \tan(A + B)\tan(A - B)}$

$\displaystyle = \frac{\frac{3}{4} + \frac{7}{24}}{1 - \frac{3}{4} \cdot \frac{7}{24}} = \frac{\frac{25}{24}}{\frac{25}{32}}$

$\displaystyle = \frac{4}{3}$

Answer: $\displaystyle \frac{4}{3}$

$\sin^2 A - \sin^2 B = \sin(A + B)\sin(A - B)$

$= \sin 90^\circ \sin 45^\circ$

$\displaystyle = 1 \cdot \frac{1}{\sqrt{2}}$

$\displaystyle = \frac{1}{\sqrt{2}}$

Answer: $\displaystyle \frac{1}{\sqrt{2}}$

In $\triangle ABC$: $\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

$3 + (1 + 3)\cot C = 1$

$\displaystyle 4\cot C = 1 - 3 \Rightarrow \cot C = -\frac{1}{2}$

Answer: $\displaystyle -\frac{1}{2}$

$\displaystyle \tan A = \frac{7}{24}$, $\displaystyle \tan^2 A = \frac{49}{576}$

$\displaystyle \cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{\frac{527}{576}}{\frac{625}{576}} = \frac{527}{625}$

$\displaystyle \tan 2A = \frac{2\tan A}{1 - \tan^2 A} = \frac{\frac{7}{12}}{\frac{527}{576}} = \frac{336}{527}$

Answer: $\displaystyle \cos 2A = \frac{527}{625}$, $\displaystyle \tan 2A = \frac{336}{527}$

Given $\displaystyle \sin\alpha = -\frac{7}{25}$ and $\alpha \in \text{IV}_Q$

Right triangle with sides $7$, $24$, $25$; with the signs of $\text{IV}_Q$: $\displaystyle \sin\alpha = -\frac{7}{25}$, $\displaystyle \cos\alpha = \frac{24}{25}$, $\displaystyle \tan\alpha = -\frac{7}{24}$

$\displaystyle \cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha = 4\left(\frac{24}{25}\right)^3 - 3\left(\frac{24}{25}\right) = \frac{55296}{15625} - \frac{72}{25} = \frac{10296}{15625}$

$\displaystyle \tan 2\alpha = \frac{2\tan\alpha}{1 - \tan^2\alpha} = \frac{-\frac{7}{12}}{1 - \frac{49}{576}} = \frac{-\frac{7}{12}}{\frac{527}{576}} = -\frac{336}{527}$

Answer: $\displaystyle \cos 3\alpha = \frac{10296}{15625}$, $\displaystyle \tan 2\alpha = -\frac{336}{527}$

Angles in A.P.: $\displaystyle a_1 = \frac{2\pi}{11}$, $\displaystyle a_n = \frac{10\pi}{11}$, $\displaystyle d = \frac{2\pi}{11}$, $n = 5$

Sum $\displaystyle = \frac{\cos\frac{a_1 + a_n}{2}\,\sin\frac{nd}{2}}{\sin\frac{d}{2}}$

$\displaystyle = \frac{\cos\frac{6\pi}{11}\sin\frac{5\pi}{11}}{\sin\frac{\pi}{11}}$, and $\displaystyle \cos\frac{6\pi}{11} = -\cos\frac{5\pi}{11}$

$\displaystyle = -\frac{\sin\frac{10\pi}{11}}{2\sin\frac{\pi}{11}} = -\frac{\sin\frac{\pi}{11}}{2\sin\frac{\pi}{11}} = -\frac{1}{2}$

Answer: $\displaystyle -\frac{1}{2}$

Use $\displaystyle \sin^2 A = \frac{1 - \cos 2A}{2}$

LHS $\displaystyle = \frac{3}{2} - \frac{1}{2}[\cos 2\theta + \cos(240^\circ + 2\theta) + \cos(240^\circ - 2\theta)]$

$\displaystyle = \frac{3}{2} - \frac{1}{2}[\cos 2\theta + 2\cos 240^\circ\cos 2\theta]$, and $\displaystyle \cos 240^\circ = -\frac{1}{2}$

$\displaystyle = \frac{3}{2} - \frac{1}{2}[\cos 2\theta - \cos 2\theta] = \frac{3}{2}$ = RHS

Put $\theta = 10^\circ$: $\displaystyle \sin^2 10^\circ + \sin^2 130^\circ + \sin^2 110^\circ = \frac{3}{2}$

Answer: $\displaystyle \frac{3}{2}$

$\displaystyle \frac{x}{y} = \frac{\tan A}{\tan B} \Rightarrow \frac{x + y}{x - y} = \frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin(A + B)}{\sin(A - B)}$

$\displaystyle = \frac{\sin 90^\circ}{\sin 60^\circ} = \frac{2}{\sqrt{3}}$

Answer: $\displaystyle \frac{2}{\sqrt{3}}$

$\displaystyle \frac{A + B}{2} = 90^\circ - \frac{C}{2}$, so $\displaystyle \sin\frac{A + B}{2} = \cos\frac{C}{2}$ and $\displaystyle \cos\frac{A + B}{2} = \sin\frac{C}{2}$

LHS $\displaystyle = 2\sin\frac{A + B}{2}\cos\frac{A - B}{2} - 2\sin\frac{C}{2}\cos\frac{C}{2} = 2\cos\frac{C}{2}\left[\cos\frac{A - B}{2} - \sin\frac{C}{2}\right]$

$\displaystyle = 2\cos\frac{C}{2}\left[\cos\frac{A - B}{2} - \cos\frac{A + B}{2}\right] = 2\cos\frac{C}{2} \cdot 2\sin\frac{A}{2}\sin\frac{B}{2}$

= RHS

Answer: Proved: LHS $\displaystyle = 4\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2}$

$\displaystyle \cos^2 x = \frac{1 + \cos 2x}{2} = \frac{1}{2} + \frac{1}{2}\cos 2x$

Period $\displaystyle = \frac{2\pi}{2} = \pi$

Answer: $\pi$