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Revision Ex 12
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(a)(i)
\begin{align} a^2 - x^2 & = a^2 - (a \sin \theta)^2 \\ & = a^2 - a^2 \sin^2 \theta \\ & = a^2 (1 - \sin^2 \theta) \\ & = a^2 (\cos^2 \theta) \phantom{00000} [\text{Since } \sin^2 A + \cos^2 A = 1 \phantom{0} \rightarrow \phantom{0} \cos^2 A = 1 - \sin^2 A] \\ & = a^2 \cos^2 \theta \end{align}
(a)(ii)
\begin{align} \require{cancel} \left(1 - {x^2 \over a^2} \right)^{3 \over 2} & = \left[ 1 - {(a \sin \theta)^2 \over a^2 } \right]^{3 \over 2} \\ & = \left[ 1 - { \cancel{a^2} \sin^2 \theta \over \cancel{a^2} } \right]^{3 \over 2} \\ & = (1 - \sin^2 \theta )^{3 \over 2} \\ & = (\cos^2 \theta)^{3 \over 2} \phantom{0000000} [\text{Since } \sin^2 A + \cos^2 A = 1 \phantom{0} \rightarrow \phantom{0} \cos^2 A = 1 - \sin^2 A] \\ & = [ (\cos \theta)^2 ]^{3 \over 2} \\ & = (\cos \theta)^{2\left(3 \over 2\right)} \phantom{00000.} [ (a^m)^n = a^{mn} ] \\ & = (\cos \theta)^3 \\ & = \cos^3 \theta \end{align}
(b)(i)
\begin{align} \sqrt{2}\cos \theta & = -\sin \theta \\ \sqrt{2} & = -{\sin \theta \over \cos \theta} \\ \sqrt{2} & = -\tan \theta \\ -\sqrt{2} & = \tan \theta \end{align}
Since $\theta$ is obtuse $(90^\circ < \theta < 180^\circ)$, the angle lies in the second quadrant.
\begin{align} \tan \theta & = {Opp \over Adj} \\ & = {\sqrt{2} \over - 1} \end{align}
\begin{align} \text{By Pyth} & \text{agora's theorem,} \\ x^2 & = (\sqrt{2})^2 + (-1)^2 \\ & = 3 \\ x & = \pm \sqrt{3} \\ & = \sqrt{3} \text{ or } -\sqrt{3} \text{ (Reject)} \\ \\ \tan (- \theta) & = -\tan \theta \\ & = - (-\sqrt{2}) \\ & = \sqrt{2} \end{align}
(b)(ii)
\begin{align} \sec \theta & = {1 \over \cos \theta} \\ & = {1 \over -{1 \over \sqrt{3}}} \\ & = 1 \div -{1 \over \sqrt{3}} \\ & = 1 \times -{\sqrt{3} \over 1} \\ & = -\sqrt{3} \end{align}
(b)(iii)
\begin{align} \text{cosec } (90^\circ - \theta) & = {1 \over \sin (90^\circ - \theta)} \\ & = {1 \over \cos \theta} \\ & = {1 \over -{1 \over \sqrt{3}}} \\ & = 1 \div -{1 \over \sqrt{3}} \\ & = 1 \times -{\sqrt{3} \over 1} \\ & = -\sqrt{3} \end{align}
(a)
\begin{align} 5\sin^2 x - 8\sin x \cos x & = 0 \\ (\sin x)(5 \sin x - 8\cos x) & = 0 \\ \\ \sin x = 0 \phantom{00}&\text{or}\phantom{00} 5\sin x - 8\cos x = 0 \end{align}
$$ \sin x = 0 $$
$$ x = 0^\circ, 180^\circ, 360^\circ $$
\begin{align} 5\sin x - 8\cos x & = 0 \\ 5\sin x & = 8\cos x \\ {5\sin x \over \cos x} & = 8 \\ {\sin x \over \cos x} & = {8 \over 5} \\ \tan x & = 1.6 \phantom{00000000} [\text{1st & 3rd quadrants since } \tan x > 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (1.6) \\ & = 57.99^\circ \end{align}
\begin{align} x & = 57.99^\circ, 180^\circ + 57.99^\circ \\ & = 57.99^\circ, 237.99^\circ \\ & \approx 58.0^\circ, 238.0^\circ \\ \\ \\ \therefore x & = 0^\circ, 58.0^\circ, 180^\circ, 238.0^\circ, 360^\circ \end{align}
(b)
\begin{align} 5\cot^2 x + 7 & = 11 \text{cosec } x \\ 5(\text{cosec}^2 x - 1) + 7 & = 11 \text{cosec }x \phantom{00000} [\text{Since } 1 + \cot^2 x = \text{cosec}^2 x \phantom{0} \rightarrow \phantom{0} \cot^2 x = \text{cosec}^2 x - 1] \\ 5\text{cosec}^2 x - 5 + 7 & = 11 \text{cosec } x \\ 5\text{cosec}^2 x - 11 \text{cosec } x + 2 & = 0 \\ (5\text{cosec } x - 1)(\text{cosec } x - 2) & = 0 \\ \\ 5\text{cosec } x - 1 = 0 \phantom{00}&\text{or}\phantom{00} \text{cosec } x - 2 = 0 \end{align}
\begin{align} 5\text{cosec } x - 1 & = 0 \\ 5\text{cosec } x & = 1 \\ \text{cosec } x & = {1 \over 5} \\ {1 \over \sin x} & = {1 \over 5} \\ \\ \sin x & = 5 \text{ (Reject, since } \sin x \le 1) \end{align}
\begin{align} \text{cosec } x - 2 & = 0 \\ \text{cosec } x & = 2 \\ {1 \over \sin x} & = 2 \\ 1 & = 2\sin x \\ \\ \sin x & = {1 \over 2} \phantom{00000000} \left[\text{1st & 2nd quadrants since } \sin x > 0 \right] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(1 \over 2\right) \\ & = 30^\circ \end{align}
\begin{align} x & = 30^\circ, 180 - 30^\circ \\ & = 30^\circ, 150^\circ \end{align}
(c)
\begin{align} 1 + 2 \cos \left({3 \over 2}x + 75^\circ \right) & = 0 \\ 2 \cos \left({3 \over 2}x + 75^\circ \right) & = -1 \\ \cos \left({3 \over 2}x + 75^\circ \right) & = -{1 \over 2} \phantom{000000} \left[ \text{2nd & 3rd quadrants since } \cos \left({3 \over 2}x + 75^\circ\right) < 0 \right] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\ & = 60^\circ \end{align}
\begin{align} \text{Since } 0^\circ \le x \le 360^\circ & \rightarrow \phantom{.} 0^\circ \le {3 \over 2}x \le 540^\circ \rightarrow \phantom{.} 75^\circ \le {3 \over 2}x + 75^\circ \le 615^\circ \\ \\ {3 \over 2}x + 75^\circ & = 180^\circ - 60^\circ, 180^\circ + 60^\circ \\ & = 120^\circ, 240^\circ \\ & = 120^\circ, 240^\circ, 120^\circ + 360^\circ, 240^\circ + 360^\circ \phantom{00000} \left[\text{Since }75^\circ \le {3 \over 2}x + 75^\circ \le 615^\circ \right] \\ & = 120^\circ, 240^\circ, 480^\circ, 600^\circ \\ \\ {3 \over 2}x & = 45^\circ, 165^\circ, 405^\circ, 525^\circ \\ \\ x & = 30^\circ, 110^\circ, 270^\circ, 350^\circ \end{align}
(a)
\begin{align} 1 + 2 \cos \left({3 \over 2}x + 75^\circ \right) & = 0 \\ 2 \cos \left({3 \over 2}x + 75^\circ \right) & = -1 \\ \cos \left({3 \over 2}x + 75^\circ \right) & = -{1 \over 2} \phantom{000000} \left[ \text{2nd & 3rd quadrants since } \cos \left({3 \over 2}x + 75^\circ\right) < 0 \right] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\ & = 60^\circ \end{align}
\begin{align} \text{Since } x > 6 & \rightarrow \phantom{.} 3x > 18 \\ \\ 3x & = \pi - 0.6435, \pi + 0.6435 \\ & = 2.4980 \text{ (Rej)}, 3.7850 \text{ (Rej)} \\ & = 2.4980 + 2\pi, 3.7850 + 2\pi \phantom{00000} [\text{Since } 3x > 18] \\ & = 8.7811 \text{ (Rej)}, 10.0681 \text{ (Rej)} \\ & = 8.7811 + 2\pi, 10.0681 + 2\pi \\ & = 15.0642 \text{ (Rej)}, 16.351 \text{ (Rej)} \\ & = 15.0642 + 2\pi, 16.351 + 2\pi \\ & = 21.3473, 22.6341 \\ \\ x & = 7.1157, 7.5447 \\ & \approx 7.12, 7.54 \\ \\ \therefore \text{Least value of } x & = 7.12 \end{align}
(b)
\begin{align} \cos (2x - 70^\circ) & = \sin 200^\circ \\ \cos (2x - 70^\circ) & = -0.34202 \phantom{000000} [\text{2nd & 3rd quadrants since } \cos (2x - 70^\circ) < 0] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.34202) \\ & = 70^\circ \end{align}
\begin{align} \text{Since } x < 800^\circ & \rightarrow \phantom{.} 2x < 1600^\circ \rightarrow \phantom{.} 2x - 70^\circ < 1530^\circ \\ \\ 2x - 70^\circ & = 180^\circ - 70^\circ, 180^\circ + 70^\circ \\ & = 110^\circ, 250^\circ, 110^\circ + 360^\circ, 250^\circ + 360^\circ \phantom{000000} [\text{Since } 2x - 70^\circ < 1530^\circ] \\ & = 110^\circ, 250^\circ, 470^\circ, 610^\circ, 470^\circ + 360^\circ, 610^\circ + 360^\circ \\ & = 110^\circ, 250^\circ, 470^\circ, 610^\circ, 830^\circ, 970^\circ, 830^\circ + 360^\circ, 970^\circ + 360^\circ \\ & = 110^\circ, 250^\circ, 470^\circ, 610^\circ, 830^\circ, 970^\circ, 1190^\circ, 1330^\circ \\ \\ 2x & = 180^\circ, 320^\circ, 540^\circ, 680^\circ, 900^\circ, 1040^\circ, 1260^\circ, 1400^\circ \\ \\ x & = 90^\circ, 160^\circ, 270^\circ, 340^\circ, 450^\circ, 520^\circ, 630^\circ, 700^\circ \\ \\ \therefore \text{Largest value of } x & = 700^\circ \end{align}
(c)
\begin{align} 3\sin \left( {3x - 2 \over 4} \right) & = - 2 \\ \sin \left(3x - 2 \over 4\right) & = -{2 \over 3} \phantom{000000} \left[ \text{3rd & 4th quadrants since } \sin \left(3x - 2\over 4\right) < 0 \right] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(2 \over 3\right) \\ & = 0.7297 \end{align}
\begin{align} \text{Since } x < 20 & \rightarrow \phantom{.} 3x < 60 \rightarrow \phantom{.} 3x - 2 < 58 \rightarrow \phantom{.} {3x - 2 \over 4} < 14.5 \\ \\ {3x - 2 \over 4} & = \pi + 0.7297, 2\pi - 0.7297 \\ & = 3.8712, 5.5534, 3.8712 + 2\pi, 5.5534 + 2\pi \phantom{00000} \left[ \text{Since } {3x - 2 \over 4} < 14.5 \right] \\ & = 3.8712, 5.5534, 10.1543, 11.8365 \\ \\ 3x - 2 & = 15.4848, 22.2136, 40.6172, 47.346 \\ \\ 3x & = 17.4848, 24.2136, 42.6172, 49.346 \\ \\ x & = 5.828, 8.071, 14.205, 16.448 \\ & \approx 5.83, 8.07, 14.2, 16.4 \\ \\ \therefore \text{Largest value of } x & = 16.4 \end{align}
(a)
\begin{align} \text{L.H.S} & = {1 - \text{cosec } x \over \cot x} \\ & = (1 - \text{cosec } x) \div \cot x \\ & = \left(1 - {1 \over \sin x} \right) \div {\cos x \over \sin x} \\ & = \left(1 - {1 \over \sin x} \right) \times {\sin x \over \cos x} \\ & = {\sin x \over \cos x} - {1 \over \sin x} \left(\sin x \over \cos x\right) \\ & = \tan x - {1 \over \cos x} \\ & = \tan x - \sec x \\ & = \text{R.H.S} \end{align}
(b)
\begin{align} \text{R.H.S} & = {1 - \cot x \over 1 + \cot x} \\ & = {1 - {1 \over \tan x} \over 1 + {1 \over \tan x}} \times {\tan x \over \tan x} \\ & = {\tan x \left(1 - {1 \over \tan x}\right) \over \tan x \left(1 + {1 \over \tan x} \right)} \\ & = {\tan x - 1 \over \tan x + 1} \\ & = \text{L.H.S} \end{align}
(c)
\begin{align} \text{L.H.S} & = \ln (\sec x + \tan x) + \ln (\sec x - \tan x) \\ & = \ln [ (\sec x + \tan x)(\sec x - \tan x)] \phantom{00000} [\text{Product law (logarithms)}] \\ & = \ln [ (\sec x)^2 - (\tan x)^2 ] \\ & = \ln (\sec^2 x - \tan^2 x ) \\ & = \ln [(1 + \tan^2 x) - \tan^2 x] \phantom{0000000000.} [\text{Identity: } 1 + \tan^2 A = \sec^2 A] \\ & = \ln (1) \\ & = 0 \\ & = \text{R.H.S} \end{align}
(i)
\begin{align} \text{L.H.S} & = \sec \theta \text{ cosec } \theta - \cot \theta \\ & = {1 \over \cos \theta} \left(1 \over \sin \theta\right) - {\cos \theta \over \sin \theta} \\ & = {1 \over \sin \theta \cos \theta} - {\cos \theta (\cos \theta) \over \sin \theta \cos \theta} \\ & = {1 - \cos^2 \theta \over \sin \theta \cos \theta} \\ & = {\sin^2 \theta \over \sin \theta \cos \theta} \phantom{00000} [\text{Since } \sin^2 A + \cos^2 A = 1 \phantom{0} \rightarrow \sin^2 A = 1 - \cos^2 A] \\ & = {\sin \theta \over \cos \theta} \\ & = \tan \theta \\ & = \text{R.H.S} \end{align}
(ii)
\begin{align} \sec 2x \text{ cosec } 2x - \cot 2x & = 1 \phantom{0000000} [\text{Use result from part i}] \\ \tan 2x & = 1 \phantom{0000000} [\text{1st & 3rd quadrants since } \tan 2x > 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (1) \\ & = 45^\circ \end{align}
\begin{align} \text{Since } 0^\circ < x < 360^\circ & \rightarrow \phantom{.} 0^\circ < 2x < 720^\circ \\ \\ 2x & = 45^\circ, 180^\circ + 45^\circ \\ & = 45^\circ, 225^\circ, 45^\circ + 360^\circ, 225^\circ + 360^\circ \phantom{00000} [\text{Since } 0^\circ < 2x < 720^\circ] \\ & = 45^\circ, 225^\circ, 405^\circ, 585^\circ \\ \\ x & = 22.5^\circ, 112.5^\circ, 202.5^\circ, 292.5^\circ \end{align}
(i)
\begin{align} y & = 4\cos 3x \\ \\ \text{Comparing with } & y = a\cos bx + c, \\ a & = 4 \\ b & = 3 \\ c & = 0 \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Maximum} & = 4 \\ \\ \text{Since the graph of } y = 4\cos 3x & \text{ starts from the maximum point,} \\ \\ \therefore P & (0^\circ, 4) \end{align}
\begin{align} y & = 4 \cos 3x \\ \\ \text{When } & y = 0, \\ 0 & = 4\cos 3x \\ 0 & = \cos 3x \end{align}
\begin{align} 3x & = 90^\circ, 270^\circ \\ x & = 30^\circ, 90^\circ \\ \\ \therefore x & \text{-intercept is } (30^\circ, 0). \end{align}
\begin{align} y & = 2 \sin x + k \\ \\ \text{Using } (30^\circ, 0), \text{when } & x = 30^\circ \text{ and } y = 0, \\ 0 & = 2\sin 30^\circ + k \\ 0 & = 2\left(1 \over 2\right) + k \\ 0 & = 1 + k \\ -1 & = k \\ \\ \therefore y & = 2\sin x - 1 \\ \\ \\ \text{Comparing with } & y = a\sin bx + c, \\ a & = 2 \\ b & = 1 \\ c & = - 1 \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Maximum} & = 2 - 1 \\ & = 1 \\ \\ \\ \text{When } & y = 1, \\ 1 & = 2\sin x - 1 \\ 2 & = 2\sin x \\ 1 & = \sin x \end{align}
$$ x = 90^\circ $$
$$ \therefore Q(90^\circ, 1) $$
(ii)
\begin{align} y & = 4 \cos 3x \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Maximum} & = 4 \\ \text{Minimum} & = -4 \\ \\ \text{Period} & = {360^\circ \over b} \\ & = {360^\circ \over 3} \\ & = 120^\circ \\ \\ \text{No. of cycles} & = {180^\circ \over 120^\circ} \\ & = 1{1 \over 2} \end{align}
\begin{align} y & = 2 \sin x - 1 \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Center line: } & y = -1 \\ \\ \text{Maximum} & = -1 + 2 = 1 \\ \text{Minimum} & = -1 - 2 = -3 \\ \\ \text{Period} & = {360^\circ \over b} \\ & = {360^\circ \over 1} \\ & = 360^\circ \\ \\ \text{No. of cycles} & = {180^\circ \over 360^\circ} \\ & = {1 \over 2} \end{align}
\begin{align} P & = 105 - 20\cos \left(16\pi t \over 7\right) \\ \\ \text{When } & P = 100, \\ 100 & = 105 - 20\cos \left(16\pi t \over 7\right) \\ 20\cos \left(16\pi t \over 7\right) & = 105 - 100 \\ 20\cos \left(16\pi t \over 7\right) & = 5 \\ \cos \left(16\pi t \over 7\right) & = {5 \over 20} \\ \cos \left(16\pi t \over 7\right) & = 0.25 \phantom{00000} \left[ \text{1st & 4th quadrants since } \cos \left(16\pi t \over 7\right) > 0 \right] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.25) \\ & = 1.3181 \end{align}
\begin{align} \text{Since } t > 10 & \rightarrow \phantom{.} 16\pi t > 160\pi \rightarrow \phantom{.} {16\pi t \over 7} > {160\pi \over 7} \rightarrow \phantom{.} {16\pi t \over 7} > 71.807 \\ \\ {16\pi t \over 7} & = 1.3181, 2\pi - 1.3181 \\ & = 1.3181, 4.9650, 1.3181 + 2\pi, 4.9650 + 2\pi, 1.3181 + 4\pi, 4.9650 + 4\pi, ..., 1.381 + 22\pi, 4.9650 + 22\pi \\ & = \phantom{.}..., 70.253 \text{ (Reject)}, 74.080 \phantom{00000} \left[ \text{Since } {16\pi t \over 7} > 71.807 \right] \\ & = 74.080 \\ \\ 16\pi t & = 518.560 \\ \\ t & = {518.560 \over 16\pi} \\ & = 10.316 \\ & \approx 10.3 \end{align}
(a)
\begin{align} \sec 2x & = 2 \\ {1 \over \cos 2x} & = 2 \\ 1 & = 2\cos 2x \\ \\ \cos 2x & = {1 \over 2} \phantom{00000} \left[ \text{1st & 4th quadrants since } \cos 2x > 0 \right] \\ \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\ & = 60^\circ \\ & = {\pi \over 3} \end{align}
\begin{align} \text{Since } 0 < x < 2\pi & \rightarrow \phantom{.} 0 < 2x < 4\pi \\ \\ 2x & = {\pi \over 3}, 2\pi - {\pi \over 3} \\ & = {\pi \over 3}, {5\pi \over 3}, {\pi \over 3} + 2\pi, {5\pi \over 3} + 2\pi \phantom{00000} [\text{Since } 0 < 2x < 4\pi ] \\ & = {\pi \over 3}, {5\pi \over 3}, {7\pi \over 3}, {11\pi \over 3} \\ \\ x & = {\pi \over 6}, {5\pi \over 6}, {7\pi \over 6}, {11\pi \over 6} \end{align}
(b)
\begin{align} 4\sin x \cos x & = \tan x \\ 4\sin x \cos x & = {\sin x \over \cos x} \\ 4\sin x \cos x(\cos x) & = \sin x \\ 4\sin x \cos^2 x - \sin x & = 0 \\ \sin x (4\cos^2 x - 1) & = 0 \\ \\ \sin x = 0 \phantom{00}&\text{or}\phantom{00} 4\cos^2 x - 1 = 0 \end{align}
\begin{align} \sin x & = 0 \end{align}
$$ x = 0, \pi, 2\pi $$
\begin{align} 4\cos^2 x - 1 & = 0 \\ 4\cos^2 x & = 1 \\ \cos^2 x & = {1 \over 4} \\ \cos x & = \pm \sqrt{1 \over 4} \\ \cos x & = \pm {1 \over 2} \phantom{00000} \left[\text{All 4 quadrants} \right] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\ & = 60^\circ \\ & = {\pi \over 3} \end{align}
\begin{align} x & = {\pi \over 3}, \pi - {\pi \over 3}, \pi + {\pi \over 3}, 2\pi - {\pi \over 3} \\ & = {\pi \over 3}, {2\pi \over 3}, {4\pi \over 3}, {5\pi \over 3} \\ \\ \\ \therefore x & = 0, {\pi \over 3}, {2\pi \over 3}, \pi, {4\pi \over 3}, {5\pi \over 3}, 2\pi \end{align}
(c)
\begin{align} (\tan x + 1)(2\tan x - 1) & = 5 \\ 2\tan^2 x - \tan x + 2\tan x - 1 & = 5 \\ 2\tan^2 x + \tan x - 6 & = 0 \\ (2\tan x - 3)(\tan x + 2) & = 0 \\ \\ 2\tan x - 3 = 0 \phantom{00}&\text{or}\phantom{00} \tan x + 2 = 0 \end{align}
\begin{align} 2\tan x - 3 & = 0 \\ 2\tan x & = 3 \\ \tan x & = {3 \over 2} \\ \tan x & = 1.5 \phantom{00000} [\text{1st & 3rd quadrants since } \tan x > 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (1.5) \\ & = 0.9827 \end{align}
\begin{align} x & = 0.9827, \pi + 0.9827 \\ & = 0.9827, 4.1242 \\ & \approx 0.983, 4.12 \end{align}
\begin{align} \tan x + 2 & = 0 \\ \tan x & = -2 \phantom{00000} [\text{2nd & 4th quadrants since } \tan x < 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (2) \\ & = 1.1071 \end{align}
\begin{align} x & = \pi - 1.1071, 2\pi - 1.1071 \\ & = 2.0344, 5.1760 \\ & \approx 2.03, 5.18 \\ \\ \\ \therefore x & = 0.983, 2.03, 4.12, 5.18 \end{align}
(i)
\begin{align} y & = 2x \\ 4\cos \theta - 3\sin \theta & = 2(4\sin \theta + 3\cos \theta) \\ 4\cos \theta - 3\sin \theta & = 8\sin \theta + 6\cos \theta \\ -3\sin \theta - 8\sin \theta & = 6\cos \theta - 4\cos \theta \\ -11\sin \theta & = 2\cos \theta \\ {-11\sin \theta \over \cos \theta} & = 2 \\ {\sin \theta \over \cos \theta} & = -{2 \over 11} \\ \tan \theta & = -{2 \over 11} \phantom{00000} \left[ \text{2nd & 4th quadrants since } \tan \theta < 0 \right] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left(2 \over 11\right) \\ & = 10.30^\circ \end{align}
\begin{align} \theta & = 180^\circ - 10.30^\circ, 360^\circ - 10.30^\circ \\ & = 169.7^\circ, 349.7^\circ \\ \\ \therefore \text{Required obtuse angle} & = 169.7^\circ \end{align}
(ii)
\begin{align} x^2 + y^2 & = (4\sin \theta + 3\cos \theta)^2 + (4\cos \theta - 3\sin \theta)^2 \\ & = (4\sin \theta)^2 + 2(4\sin \theta)(3\cos \theta) + (3\cos \theta)^2 + (4\cos \theta)^2 - 2(4\cos \theta)(3\sin \theta) + (3\sin \theta)^2 \\ & = 16\sin^2 \theta + 24\sin \theta \cos \theta + 9\cos^2 \theta + 16\cos^2 \theta - 24\sin \theta \cos \theta + 9\sin^2 \theta \\ & = 25\sin^2 \theta + 25\cos^2 \theta \\ & = 25(\sin^2 \theta + \cos^2 \theta) \\ & = 25(1) \phantom{0000000000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = 25 \text{ (Shown)} \end{align}
(a)
\begin{align} \text{L.H.S} & = {1 - \tan^2 x \over 1 + \tan^2 x} \\ & = {1 - \tan^2 x \over \sec^2 x} \phantom{0000000000} [\text{Identity: } 1 + \tan^2 A = \sec^2 A] \\ & = {1 - (\sec^2 x - 1) \over \sec^2 x} \phantom{00000} [\text{Since } 1 + \tan^2 A = \sec^2 A \phantom{0} \rightarrow \tan^2 A = \sec^2 A - 1 ] \\ & = {1 - \sec^2 x + 1 \over \sec^2 x} \\ & = {2 - \sec^2 x \over \sec^2 x} \\ & = {2 \over \sec^2 x} - {\sec^2 x \over \sec^2 x} \\ & = {2 \over {1 \over \cos^2 x}} - 1 \\ & = 2\cos^2 x - 1 \\ & = \text{R.H.S} \end{align}
(b)
\begin{align} \require{cancel} \text{L.H.S} & = {\cos x \over 1 - \sin x} - {1 \over \cos x} \\ & = {\cos x (\cos x) \over (1 - \sin x)(\cos x)} - {(1 - \sin x) \over \cos x(1 - \sin x)} \\ & = {\cos^2 x - (1 - \sin x) \over \cos x(1 - \sin x)} \\ & = {\cos^2 x - 1 + \sin x \over \cos x(1 - \sin x)} \\ & = {(1 - \sin^ 2 x) - 1 + \sin x \over \cos x(1 - \sin x)} \phantom{00000} [\text{Since } \sin^2 A + \cos^2 A = 1 \phantom{0} \rightarrow \cos^2 A = 1 - \sin^2 A ] \\ & = {1 - \sin^2 x - 1 + \sin x \over \cos x(1 - \sin x)} \\ & = {\sin x - \sin^2 x \over \cos x(1 - \sin x)} \\ & = {\sin x \cancel{(1 - \sin x)} \over \cos x \cancel{(1 - \sin x)} } \\ & = {\sin x \over \cos x} \\ & = \tan x \\ & = \text{R.H.S} \end{align}
(c)
\begin{align} \text{L.H.S} & = \ln (1 + \cos x) + \ln (1 - \cos x) \\ & = \ln [(1 + \cos x)(1 - \cos x)] \phantom{00000} [\text{Product law (logarithms)]} \\ & = \ln (1 - \cos^2 x) \\ & = \ln (\sin^2 x) \phantom{00000000000000000.} [\text{Since } \sin^2 A + \cos^2 A = 1 \phantom{0} \rightarrow \sin^2 A = 1 - \cos^2 A] \\ & = \ln (\sin x)^2 \\ & = 2\ln (\sin x) \phantom{0000000000000000..} [\text{Power law (logarithms)}] \\ & = \text{R.H.S} \end{align}
(a)
\begin{align} \tan {5x \over 2} & = 0.6 \phantom{00000} \left[ \text{1st & 3rd quadrants since } \tan {5x \over 2} > 0 \right] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (0.6) \\ & = 0.5404 \end{align}
\begin{align} \text{Since } x > 4 & \rightarrow \phantom{.} 5x > 20 \rightarrow \phantom{.} {5x \over 2} > 10 \\ \\ {5x \over 2} & = 0.5404, \pi + 0.5404 \\ & = 0.5404 \text{ (Rej)}, 3.6819 \text{ (Rej)} \phantom{00000} \left[\text{Since } {5x \over 2} > 10 \right] \\ & = 0.5404 + 2\pi, 3.6819 + 2\pi \\ & = 6.8235 \text{ (Rej)}, 9.9650 \text{ (Rej)} \\ & = 6.8235 + 2\pi \\ & = 13.1067 \\ \\ 5x & = 13.1067 \times 2 \\ & = 26.2134 \\ \\ x & = 5.24268 \\ & \approx 5.24 \end{align}
(b)
\begin{align} \text{cosec } 2x & = 6 \\ {1 \over \sin 2x} & = 6 \\ 1 & = 6\sin 2x \\ \\ \sin 2x & = {1 \over 6} \phantom{00000} \left[ \text{1st & 2nd quadrants since } \sin 2x > 0 \right] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(1 \over 6\right) \\ & = 0.1674 \end{align}
\begin{align} \text{Since } x < 5 & \rightarrow \phantom{.} 2x < 10 \\ \\ 2x & = 0.1674, \pi - 0.1674 \\ & = 0.1674, 2.9741, 0.1674 + 2\pi, 2.9741 + 2\pi \phantom{00000} [\text{Since } 2x < 10] \\ & = 0.1674, 2.9741, 6.4505, 9.2572 \\ \\ x & = 0.0837, 1.48705, 3.22525, 4.6286 \\ \\ \\ \therefore \text{Greatest value of } x & = 4.6286 \\ & \approx 4.63 \end{align}
(c)
\begin{align} 3\sec \left(2x \over 3\right) + 4 & = 0 \\ 3\sec \left(2x \over 3\right) & = -4 \\ \sec \left(2x \over 3\right) & = -{4 \over 3} \\ {1 \over \cos \left(2x \over 3\right) } & = -{4 \over 3} \\ -3 & = 4\cos \left(2x \over 3\right) \\ -{3 \over 4} & = \cos \left(2x \over 3\right) \\ \\ \cos \left(2x \over 3\right) & = -0.75 \phantom{00000} \left[ \text{2nd & 3rd quadrants since } \cos \left(2x \over 3\right)< 0 \right] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.75) \\ & = 41.40^\circ \end{align}
\begin{align} \text{Since } x > 1000^\circ & \rightarrow \phantom{.} 2x > 2000^\circ \rightarrow \phantom{.} {2x \over 3} > 666.6667^\circ \\ \\ {2x \over 3} & = 180^\circ - 41.40^\circ, 180^\circ + 41.40^\circ \\ & = 138.6^\circ, 221.4^\circ, 138.6^\circ + 360^\circ, 221.4^\circ + 360^\circ, 138.6^\circ + 2(360^\circ) \\ & = 138.6^\circ \text{ (Rej)}, 221.4^\circ \text{ (Rej)}, 498.6^\circ \text{ (Rej)}, 581.4^\circ \text{ (Rej)}, 858.6^\circ \phantom{00000} \left[ {2x \over 3} > 666.6667^\circ \right] \\ \\ 2x & = 858.6^\circ \times 3 \\ & = 2575.8^\circ \\ \\ x & = 1287.9^\circ \end{align}
(i)
\begin{align} \text{Let } f(x) & = 6x^3 + 11x^2 - 3x - 2 \\ \\ f(-2) & = 6(-2)^3 + 11(-2)^2 - 3(-2) - 2 \\ & = 0 \\ \\ \therefore \text{By Factor} & \text{ theorem, } x + 2 \text{ is a factor} \end{align}
$$ \require{enclose} \begin{array}{rll} 6x^2 - x - 1\phantom{00000000}\\ x + 2 \enclose{longdiv}{6x^3 + 11x^2 - 3x - 2\phantom{0}}\kern-.2ex \\ -\underline{(6x^3 + 12x^2){\phantom{00000000}}} \\ -x^2 - 3x - 2\phantom{00} \\ -\underline{(- x^2 - 2x){\phantom{0000}}} \\ - x - 2\phantom{0.} \\ -\underline{(-x - 2)}\phantom{.} \\ 0\phantom{0} \end{array} $$
\begin{align} \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \\ \\ f(x) & = (x + 2)(6x^2 - x - 1) \\ & = (x + 2)(2x - 1)(3x + 1) \\ \\ \\ \therefore 0 & = (x + 2)(2x - 1)(3x + 1) \\ \\ x + 2 = 0 \phantom{00000} 2x - 1 & = 0 \phantom{00000} 3x + 1 = 0 \\ x = - 2 \phantom{0000000} 2x & = 1 \phantom{00000+1} 3x = - 1 \\ x & = {1 \over 2} \phantom{00000+.2} x = -{1 \over 3} \end{align}
(ii)
\begin{align} 6\tan^2 \theta + 11\tan \theta - 3 & = 2 \cot \theta \\ 6\tan^2 \theta + 11\tan \theta - 3 & = 2 \left(1 \over \tan \theta\right) \\ 6\tan^3 \theta + 11\tan^2 \theta - 3\tan \theta & = 2 \\ 6\tan^3 \theta + 11\tan^2 \theta - 3\tan \theta - 2 & = 0 \\ \\ \\ \text{Let } & \tan \theta = x , \\ 6x^3 + 11x^2 - 3x - 2 & = 0 \phantom{00000} \text{ [Equation from part (i)]} \\ \\ \therefore x & = -2, {1 \over 2}, -{1 \over 3} \\ \\ \therefore \tan \theta & = -2, {1 \over 2}, -{1 \over 3} \end{align}
\begin{align} \tan \theta & = -2 \phantom{00000} [\text{2nd & 4th quadrants since } \tan \theta < 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (2) \\ & = 63.43^\circ \end{align}
\begin{align} \theta & = 180^\circ - 63.43^\circ, 360^\circ - 63.43^\circ \\ & = 116.57^\circ, 296.57^\circ \text{ (Reject)} \\ & \approx 116.6^\circ \end{align}
\begin{align} \tan \theta & = {1 \over 2} \phantom{00000} \left[ \text{1st & 3rd quadrants since } \tan \theta > 0 \right] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left(1 \over 2\right) \\ & = 26.56^\circ \end{align}
\begin{align} \theta & = 26.56^\circ, 180^\circ + 26.56^\circ \\ \\ & = 26.56^\circ, 206.56^\circ \text{ (Reject)} \\ & \approx 26.6^\circ \end{align}
\begin{align} \tan \theta & = -{1 \over 3} \phantom{00000} \left[ \text{2nd & 4th quadrants since } \tan \theta < 0 \right] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left(1 \over 3\right) \\ & = 18.43^\circ \end{align}
\begin{align} \theta & = 180^\circ - 18.43^\circ, 360^\circ - 18.43^\circ \\ & = 161.57^\circ, 341.57^\circ \text{ (Reject)} \\ & \approx 161.6^\circ \\ \\ \\ \therefore \theta & = 26.6^\circ, 116.6^\circ, 161.6^\circ \end{align}
(i)
\begin{align} \require{cancel} f(x) & = {1 \over \cos 2x} - {\cos 2x \over 1 + \sin 2x} \\ & = {1 + \sin 2x \over \cos 2x(1 + \sin 2x)} - {\cos 2x (\cos 2x) \over \cos 2x(1 + \sin 2x)} \\ & = {1 + \sin 2x - \cos^2 x \over \cos 2x(1 + \sin 2x)} \\ & = {1 + \sin 2x - (1 - \sin^2 2x) \over \cos 2x(1 + \sin 2x)} \phantom{00000} [\text{Since } \sin^2 A + \cos^2 A = 1 \phantom{0} \rightarrow \cos^2 A = 1 - \sin^2 A] \\ & = {1 + \sin 2x - 1 + \sin^2 2x \over \cos 2x(1 + \sin 2x)} \\ & = {\sin 2x + \sin^2 2x \over \cos 2x(1 + \sin 2x)} \\ & = {\sin 2x \cancel{(1 + \sin 2x)} \over \cos 2x \cancel{(1 + \sin 2x)} } \\ & = \tan 2x \text{ (Shown)} \end{align}
(ii)
\begin{align} y & = f(x) \\ y & = \tan 2x \\ \\ \text{Comparing with } & y = a\tan bx, \\ a & = 1 \\ b & = 2 \\ \\ \text{Period} & = {180^\circ \over b} \\ & = {180^\circ \over 2} \\ & = 90^\circ \\ \\ \text{No. of cycles} & = {270^\circ \over 90^\circ} \\ & = 3 \\ \\ \text{Asymptotes are } x = 45^\circ, x & = 135^\circ, x = 225^\circ \end{align}
(i)
\begin{align} v & = 1.6 \sin {\pi t \over 2} \\ & = 1.6 \sin \left( {\pi \over 2} t\right) \\ \\ \text{Comparing with } & v = a \sin bt + c \\ a & = 1.6 \\ b & = {\pi \over 2} \\ c & = 0 \\ \\ \text{Period} & = {2\pi \over b} \\ & = {2\pi \over {\pi \over 2}} \\ & = 4 \\ \\ \therefore \text{Time taken for one full cycle} & = 4 \text{ seconds} \end{align}
(ii)
\begin{align} \text{When } & v = 1.3, \\ 1.3 & = 1.6 \sin {\pi t \over 2} \\ {1.3 \over 1.6} & = \sin {\pi t \over 2} \\ \\ \sin {\pi t \over 2} & = {13 \over 16} \phantom{00000} \left[ \text{1st & 2nd quadrants since } \sin {\pi t \over 2} > 0 \right] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(13 \over 16\right) \\ & = 0.9484 \end{align}
\begin{align} {\pi t \over 2} & = 0.9484, \pi - 0.9484 \\ & = 0.9484, 2.1931 \\ \\ \pi t & = 1.8968, 4.3862 \\ \\ t & = 0.6037, 1.3961 \\ & \approx 0.604 \text{ s}, 1.40 \text{ s} \end{align}