Question paper, Set 3

$\displaystyle \sec\theta = \frac{\text{hyp}}{\text{adj}} = \frac{17}{15}$: take hyp $= 17$, adj $= 15$

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

$\displaystyle \sin\theta = \frac{\text{opp}}{\text{hyp}} = \frac{8}{17}$, $\displaystyle \cot\theta = \frac{\text{adj}}{\text{opp}} = \frac{15}{8}$

Answer: $\displaystyle \sin\theta = \frac{8}{17}$, $\displaystyle \cot\theta = \frac{15}{8}$

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

$\displaystyle 135^\circ = 135 \times \frac{10}{9} = 150^g$

Answer: $150^g$

$160^\circ \in \text{II}_Q \Rightarrow \cot 160^\circ$ is -ve

Answer: Negative

$= 2(2)^2 - 3(1)^2 + 1$

$= 8 - 3 + 1$

$= 6$

Answer: $6$

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

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

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

Answer: $-1$

$\tan 1215^\circ = \tan(1080^\circ + 135^\circ) = \tan 135^\circ$

$135^\circ \in \text{II}_Q$, where $\tan$ is -ve

$\tan 135^\circ = \tan(180^\circ - 45^\circ) = -\tan 45^\circ = -1$

Answer: $-1$

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

$\Rightarrow -5 \le 5\sin x \le 5$

$\Rightarrow -1 \le 4 + 5\sin x \le 9$

Least value $= -1$ (when $\sin x = -1$)

Answer: $-1$

$\cos^2 10^\circ + \cos^2 80^\circ = \cos^2 10^\circ + \sin^2 10^\circ = 1$

The 8 terms make 4 pairs: $(10^\circ, 80^\circ), (20^\circ, 70^\circ), (30^\circ, 60^\circ), (40^\circ, 50^\circ)$

G.E $= 4(1) = 4$

Answer: $4$

$\sin^6\theta + \cos^6\theta = (\sin^2\theta + \cos^2\theta)^3 - 3\sin^2\theta\cos^2\theta(\sin^2\theta + \cos^2\theta) = 1 - 3(\sin\theta\cos\theta)^2$

$\displaystyle = 1 - 3\left(\frac{2}{5}\right)^2 = 1 - \frac{12}{25} = \frac{13}{25}$

Answer: $\displaystyle \frac{13}{25}$

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

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

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

$\displaystyle = \frac{7}{25} \cdot \frac{3}{5} - \frac{24}{25} \cdot \frac{4}{5} = \frac{21}{125} - \frac{96}{125} = -\frac{3}{5}$

Answer: $\displaystyle -\frac{3}{5}$

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

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

$\displaystyle \cos(A - B) = \frac{15}{17}$ and $\displaystyle A - B < 0 \Rightarrow \sin(A - B) = -\frac{8}{17}$, $\displaystyle \tan(A - B) = -\frac{8}{15}$

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

$\displaystyle = \frac{\frac{20}{21} - \frac{8}{15}}{1 + \frac{20}{21} \cdot \frac{8}{15}} = \frac{\frac{44}{105}}{\frac{95}{63}}$

$\displaystyle = \frac{132}{475}$

Answer: $\displaystyle \frac{132}{475}$

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

$1 + 2 + \tan C = 2\tan C$

$(2 - 1)\tan C = 3 \Rightarrow \tan C = 3$

Answer: $3$

$26^\circ + 19^\circ = 45^\circ$

If $A + B = 45^\circ$ then $(1 - \cot A)(1 - \cot B) = 2$ (replace $\tan$ by $\cot$ in the result)

Answer: $2$

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

$\displaystyle \cos 2A = 2\cos^2 A - 1 = 2 \cdot \frac{225}{289} - 1 = \frac{161}{289}$

$\displaystyle \sin 2A = 2\sin A\cos A = 2 \cdot \frac{8}{17} \cdot \frac{15}{17} = \frac{240}{289}$

Answer: $\displaystyle \cos 2A = \frac{161}{289}$, $\displaystyle \sin 2A = \frac{240}{289}$

Given $\displaystyle \tan\alpha = -\frac{8}{15}$ and $\alpha \in \text{IV}_Q$

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

$\displaystyle \tan 2\alpha = \frac{2\tan\alpha}{1 - \tan^2\alpha} = \frac{-\frac{16}{15}}{1 - \frac{64}{225}} = \frac{-\frac{16}{15}}{\frac{161}{225}} = -\frac{240}{161}$

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

Answer: $\displaystyle \tan 2\alpha = -\frac{240}{161}$, $\displaystyle \cos 2\alpha = \frac{161}{289}$

$\displaystyle \tan^2\frac{A}{2} = \frac{1 - \cos A}{1 + \cos A}$ with $A = 135^\circ$, $\displaystyle \cos 135^\circ = -\frac{1}{\sqrt{2}}$

$\displaystyle \tan^2 67\tfrac{1}{2}^\circ = \frac{1 + \frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} = (\sqrt{2} + 1)^2$

$\displaystyle \tan 67\tfrac{1}{2}^\circ = \sqrt{2} + 1$ (positive, as $\displaystyle 67\tfrac{1}{2}^\circ$ is acute)

Answer: $\sqrt{2} + 1$

$\displaystyle \cos\theta\cos 2\theta\cos 4\theta\cos 8\theta = \frac{\sin 16\theta}{16\sin\theta}$ with $\theta = 12^\circ$

G.E $\displaystyle = \frac{\sin 192^\circ}{16\sin 12^\circ}$

$\sin 192^\circ = \sin(180^\circ + 12^\circ) = -\sin 12^\circ$

G.E $\displaystyle = -\frac{1}{16}$

Answer: $\displaystyle -\frac{1}{16}$

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

LHS $\displaystyle = \frac{3}{2} + \frac{1}{2}[\cos 2\theta + \cos(120^\circ + 2\theta) + \cos(120^\circ - 2\theta)]$

$\displaystyle = \frac{3}{2} + \frac{1}{2}[\cos 2\theta + 2\cos 120^\circ\cos 2\theta] = \frac{3}{2} + \frac{1}{2}[\cos 2\theta - \cos 2\theta] = \frac{3}{2}$ = RHS

Put $\theta = 20^\circ$: $\displaystyle \cos^2 20^\circ + \cos^2 80^\circ + \cos^2 40^\circ = \frac{3}{2}$

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

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

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

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

= RHS

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

Period of $\displaystyle \cos\frac{2x}{3} = \frac{2\pi}{2/3} = 3\pi$, period of $\displaystyle \sin\frac{x}{2} = 4\pi$

Period $=$ L.C.M$\{3\pi, 4\pi\} = 12\pi$

Answer: $12\pi$