Question paper, Set 5

$\displaystyle \cot\theta = \frac{\text{adj}}{\text{opp}} = \frac{21}{20}$: take adj $= 21$, opp $= 20$

hyp $= \sqrt{20^2 + 21^2} = \sqrt{841} = 29$

$\displaystyle \sin\theta = \frac{\text{opp}}{\text{hyp}} = \frac{20}{29}$, $\displaystyle \cos\theta = \frac{\text{adj}}{\text{hyp}} = \frac{21}{29}$

Answer: $\displaystyle \sin\theta = \frac{20}{29}$, $\displaystyle \cos\theta = \frac{21}{29}$

$\pi = 200^g$

$\displaystyle \frac{3\pi}{5} = \frac{3}{5} \times 200^g = 120^g$

Answer: $120^g$

$250^\circ \in \text{III}_Q \Rightarrow \sin 250^\circ$ is -ve

$250^\circ \in \text{III}_Q \Rightarrow \tan 250^\circ$ is +ve

So the product is negative

Answer: Negative

$\displaystyle = 6\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{\sqrt{2}}\right)^2 + 3\left(\frac{1}{\sqrt{3}}\right)^2$

$\displaystyle = \frac{3}{2} - 2 + 1$

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

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

Numerator: $\cos(90^\circ - \theta) = \sin\theta$, $\csc(360^\circ - \theta) = -\csc\theta$, $\csc(90^\circ - \theta) = \sec\theta$

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

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

Answer: $1$

$\csc 1230^\circ = \csc(1080^\circ + 150^\circ) = \csc 150^\circ$

$150^\circ \in \text{II}_Q$, where $\csc$ is +ve

$\csc 150^\circ = \csc(180^\circ - 30^\circ) = +\csc 30^\circ = 2$

Answer: $2$

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

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

$\Rightarrow -11 \le 2\sin x - 9 \le -7$

Greatest value $= -7$ (when $\sin x = 1$)

Answer: $-7$

$\sin^2 6^\circ + \sin^2 84^\circ = \sin^2 6^\circ + \cos^2 6^\circ = 1$

There are 14 terms ($6^\circ$ to $84^\circ$ in steps of $6^\circ$): 7 pairs, each $= 1$

G.E $= 7(1) = 7$

Answer: $7$

$\sin^2\theta + \cos^4\theta = \sin^2\theta + \cos^2\theta(1 - \sin^2\theta) = 1 - \sin^2\theta\cos^2\theta$

$\displaystyle = 1 - \left(\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16}$

Answer: $\displaystyle \frac{15}{16}$

$\displaystyle \cos A = \frac{21}{29} \Rightarrow \sin A = \frac{20}{29}$ (A acute)

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

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

$\displaystyle = \frac{20}{29} \cdot \frac{5}{13} - \frac{21}{29} \cdot \frac{12}{13} = \frac{100}{377} - \frac{252}{377} = -\frac{152}{377}$

Answer: $\displaystyle -\frac{152}{377}$

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

$\displaystyle \cos(A + B) = \frac{15}{17} \Rightarrow \sin(A + B) = \frac{8}{17}$, $\displaystyle \tan(A + B) = \frac{8}{15}$

$\displaystyle \cos(A - B) = \frac{40}{41}$ and $\displaystyle A - B < 0 \Rightarrow \sin(A - B) = -\frac{9}{41}$, $\displaystyle \tan(A - B) = -\frac{9}{40}$

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

$\displaystyle = \frac{\frac{8}{15} - \frac{9}{40}}{1 + \frac{8}{15} \cdot \frac{9}{40}} = \frac{\frac{37}{120}}{\frac{28}{25}}$

$\displaystyle = \frac{185}{672}$

Answer: $\displaystyle \frac{185}{672}$

In $\triangle ABC$: $\tan A + \tan B + \tan C = \tan A\tan B\tan C$

$3 + 4 + \tan C = 12\tan C$

$\displaystyle (12 - 1)\tan C = 7 \Rightarrow \tan C = \frac{7}{11}$

Answer: $\displaystyle \frac{7}{11}$

$105^\circ + 120^\circ = 225^\circ$, $\tan 225^\circ = 1$

$\displaystyle \frac{\tan 105^\circ + \tan 120^\circ}{1 - \tan 105^\circ\tan 120^\circ} = 1$

$\Rightarrow \tan 105^\circ + \tan 120^\circ + \tan 105^\circ\tan 120^\circ = 1$

Answer: $1$

$\displaystyle \sin A = \frac{20}{29} \Rightarrow \cos A = \frac{21}{29}$, $\displaystyle \tan A = \frac{20}{21}$ ($A$ acute)

$\displaystyle \sin 2A = 2\sin A\cos A = 2 \cdot \frac{20}{29} \cdot \frac{21}{29} = \frac{840}{841}$

$\displaystyle \cos 2A = 1 - 2\sin^2 A = 1 - 2 \cdot \frac{400}{841} = \frac{41}{841}$

Answer: $\displaystyle \sin 2A = \frac{840}{841}$, $\displaystyle \cos 2A = \frac{41}{841}$

Given $\displaystyle \cos\alpha = -\frac{15}{17}$ and $\alpha \in \text{II}_Q$

Right triangle with sides $8$, $15$, $17$; with the signs of $\text{II}_Q$: $\displaystyle \sin\alpha = \frac{8}{17}$, $\displaystyle \cos\alpha = -\frac{15}{17}$, $\displaystyle \tan\alpha = -\frac{8}{15}$

$\displaystyle \sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha = 3\left(\frac{8}{17}\right) - 4\left(\frac{8}{17}\right)^3 = \frac{24}{17} - \frac{2048}{4913} = \frac{4888}{4913}$

$\displaystyle \cos 2\alpha = \cos^2\alpha - \sin^2\alpha = \frac{225}{289} - \frac{64}{289} = \frac{161}{289}$

Answer: $\displaystyle \sin 3\alpha = \frac{4888}{4913}$, $\displaystyle \cos 2\alpha = \frac{161}{289}$

$\displaystyle = 4 \cdot \frac{\sqrt{5}-1}{4} \cdot \frac{\sqrt{5}+1}{4}$

$\displaystyle = \frac{(\sqrt{5})^2 - 1^2}{4} = \frac{4}{4} = 1$

Answer: $1$

$\cot A - \tan A = 2\cot 2A \Rightarrow \tan A = \cot A - 2\cot 2A$

$\tan 9^\circ = \cot 9^\circ - 2\cot 18^\circ$, $2\tan 18^\circ = 2\cot 18^\circ - 4\cot 36^\circ$

G.E $= \cot 9^\circ - 2\cot 18^\circ + 2\cot 18^\circ - 4\cot 36^\circ + 4\cot 36^\circ = \cot 9^\circ$

Answer: $\cot 9^\circ$

$\displaystyle \sin^2 A + \sin^2 B = 1 - (\cos^2 A - \sin^2 B) = 1 - \cos(A + B)\cos(A - B) = 1 - \frac{1}{2}\cos(A - B)$

$\displaystyle \sin A\sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] = \frac{1}{2}\cos(A - B) - \frac{1}{4}$

LHS $\displaystyle = 1 - \frac{1}{2}\cos(A - B) + \frac{1}{2}\cos(A - B) - \frac{1}{4} = \frac{3}{4}$ = RHS

$25^\circ + 35^\circ = 60^\circ$, so the value is $\displaystyle \frac{3}{4}$

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

Given $A + B + C = 180^\circ$, so $\sin(A + B) = \sin C$ and $\cos(A + B) = -\cos C$

LHS $= 1 + (\cos^2 A - \sin^2 B) + \cos^2 C = 1 + \cos(A + B)\cos(A - B) + \cos^2 C$

$= 1 - \cos C[\cos(A - B) - \cos C] = 1 - \cos C[\cos(A - B) + \cos(A + B)]$

$= 1 - \cos C \cdot 2\cos A\cos B = 1 - 2\cos A\cos B\cos C$

= RHS

Answer: Proved: LHS $= 1 - 2\cos A\cos B\cos C$

Period of $\sec(ax + b)$ is $\displaystyle \frac{2\pi}{|a|}$

$\displaystyle = \frac{2\pi}{3}$

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