Question paper, Set 9

$\displaystyle \cos\theta = \frac{\text{adj}}{\text{hyp}} = \frac{45}{53}$: take adj $= 45$, hyp $= 53$

opp $= \sqrt{53^2 - 45^2} = \sqrt{784} = 28$

$\displaystyle \sin\theta = \frac{\text{opp}}{\text{hyp}} = \frac{28}{53}$, $\displaystyle \cot\theta = \frac{\text{adj}}{\text{opp}} = \frac{45}{28}$

Answer: $\displaystyle \sin\theta = \frac{28}{53}$, $\displaystyle \cot\theta = \frac{45}{28}$

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

$\displaystyle 250^g = 250 \times \frac{9}{10} = 225^\circ$

Answer: $225^\circ$

$100^\circ \in \text{II}_Q \Rightarrow \tan 100^\circ$ is -ve

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

So the product is positive

Answer: Positive

$\displaystyle = \frac{2/\sqrt{3}}{1 - \frac{1}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \frac{3}{\sqrt{3}}$

$= \sqrt{3}$

Answer: $\sqrt{3}$

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

Denominator: $\csc(180^\circ - \theta) = \csc\theta$, $\cos(270^\circ - \theta) = -\sin\theta$, $\sin(270^\circ - \theta) = -\cos\theta$

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

Answer: $-1$

$\displaystyle \frac{17\pi}{3} = 17 \times 60^\circ = 1020^\circ$

$\tan 1020^\circ = \tan(720^\circ + 300^\circ) = \tan 300^\circ$

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

$\tan 300^\circ = \tan(360^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3}$

Answer: $-\sqrt{3}$

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

$\Rightarrow -8 \le 8\sin x \le 8$

$\Rightarrow -5 \le 3 + 8\sin x \le 11$

Greatest value $= 11$ (when $\sin x = 1$)

Answer: $11$

$\sin^2 2^\circ + \sin^2 88^\circ = \sin^2 2^\circ + \cos^2 2^\circ = 1$

There are 44 terms: 22 pairs $(2^\circ, 88^\circ), \ldots, (44^\circ, 46^\circ)$

G.E $= 22(1) = 22$

Answer: $22$

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

$\displaystyle = 1 - 2\left(\frac{2}{7}\right)^2 = 1 - \frac{8}{49} = \frac{41}{49}$

Answer: $\displaystyle \frac{41}{49}$

$\displaystyle \sin A = \frac{12}{37} \Rightarrow \cos A = \frac{35}{37}$ (A acute)

$\displaystyle \sin B = \frac{4}{5} \Rightarrow \cos B = \frac{3}{5}$ (B acute)

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

$\displaystyle = \frac{35}{37} \cdot \frac{3}{5} - \frac{12}{37} \cdot \frac{4}{5} = \frac{21}{37} - \frac{48}{185} = \frac{57}{185}$

Answer: $\displaystyle \frac{57}{185}$

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

$\displaystyle \tan(A + B) = \frac{20}{21}$

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

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

$\displaystyle = \frac{\frac{20}{21} + \frac{9}{40}}{1 - \frac{20}{21} \cdot \frac{9}{40}} = \frac{\frac{989}{840}}{\frac{11}{14}}$

$\displaystyle = \frac{989}{660}$

Answer: $\displaystyle \frac{989}{660}$

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

$20 + (4 + 5)\cot C = 1$

$\displaystyle 9\cot C = 1 - 20 \Rightarrow \cot C = -\frac{19}{9}$

Answer: $\displaystyle -\frac{19}{9}$

$\displaystyle 18^\circ + 27^\circ = 45^\circ \Rightarrow \frac{\tan 18^\circ + \tan 27^\circ}{1 - \tan 18^\circ\tan 27^\circ} = 1$

$\Rightarrow \tan 18^\circ + \tan 27^\circ + \tan 18^\circ\tan 27^\circ = 1$

Answer: $1$

$\displaystyle \cos A = \frac{35}{37} \Rightarrow \sin A = \frac{12}{37}$, $\displaystyle \tan A = \frac{12}{35}$ ($A$ acute)

$\displaystyle \cos 2A = 2\cos^2 A - 1 = 2 \cdot \frac{1225}{1369} - 1 = \frac{1081}{1369}$

$\displaystyle \sin 2A = 2\sin A\cos A = 2 \cdot \frac{12}{37} \cdot \frac{35}{37} = \frac{840}{1369}$

Answer: $\displaystyle \cos 2A = \frac{1081}{1369}$, $\displaystyle \sin 2A = \frac{840}{1369}$

Given $\displaystyle \tan\alpha = -\frac{12}{5}$ and $\alpha \in \text{II}_Q$

Right triangle with sides $12$, $5$, $13$; with the signs of $\text{II}_Q$: $\displaystyle \sin\alpha = \frac{12}{13}$, $\displaystyle \cos\alpha = -\frac{5}{13}$, $\displaystyle \tan\alpha = -\frac{12}{5}$

$\displaystyle \cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha = 4\left(-\frac{5}{13}\right)^3 - 3\left(-\frac{5}{13}\right) = -\frac{500}{2197} + \frac{15}{13} = \frac{2035}{2197}$

$\displaystyle \sin 2\alpha = 2\sin\alpha\cos\alpha = 2\left(\frac{12}{13}\right)\left(-\frac{5}{13}\right) = -\frac{120}{169}$

Answer: $\displaystyle \cos 3\alpha = \frac{2035}{2197}$, $\displaystyle \sin 2\alpha = -\frac{120}{169}$

$\displaystyle \sin 18^\circ = \frac{\sqrt{5}-1}{4}$

$\displaystyle \csc 18^\circ = \frac{4}{\sqrt{5} - 1} = \frac{4(\sqrt{5} + 1)}{5 - 1} = \sqrt{5} + 1$

Answer: $\sqrt{5} + 1$

$\displaystyle \tan\frac{\theta}{2}(1 + \sec\theta) = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\cdot\frac{2\cos^2\frac{\theta}{2}}{\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta$

$\tan 5^\circ(1 + \sec 10^\circ) = \tan 10^\circ$

$\tan 10^\circ(1 + \sec 20^\circ) = \tan 20^\circ$

Answer: $\tan 20^\circ$

Let $a = \sin\theta$, $b = \sin(120^\circ + \theta)$, $c = \sin(240^\circ + \theta)$

$b + c = 2\sin(180^\circ + \theta)\cos 60^\circ = -\sin\theta$, so $a + b + c = 0$ and $a^3 + b^3 + c^3 = 3abc$

$b = \sin(60^\circ - \theta)$, $c = -\sin(60^\circ + \theta)$, so $\displaystyle abc = -\sin\theta\sin(60^\circ - \theta)\sin(60^\circ + \theta) = -\frac{1}{4}\sin 3\theta$

LHS $\displaystyle = 3abc = -\frac{3}{4}\sin 3\theta$ = RHS

Put $\theta = 10^\circ$: value $\displaystyle = -\frac{3}{4}\sin 30^\circ = -\frac{3}{8}$

Answer: $\displaystyle -\frac{3}{8}$

$A + B = 270^\circ - C \Rightarrow \cos(A + B) = \cos(270^\circ - C) = -\sin C$

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

$= 1 - 2\sin C[\cos(A - B) - \cos(A + B)] = 1 - 2\sin C \cdot 2\sin A\sin B = 1 - 4\sin A\sin B\sin C$

= RHS

Answer: Proved: LHS $= 1 - 4\sin A\sin B\sin C$

For $a|\sin x| + b|\cos x|$ the period is $\pi$ when $a \ne b$

Here $a = 2 \ne b = 3$, so period $= \pi$

Answer: $\pi$