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Revision Ex 13
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Solutions
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(a)
\begin{align} \cos 105^\circ & = \cos (60^\circ + 45^\circ) \\ & = \cos 45^\circ \cos 60^\circ - \sin 45^\circ \sin 60^\circ \phantom{000000} [\text{Addition formula: } \cos (A + B)] \\ & = \left( {1 \over \sqrt{2}} \right) \left( {1 \over 2} \right) - \left( 1 \over \sqrt{2} \right) \left( \sqrt{3} \over 2 \right) \phantom{00000} [\text{Special values}] \\ & = {1 \over 2\sqrt{2}} - {\sqrt{3} \over 2\sqrt{2}} \\ & = {1 - \sqrt{3} \over 2\sqrt{2}} \end{align}
(b)
\begin{align} \text{L.H.S } & = \cos \left( {3\pi \over 2} - 2x \right) \\ & = \cos {3\pi \over 2} \cos 2x + \sin {3\pi \over 2} \sin 2x \phantom{000000} [\text{Addition formula: } \cos (A - B)] \\ & = \cos 270^\circ \cos 2x + \sin 270^\circ \sin 2x \\ & = (0)\cos 2x + (-1)\sin 2x \\ & = - \sin 2x \\ & = \text{R.H.S} \end{align}
Sketch graph
\begin{align} y & = 2\left[ \cos \left( {3\pi \over 2} - 2x \right) \right] + 3 \\ y & = 2(-\sin 2x) + 3 \phantom{000000} [\text{Use result from part b}]\\ y & = -2\sin 2x + 3 \\ \\ \text{Comparing} & \text{ with } y = a \sin bx + c, \\ a & = - 2 \text{ (As } a < 0, \text{ shape of graph is inverted}) \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Center line: } & y = 3 \\ \\ \text{Maximum} & = 3 + 2 = 5 \\ \\ \text{Minimum} & = 3 - 2 = 1 \\ \\ \text{Period} & = {2\pi \over b} \\ & = {2\pi \over 2} \\ & = \pi \end{align}
(i)
\begin{align} \text{By Py} & \text{thagoras theorem,} \\ 5^2 & = x^2 + (-4)^2 \\ 25 & = x^2 + 16 \\ x^2 & = 9 \\ x & = \pm\sqrt{9} \\ x & = 3 \text{ or } - 3 \text{ (Reject)} \\ \\ \tan (\alpha - 45^\circ) & = {\tan \alpha - \tan 45^\circ \over 1 + \tan \alpha \tan 45^\circ} \phantom{000000} [\text{Addition formula: } \tan (A - B)] \\ & = {\left(-{3 \over 4}\right) - (1) \over 1 + \left(-{3 \over 4}\right)(1)} \\ & = -7 \end{align}
(ii)
\begin{align} \sin 2\alpha & = 2\sin \alpha \cos \alpha \phantom{000000} [\text{Double angle formula}] \\ & = 2\left(3 \over 5\right)\left(-{4 \over 5}\right) \\ & = -{24 \over 25} \end{align}
(iii)
\begin{align} \text{By Py} & \text{thagoras theorem,} \\ 13^2 & = x^2 + (-5)^2 \\ 169 & = x^2 + 25 \\ x^2 & = 144 \\ x & = \pm\sqrt{144} \\ x & = 12 \text{ (Reject) or } -12 \\ \\ \cos 2\beta & = \cos^2 \beta - \sin^2 \beta \phantom{0000000000} [\text{Double angle formula}] \\ & = \left(-{12 \over 13}\right)^2 - \left(-{5 \over 13}\right)^2 \\ & = {144 \over 169} - {25 \over 169} \\ & = {119 \over 169} \end{align}
(iv)
\begin{align} \cos 2\beta & = 2\cos^2 \beta - 1 \\ \\ \text{Let } & \beta = {1 \over 2}\beta, \\ \cos \left[2 \left({1 \over 2}\beta\right) \right] & = 2 \cos^2 {1 \over 2}\beta - 1 \\ \cos \beta & = 2 \cos^2 {1 \over 2}\beta - 1 \\ \cos \beta + 1 & = 2\cos^2 {1 \over 2}\beta \\ \\ \cos^2 {1 \over 2}\beta & = {\cos \beta + 1 \over 2} \\ \cos {1 \over 2}\beta & = \pm \sqrt{\cos \beta + 1 \over 2} \\ \\ \text{Since } 180^\circ < \phantom{.} & \beta < 270^\circ \\ 90^\circ < \phantom{.} & {1 \over 2} \beta < 135^\circ \\ \\ \text{Since } {1 \over 2} \beta \text{ is in 2nd} & \text{ quadrant, } \cos {1 \over 2} \beta < 0 \\ \\ \\ \therefore \cos {1 \over 2}\beta & = -\sqrt{\cos \beta + 1 \over 2} \\ & = -\sqrt{\left(-{12 \over 13}\right) + 1 \over 2} \\ & = -\sqrt{1 \over 26} \\ & = -{\sqrt{1} \over \sqrt{26}} \\ & = -{1 \over \sqrt{26}} \end{align}
(a)
\begin{align}
4\cos 2x + 2\sin x & = 3 \\
4(1 - 2\sin^2 x) + 2\sin x & = 3 \\
4 - 8\sin^2 x + 2\sin x & = 3 \\
-8\sin^2 x + 2\sin x - 1 & = 0 \\
8\sin^2 x -2\sin x + 1 & = 0 \\
(4\sin x + 1)(2\sin x - 1) & = 0 \\
\\
4\sin x + 1 = 0 \phantom{00} & \text{or} \phantom{000} 2\sin x - 1 = 0
\end{align}
\begin{align}
4\sin x + 1 & = 0 \\
4\sin x & = - 1 \\
\sin x & = {-1 \over 4} \\
\sin x & = - 0.25 \phantom{000000} [\text{3rd & 4th quadrants since } \sin x < 0] \\
\\
\text{Basic angle, } \alpha & = \sin^{-1} (0.25) \\
& = 14.48^\circ
\end{align}
\begin{align} x & = 180^\circ + 14.48^\circ, 360^\circ - 14.48^\circ \\ & = 194.48^\circ, 345.52^\circ \\ & \approx 194.5^\circ, 345.5^\circ \end{align}
\begin{align} 2\sin x - 1 & = 0 \\ 2\sin x & = 1 \\ \sin x & = {1 \over 2} \\ \sin x & = 0.5 \phantom{000000} [\text{1st & 2nd quadrants since } \sin x > 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} (0.5) \\ & = 30^\circ \end{align}
\begin{align} x & = 30^\circ, 180^\circ - 30^\circ \\ & = 30^\circ, 150^\circ \\ \\ \\ \therefore x & = 30^\circ, 150^\circ, 194.5^\circ, 345.5^\circ \end{align}
(b)
\begin{align} 7\sin x \cos x & = 2 \\ \sin x \cos x & = {2 \over 7} \\ 2\sin x \cos x & = {4 \over 7} \\ \sin 2x & = {4 \over 7} \phantom{000000} [\text{1st & 2nd quadrants since } \sin 2x > 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left( {4 \over 7} \right) \\ & = 34.85^\circ \end{align}
\begin{align} \text{Since } 0^\circ < \phantom{.} & x < 360^\circ \\ 0^\circ < \phantom{.} & 2x < 720^\circ \\ \\ 2x & = 34.85^\circ, 180^\circ - 34.85^\circ, 34.85^\circ + 360^\circ, (180^\circ - 34.85^\circ) + 360^\circ \\ & = 34.85^\circ, 145.15^\circ, 394.85^\circ, 505.15^\circ \\ \\ x & = 17.425^\circ, 72.575^\circ, 197.425^\circ, 252.575^\circ \\ & \approx 17.4^\circ, 72.6^\circ, 197.4^\circ, 252.6^\circ \end{align}
(c)
\begin{align} \sin 2x & = \sin^2 x \\ 2\sin x \cos x & = \sin^2 x \\ 2\sin x \cos x - \sin^2 x & = 0 \\ \sin x (2\cos x - \sin x) & = 0 \\ \\ \sin x = 0 \phantom{00} & \text{or} \phantom{000} 2\cos x - \sin x = 0 \end{align}
$$ \sin x = 0 $$
\begin{align} x & = 0^\circ \text{ (N.A.)}, 180^\circ, 360^\circ \text{ (N.A.)} \\ & = 180^\circ \end{align}
\begin{align} 2\cos x - \sin x & = 0 \\ -\sin x & = -2\cos x \\ \sin x & = 2\cos x \\ {\sin x \over \cos x} & = 2 \\ \tan x & = 2 \phantom{000000} [\text{1st & 3rd quadrants since } \tan x > 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (2) \\ & = 63.43^\circ \end{align}
\begin{align} x & = 63.43^\circ, 180^\circ + 63.43^\circ \\ & = 63.43^\circ, 243.43^\circ \\ & \approx 63.4^\circ, 243.4^\circ \\ \\ \\ \therefore x & = 63.4^\circ, 180^\circ, 243.4^\circ \end{align}
(d)
\begin{align} \cot 2x & = 1 + \cot x \\ {1 \over \tan 2x} & = 1 + {1 \over \tan x} \\ [\text{Double angle formula}] \phantom{000000} {1 \over {2\tan x \over 1 - \tan^2 x}} & = 1 + {1 \over \tan x} \\ {1 - \tan^2 x \over 2\tan x} & = 1 + {1 \over \tan x} \\ 1 - \tan^2 x & = 2\tan x \left( 1 + {1 \over \tan x} \right) \\ 1 - \tan^2 x & = 2\tan x + 2 \\ -\tan^2 x - 2\tan x - 1 & = 0 \\ \tan^2 x + 2\tan x + 1 & = 0 \\ (\tan x + 1)^2 & = 0 \\ \tan x + 1 & = \pm \sqrt{0} \\ \tan x + 1 & = 0 \\ \tan x & = - 1 \phantom{00000000} [\text{2nd & 4th quadrants since } \tan x < 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (1) \\ & = 45^\circ \end{align}
\begin{align} x & = 180^\circ - 45^\circ, 360^\circ - 45^\circ \\ & = 135^\circ, 315^\circ \end{align}
(a)
\begin{align} \text{R.H.S} & = \cos 2A \tan A \\ & = (2\cos^2 A - 1)(\tan A) \phantom{0000000000} [\text{Double angle formula: } \cos 2A] \\ & = 2\cos^2 A \tan A - \tan A \\ & = 2\cos^2 A \left(\sin A \over \cos A\right) - \tan A \\ & = 2\sin A \cos A - \tan A \\ & = \sin 2A - \tan A \phantom{000000000000000} [\text{Double angle formula: } \sin 2A] \\ & = \text{L.H.S} \end{align}
(b)
\begin{align}
\cos 2A & = \cos^2 A - \sin^2 A \\
\\
\text{Let } & A = 2A, \\
\cos [2(2A)] & = \cos^2 2A - \sin^2 2A \\
\cos 4A & = \cos^2 2A - \sin^2 2A
\end{align}
\begin{align}
\text{L.H.S } & = \cos 4A \sec^2 2A \\
& = (\cos^2 2A - \sin^2 2A)\left( {1 \over \cos^2 2A} \right) \\
& = {\cos^2 2A \over \cos^2 2A} - {\sin^2 2A \over \cos^2 2A} \\
& = 1 - \tan^2 2A \\
& = \text{R.H.S}
\end{align}
(c)
\begin{align} \require{cancel} \text{L.H.S } & = {\sin (A + B) \over \sin A \cos B} \\ & = {\sin A \cos B + \cos A \sin B \over \sin A \cos B} \phantom{00000000} [\text{Addition formula}] \\ & = {\cancel{\sin A \cos B} \over \cancel{\sin A \cos B}} + {\cos A \sin B \over \sin A \cos B} \\ & = {1 \over 1} + \left( {\cos A \over \sin A} \right) \left( {\sin B \over \cos B} \right) \\ & = 1 + \cot A \tan B \\ & = \text{R.H.S} \end{align}
First part
\begin{align} 7\cos x + 24\sin x & = a\cos x + b\sin x \\ & = R \cos (x - \alpha) \\ \\ a = 7, & \phantom{0} b = 24 \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{7^2 + 24^2} \\ & = 25 \\ \\ \alpha & = \tan^{-1} {b \over a} \\ & = \tan^{-1} {24 \over 7} \\ & = 73.7^\circ \\ \\ \\ \therefore 7\cos x + 24\sin x & = 25\cos (x - 73.7^\circ) \end{align}
(i)
$$(7\cos x + 24\sin x) - 12 = 25\cos (x - 73.7^\circ) - 12$$
\begin{align}
\text{Maximum } & = 25(1) -12 \\
& = 13 \\
\\
x - 73.7^\circ = 0^\circ \phantom{00^\circ} & \text{or} \phantom{000} x - 73.7^\circ = 360^\circ \\
x = 73.7^\circ & \phantom{or00-73.37^\circ} x = 360^\circ + 73.7^\circ \\
& \phantom{or00-73.37^\circ} x = 433.7^\circ \text{ (Reject, } 0^\circ < x < 360^\circ)
\end{align}
\begin{align}
\text{Minimum } & = 25(-1) -12 \\
& = -37 \\
\\
x - 73.7^\circ & = 180^\circ \\
x & = 253.7^\circ
\end{align}
(ii)
\begin{align} 7\cos x & = 10 - 24\sin x \\ 7\cos x + 24\sin x & = 10 \\ 25\cos (x - 73.7^\circ) & = 10 \\ \cos (x - 73.7^\circ) & = {10 \over 25} \\ \cos (x - 73.7^\circ) & = 0.4 \phantom{00000000} [\text{1st & 4th quadrants since } \cos (x - 73.7^\circ) > 0] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.4) \\ & = 66.42^\circ \end{align}
\begin{align} \text{Since } 0^\circ < \phantom{.} & x < 360^\circ, \\ -73.7^\circ < x & - 73.7^\circ < 286.3^\circ \\ \\ x - 73.7^\circ & = 66.42^\circ, 360^\circ - 66.42^\circ \\ & = 66.42^\circ, 293.58^\circ \text{ (N.A.)}, 293.58^\circ - 360^\circ \\ & = 66.42^\circ, -66.42^\circ \\ \\ x & = 140.12^\circ, 7.28^\circ \\ & \approx 140.1^\circ, 7.3^\circ \\ \\ \\ \therefore \text{Smallest } & \text{positive angle} = 7.3^\circ \end{align}
(i)
\begin{align}
\cos \theta & = 2\cos^2 {\theta \over 2} - 1 = 1 - 2\sin^2 {\theta \over 2} \\
\\
\sin \theta & = 2\sin {\theta \over 2} \cos {\theta \over 2}
\end{align}
\begin{align}
\text{L.H.S } & = {1 - \cos \theta + \sin \theta \over 1 + \cos \theta + \sin \theta} \\
& = {1 - \left(1 - 2\sin^2 {\theta \over 2}\right) + 2\sin {\theta \over 2} \cos {\theta \over 2} \over
1 + \left(2\cos^2 {\theta \over 2} - 1\right) + 2\sin {\theta \over 2} \cos {\theta \over 2}} \\
& = {2 \sin^2 {\theta \over 2} + 2\sin {\theta \over 2} \cos {\theta \over 2} \over
2\cos^2 {\theta \over 2} + 2\sin {\theta \over 2} \cos {\theta \over 2}} \\
& = {2\sin {\theta \over 2} \left(\sin {\theta \over 2} + \cos {\theta \over 2}\right) \over
2\cos {\theta \over 2} \left(\cos {\theta \over 2} + \sin {\theta \over 2}\right)} \\
& = {2\sin {\theta \over 2} \over 2\cos {\theta \over 2}} \\
& = {\sin {\theta \over 2} \over \cos {\theta \over 2}} \\
& = \tan {\theta \over 2} \\
& = \text{R.H.S}
\end{align}
(ii)
\begin{align} 1 + \sin \theta & = 2\cos \theta \\ 2 + 2\sin \theta & = 4\cos \theta \\ 2 + 2\sin \theta - 3 \cos \theta & = \cos \theta \\ 2 + 2\sin \theta - 3 \cos \theta + 1 + \sin \theta & = \cos \theta + 1 + \sin \theta \\ 3 - 3\cos \theta + 3\sin \theta & = 1 + \cos \theta + \sin \theta \\ 3(1 - \cos \theta + \sin \theta) & = 1 + \cos \theta + \sin \theta \\ {3(1 - \cos \theta + \sin \theta) \over 1 + \cos \theta + \sin \theta} & = 1 \\ {1 - \cos \theta + \sin \theta \over 1 + \cos \theta + \sin \theta} & = {1 \over 3} \\ \\ \tan {\theta \over 2} & = {1 \over 3} \phantom{00000000} \left[ \text{1st & 3rd quadrants since } \tan {\theta \over 2} > 0 \right] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left(1 \over 3\right) \\ & = 0.32175 \end{align}
\begin{align} \text{Since } 0 < \phantom{.} & \theta < \pi \\ 0 < \phantom{.} & {\theta \over 2} < {\pi \over 2} \\ 0 < \phantom{.} & {\theta \over 2} < 1.5708 \\ \\ {\theta \over 2} & = 0.32175, \pi + 0.32175 \\ & = 0.32175, 3.46334 \text{ (N.A.)} \\ \\ \theta & = 0.32175 \times 2 \\ & = 0.64350 \\ & \approx 0.644 \end{align}
(i)
\begin{align} \sin \angle DAF & = {Opp \over Hyp} \\ \sin \theta & = {DF \over DA} \\ \sin \theta & = {DF \over 2.5} \\ DF & = 2.5 \sin \theta \\ \\ \angle ADF & = 90^\circ - \theta \\ \angle CDE & = 180^\circ - (90^\circ - \theta) - 90^\circ \\ & = \theta \\ \\ \cos \angle CDE & = {Adj \over Hyp} \\ \cos \theta & = {ED \over CD} \\ \cos \theta & = {ED \over 1} \\ ED & = \cos \theta \\ \\ \therefore EF & = ED + DF \\ 1.5 & = \cos \theta + 2.5\sin \theta \end{align}
(ii)
\begin{align} \cos \theta + 2.5\sin \theta & = 1.5 \\ a\cos \theta + b \sin \theta & = 1.5 \\ R\cos (\theta - \alpha) & = 1.5 \\ \\ a = 1, & \phantom{0} b = 2.5 \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{1^2 + 2.5^2} \\ & = \sqrt{7.25} \\ & = 2.6926 \\ \\ \alpha & = \tan^{-1} \left( {b \over a} \right) \\ & = \tan^{-1} \left( {2.5 \over 1} \right) \\ & = 68.2^\circ \\ \\ \\ \therefore 2.6926 \cos (\theta - 68.2^\circ) & = 1.5 \\ \cos (\theta - 68.2^\circ) & = 1.5 \div 2.6926 \\ \cos (\theta - 68.2^\circ) & = 0.55708 \phantom{00000000} [\text{1st & 4th quadrants since } \cos (\theta - 68.2^\circ) > 0] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.55708) \\ & = 56.15^\circ \end{align}
\begin{align} 0^\circ < \phantom{.} & \theta < 90^\circ \\ -68.2^\circ < \theta & - 68.2^\circ < 21.8^\circ \\ \\ \\ \theta -68.2^\circ & = 56.15^\circ, 360^\circ - 56.15^\circ \\ & = 56.15^\circ \text{ (N.A.)}, 303.85^\circ \text{ (N.A.)}, 56.15^\circ - 360^\circ, 303.85^\circ - 360^\circ \\ & = -303.85^\circ \text{ (N.A.)}, -56.15^\circ \\ \\ \theta & = -56.15^\circ + 68.2^\circ \\ & = 12.05^\circ \\ & \approx 12.1^\circ \end{align}
(i)
\begin{align}
{\cos (A - B) \over \cos (A + B)} & = {7 \over 5} \\
{\cos A \cos B + \sin A \sin B \over \cos A \cos B - \sin A \sin B} & = {7 \over 5} \\
5(\cos A \cos B + \sin A \sin B) & = 7(\cos A \cos B - \sin A \sin B) \\
5\cos A \cos B + 5\sin A \sin B & = 7\cos A \cos B - 7\sin A \sin B \\
5\sin A \sin B + 7\sin A \sin B & = 7\cos A \cos B - 5\cos A \cos B \\
12\sin A \sin B & = 2\cos A \cos B \\
6\sin A \sin B & = \cos A \cos B \text{ (Shown)}
\end{align}
\begin{align}
6\sin A \sin B & = \cos A \cos B \\
{6\sin A \sin B \over \cos A \cos B} & = 1 \\
\left(6 \over 1\right)\left(\sin A \over \cos A\right)\left(\sin B \over \cos B\right) & = 1 \\
6\tan A \tan B & = 1 \\
\tan A \tan B & = {1 \over 6}
\end{align}
(ii)
\begin{align} \text{Since } & A + B = 45^\circ, \\ \tan (A + B) & = \tan 45^\circ \\ \tan (A + B) & = 1 \\ [\text{Addition formula}] \phantom{000000} {\tan A + \tan B \over 1 - \tan A \tan B} & = 1 \\ \tan A + \tan B & = 1 - \tan A \tan B \\ & = 1 - \left({1 \over 6}\right) \phantom{00000000} [\text{From part i}] \\ & = {5 \over 6} \end{align}
(i)
\begin{align} \cos^2 A & = {1 \over 4} \\ \cos A & = \pm\sqrt{1 \over 4} \\ \cos A & = \pm{1 \over 2} \\ \cos A & = {1 \over 2} \text{ (Reject ) or } -{1 \over 2} \end{align}
\begin{align} \text{By Py} & \text{thagoras theorem,} \\ 2^2 & = (-1)^2 + x^2 \\ x^2 & = 4 - 1 \\ x^2 & = 3 \\ x & = \pm \sqrt{3} \\ x & = \sqrt{3} \text{ or } -\sqrt{3} \text{ (Reject)} \\ \\ \cos (A + 60^\circ) & = \cos A \cos 60^\circ - \sin A \sin 60^\circ \phantom{000000000000} [\text{Addition formula: } \cos (A + B)]\\ & = \left(-{1 \over 2}\right)\cos 60^\circ - \left(\sqrt{3} \over 2\right)\sin 60^\circ \\ & = \left(-{1 \over 2}\right)\left(1 \over 2\right) - \left(\sqrt{3} \over 2\right)\left(\sqrt{3} \over 2\right) \phantom{00000000} [\text{Special values: } \cos 60^\circ, \sin 60^\circ] \\ & = -{1 \over 4} - {3 \over 4} \\ & = - 1 \end{align}
(ii)
\begin{align} \sin 2A & = 2\sin A \cos A \\ & = 2\left(\sqrt{3} \over 2\right) \left(-{1 \over 2}\right) \\ & = -{\sqrt{3} \over 2} \end{align}
(iii)
\begin{align} \cos 2A & = \cos^2 A - \sin^2 A \\ & = \left(-{1 \over 2}\right)^2 - \left(\sqrt{3} \over 2\right)^2 \\ & = -{1 \over 2} \end{align}
(iv)
\begin{align} \sin 3A & = \sin (2A + A) \\ & = \sin 2A \cos A + \cos 2A \sin A \\ & = (\sin 2A)\left(-{1 \over 2}\right) + (\cos 2A)\left(\sqrt{3} \over 2\right) \\ & = \left(-{\sqrt{3} \over 2}\right)\left(-{1 \over 2}\right) + \left(-{1 \over 2}\right)\left(\sqrt{3} \over 2\right) \phantom{00000000} [\text{From parts ii & iii}]\\ & = 0 \end{align}
(i)
\begin{align} 4\sin 3x \cos x - 4\cos 3x \sin x + 1 & = 0 \\ 4\sin 3x \cos x - 4\cos 3x \sin x & = - 1 \\ 4(\sin 3x \cos x - \cos 3x \sin x) & = -1 \\ \sin 3x \cos x - \cos 3x \sin x & = -{1 \over 4} \\ \sin 3x \cos x - \cos 3x \sin x & = -0.25 \\ [\text{Addition formula: } \sin (A - B)] \phantom{0000000} \sin (3x - x) & = -0.25 \\ \sin 2x & = -0.25 \phantom{00000000} [\text{3rd & 4th quadrants since } \sin 2x < 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} (0.25) \\ & = 14.48^\circ \end{align}
\begin{align} 0^\circ < \phantom{.} & x < 180^\circ \\ 0^\circ < \phantom{.} & 2x < 360^\circ \\ \\ 2x & = 14.48^\circ + 180^\circ, 360^\circ - 14.48^\circ \\ & = 194.48^\circ, 345.52^\circ \\ \\ x & = 97.24^\circ, 162.76^\circ \\ & \approx 97.2^\circ, 172.8^\circ \end{align}
(ii)
\begin{align} \sin 2x & = 2\cos x \\ [\text{Double angle formula}] \phantom{000000} 2\sin x \cos x & = 2\cos x \\ 2\sin x \cos x - 2\cos x & = 0 \\ 2\cos x (\sin x - 1) & = 0 \\ 2\cos x = 0 \phantom{00} & \text{or} \phantom{000} \sin x - 1 = 0 \end{align}
\begin{align} 2\cos x & = 0 \\ \cos x & = 0 \end{align}
\begin{align} x & = 90^\circ, 270^\circ \end{align}
\begin{align} \sin x - 1 & = 0 \\ \sin x & = 1 \end{align}
\begin{align} x & = 90^\circ \\ \\ \\ \therefore x & = 90^\circ, 270^\circ \end{align}
(iii)
\begin{align} \sin 2x & = 2\cos 2x + 1 \\ 2\sin x \cos x & = 2\cos 2x + 1 \phantom{0000000000000000000000(} [\text{Double angle formula: } \sin 2A] \\ 2\sin x \cos x & = 2(\cos^2 x - \sin^2 x) + 1 \phantom{000000000000000} [\text{Double angle formula: } \cos 2A] \\ 2\sin x \cos x & = 2(\cos^2 x - \sin^2 x) + (\sin^2 x + \cos^2 x) \phantom{000} [\text{Identity: } \sin^2 A + \cos^2 A = 1]\\ 2\sin x \cos x & = 2\cos^2 x - 2\sin^2 x + \sin^2 x + \cos^2 x \\ 2\sin x \cos x & = 3\cos^2 x - \sin^2 x \\ 0 & = 3\cos^2 x - 2\sin x \cos x - \sin^2 x \\ 0 & = (3\cos x + \sin x)(\cos x - \sin x) \\ \\ 3\cos x + \sin x & = 0 \phantom{000}\text{ or }\phantom{000} \cos x - \sin x = 0 \end{align}
\begin{align} 3\cos x + \sin x & = 0 \\ \sin x & = -3\cos x \\ {\sin x \over \cos x} & = -3 \\ \tan x & = - 3 \phantom{00000000} [\text{2nd & 4th quadrants since } \tan x < 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (3) \\ & = 71.57^\circ \end{align}
\begin{align} x & = 180^\circ - 71.57^\circ, 360^\circ - 71.57^\circ \\ & = 108.43^\circ, 288.43^\circ \\ & \approx 108.4^\circ, 288.4^\circ \end{align}
\begin{align} \cos x - \sin x & = 0 \\ -\sin x & = -\cos x \\ \sin x & = \cos x \\ {\sin x \over \cos x} & = 1 \\ \tan x & = 1 \phantom{00000000} [\text{1st & 3rd quadrants since } \tan x > 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (1) \\ & = 45^\circ \end{align}
\begin{align} x & = 45^\circ, 180^\circ + 45^\circ \\ & = 45^\circ, 225^\circ \\ \\ \\ \therefore x & = 45^\circ, 108.4^\circ, 225^\circ, 288.4^\circ \end{align}
(iv)
\begin{align} \cos 2x - 3\cos x & = 1 \\ (2\cos^2 x - 1) - 3\cos x & = 1 \\ 2\cos^2 x - 3\cos x - 2 & = 0 \\ (2\cos x + 1)(\cos x - 2) & = 0 \\ \\ 2\cos x + 1 = 0 \phantom{00} & \text{or} \phantom{000} \cos x - 2 = 0 \end{align}
\begin{align} 2\cos x + 1 & = 0 \\ 2\cos x & = - 1 \\ \cos x & = {-1 \over 2} \\ \cos x & = -0.5 \phantom{00000000} [\text{2nd & 3rd quadrants since } \cos x < 0] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.5) \\ & = {\pi \over 3} \end{align}
\begin{align} x & = \pi - {\pi \over 3}, \pi + {\pi \over 3} \\ & = {2\pi \over 3}, {4\pi \over 3} \text{ (N.A.)} \end{align}
\begin{align} \cos x - 2 & = 0 \\ \cos x & = 2 \text{ (Reject, since } -1 \le \cos x \le 1) \\ \\ \\ \therefore x & = {2\pi \over 3} \end{align}
(v)
\begin{align} 3 \sec 2x & = 8 \sin 2x \\ 3 \left(1 \over \cos 2x\right) & = 8 \sin 2x \\ {3 \over \cos 2x} & = 8 \sin 2x \\ 3 & = 8 \sin 2x \cos 2x \\ {3 \over 4} & = 2\sin 2x \cos 2x \\ {3 \over 4} & = \sin [2(2x)] \phantom{000000000} [\text{Double angle formula}] \\ {3 \over 4} & = \sin 4x \\ \therefore \sin 4x & = 0.75 \phantom{0000000000000} [\text{1st & 2nd quadrants since } \sin 4x > 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} (0.75) \\ & = 0.84806 \end{align}
\begin{align} \text{Since } 0 \le \phantom{.} & x \le \pi \\ 0 \le \phantom{.} & 4x \le 4\pi \\ 0 \le \phantom{.} & 4x \le 12.56637 \\ \\ 4x & = 0.84806, \pi - 0.84806 \\ & = 0.84806, 2.29353, 0.84806 + 2\pi, 2.29353 + 2\pi \\ & = 0.84806, 2.29353, 7.13125, 8.57672 \\ \\ x & = 0.212015, 0.5733825, 1.7828125, 2.14418 \\ & \approx 0.212, 0.573, 1.78, 2.14 \end{align}
(a)(i)
\begin{align} \sin (x - y) & = \sin x \cos y - \cos x \sin y \phantom{000000} [\text{Addition formula: } \sin (A - B)] \\ p & = q - \cos x \sin y \\ \cos x \sin y & = q - p \\ \\ \\ \sin (x + y) & = \sin x \cos y + \cos x \sin y \phantom{000000} [\text{Addition formula: } \sin (A + B)] \\ & = (q) + (q - p) \\ & = 2q - p \end{align}
(a)(ii)
\begin{align} {\tan x \over \tan y} & = \tan x \div \tan y \\ & = {\sin x \over \cos x} \div {\sin y \over \cos y} \\ & = {\sin x \over \cos x} \times {\cos y \over \sin y} \\ & = {\sin x \cos y \over \cos x \sin y} \\ & = {q \over q - p} \end{align}
(a)(iii)
\begin{align} \sin 2x \sin 2y & = (2\sin x \cos x)(2\sin y \cos y) \\ & = 4\sin x \cos y \cos x\sin y \\ & = 4(\sin x \cos y)(\cos x \sin y) \\ & = 4(q)(q - p) \\ & = 4q (q - p) \end{align}
(b)
\begin{align}
A + B + C & = 90^\circ \\
A + B & = 90^\circ - C \\
\\
\therefore \tan (A + B) & = \tan (90^\circ - C) \\
\tan (A + B) & = \cot C \\
[\text{Addition formula}] \phantom{000000} {\tan A + \tan B \over 1 - \tan A \tan B} & = \cot C \\
{\tan A + \tan B \over 1 - \tan A \tan B} & = {1 \over \tan C} \\
\tan C(\tan A + \tan B) & = 1 - \tan A \tan B \\
\tan C \tan A + \tan B \tan C & = 1 - \tan A \tan B
\end{align}
\begin{align}
\text{L.H.S } & = \tan A \tan B + (\tan B \tan C + \tan C \tan A) \\
& = \tan A \tan B + (1 - \tan A \tan B) \\
& = \tan A \tan B + 1 - \tan A \tan B \\
& = 1 \\
& = \text{R.H.S}
\end{align}
(a)
\begin{align}
2\cos x - 3\sin x & = a \cos x - b \sin x \\
& = R \cos (x + \alpha) \\
\\
a = 2 &, b = 3 \\
\\
R & = \sqrt{a^2 + b^2} \\
& = \sqrt{2^2 + 3^2} \\
& = \sqrt{13} \\
\\
\alpha & = \tan^{-1} \left(b \over a\right) \\
& = \tan^{-1} \left(3 \over 2\right) \\
& = 56.31^\circ \\
\\
\therefore 2\cos x - 3\sin x & = \sqrt{13}\cos (x + 56.31^\circ)
\end{align}
\begin{align}
-\sqrt{13} \le \sqrt{13}\cos (x + 56.31^\circ) & \le \sqrt{13} \\
\\
[\sqrt{13}\cos (x + 56.31^\circ)]^2 & \le (\sqrt{13})^2 \\
[\sqrt{13}\cos (x + 56.31^\circ)]^2 & \le 13 \\
\\
\therefore (2\cos x - 3\sin x)^2 & \le 13
\end{align}
\begin{align}
(2\cos x - 3\sin x)^2 & = 13 \\
\sqrt{(2\cos x - 3\sin x)^2} & = \pm \sqrt{13} \\
2\cos x - 3\sin x & = \pm \sqrt{13} \\
\\
\therefore \sqrt{13}\cos (x + 56.31^\circ) & = \pm \sqrt{13}
\end{align}
\begin{align}
\sqrt{13}\cos (x + 56.31^\circ) & = \sqrt{13} &&&
\sqrt{13}\cos (x + 56.31^\circ) & = -\sqrt{13} \\
\\
x + 56.31^\circ & = 0^\circ, 360^\circ &&&
x + 56.31^\circ & = 180^\circ \\
x & = -56.3^\circ \text{ (N.A.)}, 303.7^\circ &&&
x & = 123.7^\circ
\end{align}
(b)
\begin{align} \cos 2x & = 2\cos^2 x - 1 \\ 2\cos^2 x & = \cos 2x + 1 \\ \cos^2 x & = {\cos 2x + 1 \over 2} \end{align}
\begin{align} y & = 4(\cos^2 x) - 1 \\ & = 4\left(\cos 2x + 1 \over 2\right) - 1 \\ & = 2(\cos 2x + 1) - 1 \\ & = 2\cos 2x + 2 - 1 \\ & = 2\cos 2x + 1 \\ \\ \text{Comparing with } & y = a\cos bx + c, \\ a & = 2 \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Period} & = {360^\circ \over b} \\ & = {360^\circ \over 2} \\ & = 180^\circ \\ \\ \text{Center line: } & y = 1 \\ \\ \text{Maximum} & = 1 + 2 = 3 \\ \\ \text{Minimum} & = 1 - 2 = -1 \end{align}
(a)
\begin{align} \require{cancel} \text{L.H.S } & = {\cos (x + y) + \cos (x - y) \over \sin (x + y) - \sin (x - y)} \\ & = {(\cos x \cos y + \sin x \sin y) + (\cos x \cos y - \sin x \sin y) \over (\sin x \cos y + \cos x \sin y) - (\sin x \cos y - \cos x \sin y)} \\ & = {2\cos x \cos y \over \sin x \cos y + \cos x \sin y - \sin x \cos y + \cos x \sin y} \\ & = {2 \cancel{\cos x} \cos y \over 2 \cancel{\cos x} \sin y} \\ & = {\cos y \over \sin y} \\ & = \cot y \\ & = \text{R.H.S} \end{align}
(b)
\begin{align} \require{cancel} \text{L.H.S } & = {1 + \sec 2\theta \over \tan 2\theta} \\ & = {1 + {1 \over \cos 2\theta} \over {\sin 2\theta \over \cos 2\theta}} \\ & = {{\cos 2\theta \over \cos 2\theta} + {1 \over \cos 2\theta} \over {\sin 2\theta \over \cos 2\theta} } \\ & = {{\cos 2\theta + 1 \over \cos 2\theta} \over {\sin 2\theta \over \cos 2\theta}} \\ & = {\cos 2\theta + 1 \over \cos 2\theta} \div {\sin 2\theta \over \cos 2\theta} \\ & = {\cos 2\theta + 1 \over \cos 2\theta} \times {\cos 2\theta \over \sin 2\theta} \\ & = {\cos 2\theta + 1 \over \sin 2\theta} \\ & = {2\cos^2 \theta - 1 + 1 \over \sin 2\theta} \phantom{00000000} [\text{Double angle formula: } \cos 2A] \\ & = {2\cos^2 \theta \over \sin 2\theta} \\ & = {\cancel{2}\cos^\cancel{2} \theta \over \cancel{2}\sin \theta \cancel{\cos \theta}} \phantom{000000000.} [\text{Double angle formula: } \sin 2A] \\ & = {\cos \theta \over \sin \theta} \\ & = \cot \theta \\ & = \text{R.H.S} \end{align}
(c)
\begin{align} \text{L.H.S } & = \sin^2 2\theta (\cot^2 \theta - \tan^2 \theta) \\ & = (\sin 2\theta)^2(\cot^2 \theta - \tan^2 \theta) \\ & = (2\sin \theta \cos \theta)^2 (\cot^2 \theta - \tan^2 \theta) \phantom{00000000} [\text{Double angle formula}] \\ & = 4\sin^2 \theta \cos^2 \theta (\cot^2 \theta - \tan^2 \theta) \\ & = 4\sin^2 \theta \cos^2 \theta \left({\cos^2 \theta \over \sin^2 \theta} - {\sin^2 \theta \over \cos^2 \theta}\right) \\ & = 4\cos^4 \theta - 4\sin^4 \theta \\ & = 4(\cos^4 \theta - \sin^4 \theta) \\ & = 4(\cos^2 \theta + \sin^2 \theta)(\cos^2 \theta - \sin^2 \theta) \\ & = 4(1)(\cos^2 \theta - \sin^2 \theta) \phantom{0000000000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = 4(\cos^2 \theta - \sin^2 \theta) \\ & = 4\cos 2 \theta \phantom{00000000000000000000000000} [\text{Double angle formula}] \\ & = \text{R.H.S} \end{align}
(i)
\begin{align} \cos \angle QOA & = {Adj \over Hyp} \\ \cos \theta & = {OA \over OQ} \\ \cos \theta & = {OA \over 8} \\ \\ OA & = 8\cos \theta \\ \\ \\ \sin \angle OPB & = {Opp \over Hyp} \\ \sin \theta & = {OB \over OP} \\ \sin \theta & = {OB \over 15} \\ \\ OB & = 15\sin \theta \\ \\ \\ AB & = OB - OA \\ & = 15\sin \theta - 8\cos \theta \\ & = a\sin \theta - b\cos \theta \\ & = R\sin (\theta - \alpha) \\ \\ a & = 15, b = 8 \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{15^2 + 8^2} \\ & = 17 \\ \\ \alpha & = \tan^{-1} \left(b \over a\right) \\ & = \tan^{-1} \left(8 \over 15\right) \\ & = 28.1^\circ \\ \\ \\ \therefore AB & = 17\sin (\theta - 28.1^\circ) \end{align}
(ii)
\begin{align} \text{When } & AB = 5, \\ 5 & = 17\sin (\theta - 28.1^\circ) \\ {5 \over 17} & = \sin (\theta - 28.1^\circ) \\ \\ \sin (\theta - 28.1^\circ) & = 0.29412 \phantom{000000} [\text{1st & 2nd quadrants since } \sin (\theta - 28.1^\circ) > 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} (0.29412) \\ & = 17.1^\circ \end{align}
\begin{align} \text{Since } 0^\circ < \phantom{.} & \theta < 90^\circ \\ -28.1^\circ < \theta & - 28.1^\circ < 61.9^\circ \\ \\ \theta - 28.1^\circ & = 17.1^\circ, 180^\circ - 17.1^\circ \\ & = 17.1^\circ, 162.9^\circ \text{ (N.A.)} \\ \\ \theta & = 17.1^\circ + 28.1^\circ \\ & = 45.2^\circ \end{align}
(iii)
\begin{align} \theta - 28.1^\circ & = 0^\circ \\ \\ \theta & = 28.1^\circ \\ \\ \\ \therefore 28.1^\circ < \phantom{.} & \theta < 90^\circ \end{align}
(iv)
\begin{align} \text{Area of }\triangle \text{ QOA } & = {1 \over 2}(QO)(OA)(\sin \angle QOA) \\ & = {1 \over 2}(8)(8\cos \theta)(\sin \theta) \\ & = 32\sin \theta \cos \theta \\ \\ \\ \cos \angle OPB & = {Adj \over Hyp} \\ \cos \theta & = {PB \over 15} \\ 15 \cos \theta & = PB \\ \\ \text{Area of }\triangle POB & = {1 \over 2}\times b \times h \\ & = {1 \over 2} \times OB \times PB \\ & = {1 \over 2} \times 15\sin \theta \times 15\cos \theta \\ & = {225 \over 2} \sin \theta \cos \theta \\ \\ \\ \text{Shaded area } & = {225 \over 2}\sin \theta \cos \theta - 32 \sin \theta \cos \theta \\ & = {161 \over 2} \sin \theta \cos \theta \\ & = {161 \over 2}\left(1 \over 2\right)(2\sin \theta \cos \theta) \\ & = {161 \over 2}\left(1 \over 2\right)(\sin 2\theta) \phantom{00000000} [\text{Double angle formula}] \\ & = {161 \over 4}\sin 2\theta \end{align}
(v)
\begin{align} \text{Maximum } & = {161 \over 4}(1) \\ & = {161 \over 4} \\ & = 40.25 \text{ cm}^2 \\ \\ \\ 2\theta & = 90^\circ \\ \theta & = 45^\circ \end{align}