(a)

\begin{align} \cos^2 22{1 \over 2}^\circ - \sin^2 22{1 \over 2}^\circ & = \cos \left[ 2\left( 22{1 \over 2}\right) \right]^\circ \phantom{000} [\text{Double angle formula}] \\ & = \cos 45^\circ \phantom{0000000000.} [\text{Special value}] \\ & = {1 \over \sqrt{2}} \end{align}

(b)

\begin{align} 2\sin {\pi \over 8} \cos {\pi \over 8} & = \sin \left[2\left({\pi \over 8}\right)\right] \phantom{000000} [\text{Double angle formula}] \\ & = \sin {\pi \over 4} \phantom{00000000000} [\text{Special value}]\\ & = {1 \over \sqrt{2}} \end{align}

(c)

\begin{align} 2\cos^2 {\pi \over 12} - 1 & = \cos \left[2\left({\pi \over 12}\right)\right] \phantom{000000} [\text{Double angle formula}] \\ & = \cos {\pi \over 6} \phantom{000000000000} [\text{Special value}] \\ & = {\sqrt{3} \over 2} \end{align}

(d)

\begin{align} (\cos 75^\circ + \sin 75^\circ)^2 & = \cos^2 75^\circ + 2\sin 75^\circ \cos 75^\circ + \sin^2 75^\circ \\ & = (\cos^2 75^\circ + \sin^2 75^\circ) + 2\sin 75^\circ \cos 75^\circ \\ & = 1 + 2\sin 75^\circ \cos 75^\circ \phantom{00000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = 1 + \sin [2(75)]^\circ \phantom{00000000000..} [\text{Double angle formula}] \\ & = 1 + \sin 150^\circ \\ & = 1 + \sin (180^\circ - 30)^\circ \\ & = 1 + \sin 30^\circ \phantom{000000000000000.} [\sin (180^\circ - \theta) = \sin \theta] \\ & = 1 + {1 \over 2} \\ & = {3 \over 2} \end{align}

(e)

\begin{align} \text{Double angle formula: } \cos 2A & = 1 - 2\sin^2 A \\ 2\sin^2 A & = 1 - \cos 2A \\ \sin^2 A & = {1 \over 2} - {1 \over 2}\cos 2A \\ \\ \\ \therefore \sin^2 67.5^\circ & = {1 \over 2} - {1 \over 2} \cos [2(67.5)]^\circ \\ & = {1 \over 2} - {1 \over 2}\cos 135^\circ \\ & = {1 \over 2} - {1 \over 2} \cos (180^\circ - 45^\circ) \\ & = {1 \over 2} - {1 \over 2}(-\cos 45^\circ) \phantom{00000000} [\cos (180^\circ - \theta) = -\cos \theta] \\ & = {1 \over 2} + {1 \over 2}\cos 45^\circ \\ & = {1 \over 2} + {1 \over 2}\left({1 \over \sqrt{2}}\right) \phantom{0000000000} [\text{Special value}] \\ & = {1 \over 2} + {1 \over 2\sqrt{2}} \\ & = {\sqrt{2} \over 2\sqrt{2}} + {1 \over 2\sqrt{2}} \\ & = {\sqrt{2} + 1 \over 2\sqrt{2}} \end{align}

(f)

\begin{align} {2\tan 15^\circ \over 1 - \tan^2 15^\circ} & = \tan [2(15)]^\circ \phantom{0000000} [\text{Double angle formula}] \\ & = \tan 30^\circ \phantom{0000000000.} [\text{Special value}] \\ & = {1 \over \sqrt{3}} \end{align}

(a)

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ 5^2 & = 3^2 + x^2 \\ 25 & = 9 + x^2 \\ x^2 & = 16 \\ x & = \pm \sqrt{16} \\ x & = 4 \text{ or } -4 \text{ (Reject)} \end{align}

\begin{align} \cos A & = {Adj \over Hyp} = {4 \over 5} \\ \\ \tan A & = {Opp \over Adj} = {3 \over 4} \\ \\ \\ \sin 2A & = 2\sin A \cos A \\ & = 2\left( {3 \over 5} \right) \left( {4 \over 5} \right) \\ & = {24 \over 25} \\ \\ \cos 2A & = \cos^2 A - \sin^2 A \\ & = \left( {4 \over 5} \right)^2 - \left( {3 \over 5} \right)^2 \\ & = {7 \over 25} \\ \\ \tan 2A & = {2\tan A \over 1 - \tan^2 A} \\ & = {2\left( {3 \over 4} \right) \over 1 - \left( {3 \over 4} \right)^2 } \\ & = {24 \over 7} \end{align}

(b)

\begin{align} A & = \cos^{-1} \left(-{1 \over 2}\right) \\ \cos A & = -{1 \over 2} \end{align}

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ 2^2 & = (-1)^2 + x^2 \\ 4 & = 1 + x^2 \\ x^2 & = 3 \\ x & = \pm \sqrt{3} \\ x & = \sqrt{3} \text{ or } -\sqrt{3} \text{ (Reject)} \\ \\ \\ \sin A & = {Opp \over Hyp} = {\sqrt{3} \over 2} \\ \\ \tan A & = {Opp \over Adj} = {\sqrt{3} \over -1} = -\sqrt{3} \\ \\ \\ \sin 2A & = 2\sin A \cos A \\ & = 2\left( {\sqrt{3} \over 2} \right) \left( -{1 \over 2} \right) \\ & = -{\sqrt{3} \over 2} \\ \\ \cos 2A & = \cos^2 A - \sin^2 A \\ & = \left( -{1 \over 2} \right)^2 - \left( {\sqrt{3} \over 2} \right)^2 \\ & = -{1 \over 2} \\ \\ \tan 2A & = {2\tan A \over 1 - \tan^2 A} \\ & = {2(-\sqrt{3}) \over 1 - (-\sqrt{3})^2 } \\ & = \sqrt{3} \end{align}

(i)

\begin{align} 3 \sin x \cos x & = 1 \\ \sin x \cos x & = {1 \over 3} \\ 2\sin x \cos x & = {2 \over 3} \\ [\text{Double angle formula}] \phantom{000000} \sin 2x & = {2 \over 3} \end{align}

(ii)

\begin{align} \cos 2x & = 1 - 2\sin^2 x \\ \\ \text{Let } & x = 2x, \\ \cos [2(2x)] & = 1 - 2\sin^2 2x \\ \cos 4x & = 1 - 2\sin^2 2x \\ & = 1 - 2\left( {2 \over 3} \right)^2 \phantom{000000} [\text{From part i}] \\ & = {1 \over 9} \end{align}

(iii)

\begin{align} \cos 2x & = 2\cos^2 x - 1 \\ \\ \text{Let } & x = 4x, \\ \cos [2(4x)] & = 2\cos^2 4x - 1 \\ \cos 8x & = 2\cos^2 4x - 1 \\ & = 2\left( {1 \over 9} \right)^2 - 1 \phantom{000000} [\text{From part ii}] \\ & = -{79 \over 81} \end{align}

(a)

\begin{align} \require{cancel} \text{L.H.S } & = {\cos 2A \over \cos A + \sin A} \\ & = {\cos^2 A - \sin^2 A \over \cos A + \sin A} \phantom{000000} [\text{Double angle formula}] \\ & = {\cancel{(\cos A + \sin A)}(\cos A - \sin A) \over \cancel{\cos A + \sin A}} \\ & = \cos A - \sin A \\ & = \text{R.H.S} \end{align}

(b) Choose the appropriate form of the double angle formula that allows us to eliminate the term '1' in the numerator

\begin{align} \require{cancel} \text{L.H.S} & = {1 - \cos 2A \over \sin 2A} \\ & = {1 - (1 - 2\sin^2 A) \over \sin 2A} \phantom{000000} [\text{Double angle formula}] \\ & = {1 - 1 + 2\sin^2 A \over \sin 2A} \\ & = {2\sin^2 A \over \sin 2A} \\ & = {\cancel{2}\sin^\cancel{2} A \over \cancel{2}\cancel{\sin A} \cos A} \phantom{000000000.} [\text{Double angle formula}] \\ & = {\sin A \over \cos A} \\ & = \tan A \\ & = \text{R.H.S} \end{align}

(c) Apply the appropriate form of the double angle formulas such that the term '1' is eliminated

\begin{align} \require{cancel} \text{L.H.S } & = {1 - \cos 2A \over 1 + \cos 2A} \\ & = {1 - (1 - 2\sin^2 A) \over 1 + \cos 2A} \phantom{000000} [\text{Double angle formula}] \\ & = {1 - 1 + 2\sin^2 A \over 1 + \cos 2A} \\ & = {2\sin^2 A \over 1 + \cos 2A} \\ & = {2\sin^2 A \over 1 + (2\cos^2 A - 1)} \phantom{000000} [\text{Double angle formula}] \\ & = {\cancel{2}\sin^2 A \over \cancel{2}\cos^2 A} \\ & = \left({\sin A \over \cos A}\right)^2 \\ & = (\tan A)^2 \\ & = \tan^2 A \\ & = \text{R.H.S} \end{align}

(d)

\begin{align} \text{L.H.S } & = 4\sin A \cos^3 A - 4\sin^3 A \cos A \\ & = 4\sin A \cos A (\cos^2 A - \sin^2 A) \\ & = 2(2\sin A \cos A)(\cos^2 A - \sin^2 A) \phantom{000000} [\text{Double angle formulas}] \\ & = 2(\sin 2A)(\cos 2A) \\ & = 2\sin 2A \cos 2A \\ & = \sin [2(2A)] \phantom{00000000000000000000000.} [\text{Double angle formula}] \\ & = \sin 4A \\ & = \text{R.H.S} \end{align}

(a)

\begin{align} \text{L.H.S} & = \sec A \text{ cosec} A \\ & = \left( {1 \over \cos A} \right) \left( {1 \over \sin A} \right) \\ & = {1 \over \sin A \cos A} \\ & = {1 \over \sin A \cos A} \times {2 \over 2} \\ & = {2 \over 2\sin A \cos A} \\ & = {2 \over \sin 2A } \phantom{0000000000} [\text{Double angle formula}] \\ & = 2 \left( {1 \over \sin 2A} \right) \\ & = 2 (\text{cosec } 2A) \\ & = 2\text{ cosec } 2A \\ & = \text{R.H.S} \end{align}

(b)

\begin{align} \require{cancel} \text{L.H.S } & = {\sec^2 A \over \tan A} \\ & = \sec^2 A \div \tan A \\ & = {1 \over \cos^2 A} \div {\sin A \over \cos A} \\ & = {1 \over \cos^\cancel{2} A} \times {\cancel{\cos A} \over \sin A} \\ & = {1 \over \sin A \cos A} \\ & = {1 \over \sin A \cos A} \times {2 \over 2} \\ & = {2 \over 2\sin A \cos A} \\ & = {2 \over \sin 2A} \phantom{0000000000} [\text{Double angle formula}] \\ & = 2\left( {1 \over \sin 2A} \right) \\ & = 2(\text{cosec }2A) \\ & = 2\text{ cosec }2A \\ & = \text{R.H.S} \end{align}

(c)

\begin{align} \cos 2A & = 2\cos^2 A - 1 \\ \\ \text{Let } & A = {A \over 2}, \\ \cos \left[2\left( {A \over 2} \right)\right] & = 2\cos^2 {A \over 2} - 1 \\ \cos A & = 2\cos^2 {A \over 2} - 1 \end{align}
\begin{align} \require{cancel} \text{R.H.S } & = {2 \over 1 + \cos A} \\ & = {2 \over 1 + \left(2\cos^2 {A \over 2} - 1 \right)} \phantom{000000} [\text{Half angle formula}] \\ & = {\cancel{2} \over \cancel{2}\cos^2 {A \over 2}} \\ & = {1 \over \cos^2 {A \over 2}} \\ & = \left( {1 \over \cos {A \over 2}} \right)^2 \\ & = \left( \sec {A \over 2} \right)^2 \\ & = \sec^2 {A \over 2} \\ & = \text{L.H.S} \end{align}

(d)

\begin{align} \text{R.H.S} & = \cot A - \tan A \\ & = {\cos A \over \sin A} - {\sin A \over \cos A} \\ & = {\cos A (\cos A) \over \sin A (\cos A)} - {\sin A (\sin A) \over \cos A (\sin A)} \\ & = {\cos^2 A \over \sin A \cos A} - {\sin^2 A \over \sin A \cos A} \\ & = {\cos^2 A - \sin^2 A \over \sin A \cos A} \\ & = {\cos 2A \over \sin A \cos A} \phantom{00000000} [\text{Double angle formula}] \\ & = {\cos 2A \over \sin A \cos A} \times {2 \over 2} \\ & = {2\cos 2A \over 2\sin A \cos A} \\ & = {2\cos 2A \over \sin 2A} \phantom{000000000.} [\text{Double angle formula}] \\ & = 2 \left(\cos 2A \over \sin 2A\right) \\ & = 2 \cot 2A \\ & = \text{L.H.S} \end{align}

(e)

\begin{align} \cos 2A & = 2\cos^2 A - 1 = 1 - 2\sin^2 A \\ \\ \text{Let } & A = {A \over 2}, \\ \cos \left[2\left( {A \over 2} \right)\right] & = 2\cos^2 {A \over 2} - 1 = 1 - 2\sin^2 {A \over 2} \\ \cos A & = 2\cos^2 {A \over 2} - 1 = 1 - 2\sin^2 {A \over 2} \end{align}
\begin{align} \require{cancel} \text{L.H.S } & = {1 + \cos A \over 1 - \cos A} \\ & = {1 + \left( 2\cos^2 {A \over 2} - 1 \right) \over 1 - \cos A} \\ & = {2\cos^2 {A \over 2} \over 1 - \cos A} \\ & = {2\cos^2 {A \over 2} \over 1 - \left( 1 - 2\sin^2 {A \over 2} \right)} \\ & = {\cos^2 {A \over 2} \over 1 - 1 + 2\sin^2 {A \over 2} } \\ & = {\cancel{2}\cos^2 {A \over 2} \over \cancel{2}\sin^2 {A \over 2}} \\ & = \left( {\cos {A \over 2} \over \sin {A \over 2}} \right)^2 \\ & = \left( \cot {A \over 2} \right)^2 \\ & = \cot^2 {A \over 2} \\ & = \text{R.H.S} \end{align}

(i)

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ (2)^2 & = (\sqrt{3})^2 + x^2 \\ 4 & = 3 + x^2 \\ x^2 & = 1 \\ x & = \pm \sqrt{1} \\ x & = 1 \text{ (Reject) or } -1 \\ \\ \\ \cos 2A & = 1 - 2\sin^2 A \\ & = 1 - 2 \left( {\sqrt{3} \over 2} \right)^2 \\ & = -{1 \over 2} \end{align}

(ii)

\begin{align} \cos 2A & = 2\cos^2 A - 1 \\ \\ \text{Let } & A = 2A, \\ \cos [2(2A)] & = 2\cos^2 2A - 1 \\ \cos 4A & = 2\cos^2 2A - 1 \\ & = 2\left( -{1 \over 2} \right)^2 - 1 \phantom{00000000} [\text{From part i}] \\ & = -{1 \over 2} \end{align}

(iii)

\begin{align} \sin 2A & = 2\sin A \cos A \\ \\ \text{Let } & A = 2A, \\ \sin [2(2A)] & = 2\sin 2A \cos 2A \\ \sin 4A & = 2\sin 2A \cos 2A \\ & = 2 (2 \sin A \cos A) \left(-{1 \over 2}\right) \\ & = 2 (2) \left(\sqrt{3} \over 2\right)\left(-{1 \over 2}\right)\left(-{1 \over 2}\right) \\ & = 4 \left(\sqrt{3} \over 2\right)\left(-{1 \over 2}\right)\left(-{1 \over 2}\right) \\ & = 2 \sqrt{3} \left(-{1 \over 2}\right) \left(-{1 \over 2}\right) \\ & = - \sqrt{3} \left(-{1 \over 2}\right) \\ & = {\sqrt{3} \over 2} \end{align}

(iv)

\begin{align} \cos 2A & = 2\cos^2 A - 1 \\ \\ \text{Let } & A = {A \over 2}, \\ \cos \left[ 2\left({A \over 2}\right) \right] & = 2\cos^2 {A \over 2} - 1 \\ \cos A & = 2\cos^2 {A \over 2} - 1 \\ \\ 2\cos^2 {A \over 2} & = \cos A + 1 \\ \cos^2 {A \over 2} & = {1 \over 2} \cos A + {1 \over 2} \\ \cos {A \over 2} & = \pm\sqrt{{1 \over 2} \cos A + {1 \over 2}} \\ \\ \\ \text{Since } 90^\circ < A < 180^\circ, & \phantom{.} 45^\circ < {A \over 2} < 90^\circ \text{ and } \cos {A \over 2} > 0, \\ \\ \therefore \cos {A \over 2} & = \sqrt{{1 \over 2} \cos A + {1 \over 2}} \\ & = \sqrt{{1 \over 2} \cos A + {1 \over 2}} \\ & = \sqrt{{1 \over 2} \cos A + {1 \over 2}} \\ & = \sqrt{{1 \over 2}\left(-{1 \over 2}\right) + {1 \over 2}} \\ & = {1 \over 2} \end{align}

(i)

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ x^2 & = (-2)^2 + (-3)^2 \\ x^2 & = 4 + 9 \\ x^2 & = 13 \\ x & = \pm \sqrt{13} \\ x & = \sqrt{13} \text{ or } -\sqrt{13} \text{ (Reject)} \\ \\ \\ \tan 2A & = {2\tan A \over 1 - \tan^2 A} \\ & = {2\left({2 \over 3}\right) \over 1 - \left({2 \over 3}\right)^2} \\ & = {12 \over 5} \end{align}

(ii)

\begin{align} \tan 3A & = \tan (A + 2A) \\ & = {\tan A + \tan 2A \over 1 - \tan A \tan 2A} \phantom{000000} [\text{Addition formula: } \tan (A + B)] \\ & = {\left({2 \over 3}\right) + \left({12 \over 5}\right) \over 1 - \left({2 \over 3}\right)\left({12 \over 5}\right)} \\ & = -{46 \over 9} \end{align}

(iii)

\begin{align} \tan 2A & = {2\tan A \over 1 - \tan^2 A} \\ \\ \text{Let } & A = {A \over 2}, \\ \tan \left[2\left({A \over 2}\right)\right] & = {2\tan {A \over 2} \over 1 - \tan^2 {A \over 2}} \\ \tan A & = {2\tan {A \over 2} \over 1 - \tan^2 {A \over 2}} \\ {2 \over 3} & = {2\tan {A \over 2} \over 1 - \tan^2 {A \over 2}} \\ 2\left(1 - \tan^2 {A \over 2}\right) & = 3\left(2\tan {A \over 2}\right) \\ 2 - 2\tan^2 {A \over 2} & = 6\tan {A \over 2} \\ -2\tan^2 {A \over 2} - 6\tan {A \over 2} + 2 & = 0 \\ 2\tan^2 {A \over 2} + 6\tan {A \over 2} -2 & = 0 \\ \tan^2 {A \over 2} + 3\tan {A \over 2} - 1 & = 0 \\ \left(\tan {A \over 2}\right)^2 + 3 \left(\tan {A \over 2}\right) - 1 & = 0 \\ \\ \\ \tan {A \over 2} & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(3) \pm \sqrt{(3)^2 - 4(1)(-1)} \over 2(1)} \\ & = {-3 \pm \sqrt{13} \over 2} \\ \\ \\ \text{Since } \pi < A < {3\pi \over 2}, & \phantom{.} {\pi \over 2} < {A \over 2} < {3\pi \over 4} \text{ and } \tan {A \over 2} < 0 \\ \\ \therefore \tan {A \over 2} & = {-3 - \sqrt{13} \over 2} \end{align}

(iv)

\begin{align} \text{cosec } 2A & = {1 \over \sin 2A} \\ & = {1 \over 2\sin A \cos A} \phantom{0000000000} [\text{Double angle formula}] \\ & = {1 \over 2\left(-{2 \over \sqrt{13}}\right)\left(-{3 \over \sqrt{13}}\right)} \\ & = {13 \over 12} \end{align}

(i)

\begin{align} \text{By Py} & \text{thagoras theorem,} \\ x^2 & = (k)^2 + (1)^2 \\ x^2 & = k^2 + 1 \\ x & = \pm \sqrt{k^2 + 1} \\ x & = \sqrt{k^2 + 1} \text{ or } -\sqrt{k^2 + 1} \text{ (Reject)} \\ \\ \\ \sin 2\theta & = 2\sin \theta \cos \theta \\ & = 2\left({1 \over \sqrt{k^2 + 1}}\right)\left({k \over \sqrt{k^2 + 1}}\right) \\ & = {2k \over k^2 + 1} \end{align}

(ii)

\begin{align} \cot (\theta + 45^\circ) & = {1 \over \tan (\theta + 45^\circ)} \\ & = {1 \over {\tan \theta + \tan 45^\circ \over 1 - \tan \theta \tan 45^\circ}} \\ & = {1 - \tan \theta \tan 45^\circ \over \tan \theta + \tan 45^\circ} \\ & = {1 - \left({1 \over k}\right)(1) \over \left({1 \over k}\right) + (1)} \\ & = {1 - {1 \over k} \over {1 \over k} + 1 } \\ & = {1 - {1 \over k} \over {1 \over k} + 1 } \times {k \over k} \\ & = {k - 1 \over 1 + k} \end{align}

(iii)

\begin{align} \sec 2\theta & = {1 \over \cos 2\theta} \\ & = {1 \over \cos^2 \theta - \sin^2 \theta} \\ & = {1 \over \left( {k \over \sqrt{k^2 + 1}} \right)^2 - \left( {1 \over \sqrt{k^2 + 1}} \right)^2} \\ & = {1 \over {k^2 \over k^2 + 1} - {1 \over k^2 + 1}} \\ & = {1 \over {k^2 - 1 \over k^2 + 1} } \\ & = {k^2 + 1 \over k^2 - 1} \end{align}

(iv)

\begin{align} \require{cancel} \tan 3\theta & = {\tan (\theta + 2\theta) } \\ & = {\tan \theta + \tan 2\theta \over 1 - \tan \theta \tan 2\theta} \phantom{00000000} [\text{Addition formula: } \tan (A + B)] \\ \\ \\ \tan 2\theta & = {2\tan \theta \over 1 - \tan^2 \theta} \\ & = {2\left({1 \over k}\right) \over 1 - \left({1 \over k}\right)^2} \\ & = {{2 \over k} \over 1 - {1 \over k^2}} \\ & = {{2 \over k} \over {k^2 \over k^2} - {1 \over k^2}} \\ & = {{2 \over k} \over {k^2 - 1 \over k^2}} \\ & = {2 \over k} \div {k^2 - 1 \over k^2} \\ & = {2 \over \cancel{k}} \times {k^\cancel{2} \over k^2 - 1} \\ & = {2k \over k^2 - 1} \\ \\ \\ \therefore \tan 3\theta & = {\tan \theta + \tan 2\theta \over 1 - \tan \theta \tan 2\theta} \\ & = {{1 \over k} + {2k \over k^2 - 1} \over 1 - \left( {1 \over k} \right) \left( {2k \over k^2 - 1} \right)} \\ & = {{k^2 + 1 \over k(k^2 + 1)} + {2k(k) \over k(k^2 + 1)} \over 1 - {2 \over k^2 - 1}} \\ & = {{k^2 - 1 + 2k^2 \over k(k^2 - 1)} \over {k^2 - 1\over k^2 - 1} - {2 \over k^2 - 1}} \\ & = {{3k^2 - 1 \over k(k^2 - 1)} \over {k^2 - 3 \over k^2 - 1}} \\ & = {3k^2 - 1 \over k(k^2 - 1)} \div {k^2 - 3 \over k^2 - 1} \\ & = {3k^2 - 1 \over k\cancel{(k^2 - 1)}} \times {\cancel{k^2 - 1} \over k^2 - 3} \\ & = {3k^2 - 1 \over k(k^2 - 3)} \end{align}

(a)

\begin{align} \sin 2A & = 2\sin A \cos A & \cos 2A & = 2\cos^2 A - 1 \\ \\ \text{Let } & A = {A \over 2}, \\ \sin \left[2 \left({A \over 2}\right) \right] & = 2\sin {A \over 2} \cos {A \over 2} & \cos \left[2 \left({A \over 2}\right) \right] & = 2\cos^2 {A \over 2} - 1 \\ \sin A & = 2\sin {A \over 2} \cos {A \over 2} & \cos A & = 2\cos^2 {A \over 2} - 1 \end{align}
\begin{align} \require{cancel} \text{L.H.S } & = {\sin A \over 1 + \cos A} \\ & = {2\sin {A \over 2} \cos {A \over 2} \over 1 + 2\cos^2 {A \over 2} - 1} \\ & = {\cancel{2}\sin {A \over 2} \cancel{\cos {A \over 2}} \over \cancel{2}\cos^\cancel{2} {A \over 2}} \\ & = {\sin {A \over 2} \over \cos {A \over 2}} \\ & = \tan {A \over 2} \\ & = \text{R.H.S} \end{align}

(b)

\begin{align} \require{cancel} \text{L.H.S } & = {1 + \tan^2 A \over 1 - \tan^2 A} \\ & = {1 + {\sin^2 A \over \cos^2 A} \over 1 - {\sin^2 A \over \cos^2 A}} \\ & = {{\cos^2 A \over \cos^2 A} + {\sin^2 A \over \cos^2 A} \over {\cos^2 A \over \cos^2 A} - {\sin^2 \over \cos^2 A}} \\ & = {{\cos^2 A + \sin^2 A \over \cos^2 A} \over {\cos^2 A - \sin^2 A \over \cos^2 A}} \\ & = {\cos^2 A + \sin^2 A \over \cos^2 A} \div {\cos^2 A - \sin^2 A \over \cos^2 A} \\ & = {\cos^2 A + \sin^2 A \over \cancel{\cos^2 A}} \times {\cancel{\cos^2 A} \over \cos^2 A - \sin^2 A} \\ & = {\cos^2 A + \sin^2 A \over \cos^2 A - \sin^2 A} \\ & = {1 \over \cos^2 A - \sin^2 A} \phantom{000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1 ] \\ & = {1 \over \cos 2A} \phantom{0000000000000} [\text{Double angle formula}] \\ & = \sec 2A \\ & = \text{R.H.S} \end{align}

(c)

\begin{align} \text{L.H.S } & = \cos^4 A - \sin^4 A \\ & = (\cos^2 A + \sin^2 A)(\cos^2 A - \sin^2 A) \phantom{000000} [ a^2 - b^2 = (a + b)(a - b)] \\ & = (1)(\cos^2 A - \sin^2 A) \phantom{00000000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = \cos^2 A - \sin^2 A \\ & = \cos 2A \phantom{000000000000000000000000000000.} [\text{Double angle formula}] \\ & = \text{R.H.S} \end{align}

(d)

\begin{align} \text{L.H.S } & = \cos 4A \\ & = \cos [2(2A)] \\ & = \cos^2 2A - \sin^2 2A \phantom{0000000000000000000000000} [\text{Double angle formula}] \\ & = (\cos 2A)^2 - (\sin 2A)^2 \\ & = (1 - 2\sin^2 A)^2 - (2 \sin A \cos A)^2 \phantom{0000000000000.} [\text{Double angle formula}] \\ & = 1 - 4\sin^2 A + 4\sin^4 A - 4\sin^2 A \cos^2 A \\ & = 1 - 4\sin^2 A + 4\sin^4 A - 4\sin^2 A (1 - \sin^2 A) \phantom{000.} [\text{Since } \sin^2 A + \cos^2 A = 1 \rightarrow \cos^2 A = 1 - \sin^2 A] \\ & = 1 - 4\sin^2 A + 4\sin^4 A - 4\sin^2 A + 4\sin^4 A \\ & = 1 - 8\sin^2 A + 8\sin^4 A \\ & = \text{R.H.S} \end{align}

(e)

\begin{align} \text{R.H.S } & = {1 \over 8}(3 - 4\cos 2A + \cos 4A) \\ \\ & = {1 \over 8} \left\{ 3 - 4\cos 2A + \cos [2(2A)] \right\} \\ \\ & = {1 \over 8}[3 - 4\cos 2A + (2\cos^2 2A - 1)] \phantom{000000000000} [\text{Double angle formula}] \\ \\ & = {1 \over 8}(3 - 4\cos 2A + 2 \cos^2 2A - 1) \\ \\ & = {1 \over 8}(2 - 4\cos 2A + 2\cos^2 2A) \\ \\ & = {1 \over 8}[2 - 4(1 - 2\sin^2 A) + 2(1 - 2\sin^2 A)^2] \phantom{000000} [\text{Double angle formula}] \\ \\ & = {1 \over 8}[2 - 4 + 8\sin^2 A + 2(1 - 4\sin^2 A + 4\sin^4 A)] \\ \\ & = {1 \over 8}[-2 + 8\sin^2 A + 2 - 8\sin^2 A + 8\sin^4 A] \\ \\ & = {1 \over 8}(8\sin^4 A) \\ \\ & = \sin^4 A \\ \\ & = \text{L.H.S} \end{align}

(f)

\begin{align} \require{cancel} \text{L.H.S } & = {1 + \cos 2A + \sin 2A \over 1 - \cos 2A + \sin 2A} \\ & = {1 + (2\cos^2 A - 1) + 2\sin A \cos A \over 1 - (1 - 2\sin^2 A) + 2\sin A \cos A} \\ & = {2\cos^2 A + 2\sin A \cos A \over 2\sin^2 A + 2\sin A \cos A} \\ & = {\cancel{2}\cos A \cancel{(\cos A + \sin A)} \over \cancel{2}\sin A \cancel{(\sin A + \cos A)} } \\ & = {\cos A \over \sin A} \\ & = \cot A \\ & = \text{R.H.S} \end{align}

(a)

\begin{align} 4\sin 2x & = \sin x \\ [\text{Double angle formula}] \phantom{000000} 4(2\sin x \cos x) & = \sin x \\ 8 \sin x \cos x & = \sin x \\ 8 \sin x \cos x - \sin x & = 0 \\ (\sin x)(8 \cos x - 1) & = 0 \\ \\ \sin x = 0 \phantom{00} & \text{or} \phantom{000} 8\cos x - 1 = 0 \end{align}

$$ \sin x = 0 $$

$$ x = 0^\circ \text{ (Reject)}, 180^\circ, 360^\circ \text{ (Reject)} $$

\begin{align} 8\cos x - 1 & = 0 \\ 8\cos x & = 1 \\ \cos x & = {1 \over 8} \phantom{000000} [\text{1st & 4th quadrants since } \cos x > 0 ] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} \left( {1 \over 8} \right) \\ & = 82.82^\circ \end{align}

\begin{align} x & = 82.82^\circ, 360^\circ - 82.82^\circ \\ & = 82.82^\circ, 277.18^\circ \\ & \approx 82.8^\circ, 277.2^\circ \\ \\ \\ \therefore x & = 82.8^\circ, 180^\circ, 277.2^\circ \end{align}

(b)

\begin{align} 4\sin x \cos x & = 1 \\ [\text{Apply double angle formula}] \phantom{000000} 2\sin x \cos x & = 0.5 \\ \sin 2x & = 0.5 \phantom{000000} [\text{1st & 2nd quadrants since } \sin 2x > 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} (0.5) \\ & = 30^\circ \end{align}

\begin{align} 0^\circ < \phantom{.} & x < 360^\circ \\ 0^\circ < \phantom{.} & 2x < 720^\circ \\ \\ 2x & = 30^\circ, 180^\circ - 30^\circ \\ & = 30^\circ, 150^\circ, 30^\circ + 360^\circ, 150^\circ + 360^\circ \\ & = 30^\circ, 150^\circ, 390^\circ, 510^\circ \\ \\ x & = 15^\circ, 75^\circ, 195^\circ, 255^\circ \end{align}

(c)

\begin{align} (\sin x - \cos x)^2 & = 2 \\ \sin^2 x - 2\sin x \cos x + \cos^2 x & = 2 \\ \sin^2 x + \cos^2 x - 2\sin x \cos x & = 2 \\ [\text{Identity: } \sin^2 A + \cos^2 A = 1] \phantom{000000} 1 - 2\sin x \cos x & = 2 \\ [\text{Double angle formula}] \phantom{000000000} 1 - (\sin 2x) & = 2 \\ - \sin 2x & = 1 \\ \sin 2x & = - 1 \end{align}

\begin{align} 0^\circ < \phantom{.} & x < 360^\circ \\ 0^\circ < \phantom{.} & 2x < 720^\circ \\ \\ 2x & = 270^\circ \\ & = 270^\circ, 270^\circ + 360^\circ \\ & = 270^\circ, 630^\circ \\ \\ x & = 135^\circ, 315^\circ \end{align}

(d)

\begin{align} \cos 2x - 3\cos x + 2 & = 0 \\ (2\cos^2 x - 1) - 3\cos x + 2 & = 0 \\ 2\cos^2 x - 3\cos x + 1 & = 0 \\ (2\cos x - 1)(\cos x - 1) & = 0 \\ \\ 2\cos x - 1 = 0 \phantom{00} & \text{or} \phantom{000} \cos x - 1 = 0 \end{align}

\begin{align} 2\cos x - 1 & = 0 \\ 2\cos x & = 1 \\ \cos x & = {1 \over 2} \\ \cos x & = 0.5 \phantom{000000} [\text{1st & 4th quadrants since } \cos x >0] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.5) \\ & = 60^\circ \end{align}

\begin{align} x & = 60^\circ, 360^\circ - 60^\circ \\ & = 60^\circ, 300^\circ \end{align}

\begin{align} \cos x - 1 & = 0 \\ \cos x & = 1 \end{align}

\begin{align} x & = 0^\circ \text{ (Reject)}, 360^\circ \text{ (Reject)} \\ \\ \\ \therefore x & = 60^\circ, 300^\circ \end{align}

(e)

\begin{align} \cos 2x - 1 & = 3 \sin x \\ (1 - 2\sin^2 x) - 1 & = 3 \sin x \\ 1 - 2\sin^2 x - 1 & = 3\sin x \\ -2 \sin^2 x & = 3 \sin x \\ -2 \sin^2 x - 3\sin x & = 0 \\ 2\sin^2 x + 3 \sin x & = 0 \\ (\sin x)(2\sin x + 3) & = 0 \\ \\ \sin x = 0 \phantom{00} & \text{or} \phantom{000} 2\sin x + 3 = 0 \\ & \phantom{or000+3} 2\sin x = -3 \\ & \phantom{or000+32} \sin x = -{3 \over 2} \text{ (Reject, since } -1 \le \sin x \le 1) \end{align}

$$ \sin x = 0 $$

\begin{align} x & = 0^\circ \text{ (Reject)}, 180^\circ, 360^\circ \text{ (Reject)} \\ \\ \\ \therefore x & = 180^\circ \end{align}

(f)

\begin{align} \tan 2x \tan x & = 3 \\ \left( {2\tan x \over 1 - \tan^2 x} \right)\tan x & = 3 \\ {2\tan^2 x \over 1 - \tan^2 x} & = 3 \\ 2\tan^2 x & = 3(1 - \tan^2 x) \\ 2\tan^2 x & = 3 - 3\tan^2 x \\ 5\tan^2 x & = 3 \\ \tan^2 x & = {3 \over 5} \\ \tan x & = \pm\sqrt{3 \over 5} \phantom{000000} [\text{All 4 quadrants}] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left( \sqrt{3 \over 5} \right) \\ & = 37.76^\circ \end{align}

\begin{align} x & = 37.76^\circ, 180^\circ - 37.76^\circ, 180^\circ + 37.76^\circ, 360^\circ - 37.76^\circ \\ & = 37.76^\circ, 142.24^\circ, 217.76^\circ, 322.24^\circ \\ & \approx 37.8^\circ, 142.2^\circ, 217.8^\circ, 322.2^\circ \end{align}

(a)

\begin{align} 2\sin x & = \sin 2x \\ 2\sin x & = 2\sin x \cos x \phantom{000000} [\text{Double angle formula}] \\ 2\sin x - 2\sin x \cos x & = 0 \\ (2\sin x)(1 - \cos x) & = 0 \\ \\ 2\sin x = 0 \phantom{00} & \text{or} \phantom{000} 1 - \cos x = 0 \end{align}

\begin{align} 2 \sin x & = 0 \\ \sin x & = 0 \end{align}

\begin{align} x & = 0, \pi, 2\pi \end{align}

\begin{align} 1 - \cos x & = 0 \\ -\cos x & = - 1 \\ \cos x & = 1 \end{align}

\begin{align} x & = 0, 2\pi \\ \\ \\ \therefore x & = 0, \pi, 2\pi \end{align}

(b)

\begin{align} 5\sin x \cos^2 x + 2\cos x & = 0 \\ (\cos x)(5\sin x \cos x + 2) & = 0 \\ \\ \cos x = 0 \phantom{00} & \text{or} \phantom{000} 5\sin x \cos x + 2 = 0 \end{align}

$$ \cos x = 0 $$

\begin{align} x & = {\pi \over 2}, {3\pi \over 2} \end{align}

\begin{align} 5\sin x \cos x + 2 & = 0 \\ 5\sin x \cos x & = -2 \\ \sin x \cos x & = {-2 \over 5} \\ \sin x \cos x & = -0.4 \\ 2\sin x \cos x & = -0.8 \\ [\text{Double angle formula}] \phantom{000000} \sin 2x & = -0.8 \phantom{000000} [\text{3rd & 4th quadrant since } \sin 2x < 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} (0.8) \\ & = 0.9273 \end{align}

\begin{align} 0 \le \phantom{.} & x \le 2\pi \\ 0 \le \phantom{.} & 2x \le 4\pi \\ 0 \le \phantom{.} & 2x \le 12.566 \\ \\ 2x & = \pi + 0.9273, 2\pi - 0.9273 \\ 2x & = 4.069, 5.356, 4.069 + 2\pi, 5.356 + 2\pi \\ 2x & = 4.069, 5.356, 10.352, 11.639 \\ \\ x & = 2.0345, 2.678, 5.176, 5.8195 \\ x & \approx 2.03, 2.68, 5.18, 5.82 \\ \\ \\ \therefore x & = {\pi \over 2}, {3\pi \over 2}, 2.03, 2.68, 5.18, 5.82 \end{align}

(c)

\begin{align} \cos 2x & = -\sin x \\ 1 - 2\sin^2 x & = -\sin x \\ -2\sin^2 x + \sin x + 1 & = 0 \\ 2\sin^2 x - \sin x - 1 & = 0 \\ (2\sin x + 1)(\sin x - 1) & = 0 \\ \\ 2\sin x + 1 = 0 \phantom{00} & \text{or} \phantom{000} \sin x - 1 = 0 \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{3rd & 4th quadrants since } \sin x < 0 ] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} (0.5) \\ & = 30^\circ \\ & = {\pi \over 6} \end{align}

\begin{align} x & = \pi + {\pi \over 6}, 2\pi - {\pi \over 6} \\ & = {7\pi \over 6}, {11\pi \over 6} \end{align}

\begin{align} \sin x - 1 & = 0 \\ \sin x & = 1 \end{align}

\begin{align} x & = {\pi \over 2} \\ \\ \\ \therefore x & = {\pi \over 2}, {7\pi \over 6}, {11\pi \over 6} \end{align}

(d)

\begin{align} 2\cos 2x & = 11 \cos x + 1 \\ 2(2\cos^2 x - 1) & = 11\cos x + 1 \\ 4\cos^2 x - 2 & = 11\cos x + 1 \\ 4\cos^2 x - 11\cos x - 3 & = 0 \\ (4\cos x + 1)(\cos x - 3) & = 0 \\ \\ 4\cos x + 1 = 0 \phantom{00} & \text{or}\phantom{000} \cos x - 3 = 0 \\ & \phantom{or000-3} \cos x = 3 \text{ (Reject, since } -1 \le \cos x \le 1] \end{align}

\begin{align} 4\cos x + 1 & = 0 \\ 4\cos x & = -1 \\ \cos x & = -{1 \over 4} \\ \cos x & = -0.25 \phantom{000000} [\text{2nd & 3rd quadrants since } \cos x < 0] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.25) \\ & = 1.318 \end{align}

\begin{align} x & = \pi - 1.138, \pi + 1.138 \\ & \approx 1.82, 4.46 \end{align}

(e)

\begin{align} \tan 2x & = 3\tan x \\ {2\tan x \over 1 - \tan^2 x} & = {3 \tan x \over 1} \\ 2\tan x & = 3\tan x(1 - \tan^2 x) \\ 2\tan x & = 3\tan x - 3\tan^3 x \\ 0 & = \tan x - 3\tan^3 x \\ 0 & = \tan x (1 - 3\tan^2 x) \\ \\ \tan x = 0 \phantom{00} & \text{or} \phantom{000} 1 - 3\tan^2 x = 0 \end{align}

$$ \tan x = 0 $$

\begin{align} x & = 0, \pi, 2\pi \end{align}

\begin{align} 1 - 3\tan^2 x & = 0 \\ -3\tan^2 x & = -1 \\ \tan^2 x & = {-1 \over -3} \\ \tan^2 x & = {1 \over 3} \\ \tan x & = \pm \sqrt{1 \over 3} \phantom{000000} [\text{All 4 quadrants}] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left( \sqrt{1 \over 3} \right) \\ & = {\pi \over 6} \end{align}

\begin{align} x & = {\pi \over 6}, \pi - {\pi \over 6}, \pi + {\pi \over 6}, 2\pi - {\pi \over 6} \\ & = {\pi \over 6}, {5\pi \over 6}, {7\pi \over 6}, {11\pi \over 6} \\ \\ \\ \therefore x & = 0, \pi, 2\pi, {\pi \over 6}, {5\pi \over 6}, {7\pi \over 6}, {11\pi \over 6} \end{align}

(i)

\begin{align} y & = 3\sin 2x + 2\cos^2 x \\ y & = 3(2\sin x \cos x) + 2\cos^2 x \\ y & = 6\sin x \cos x + 2\cos^2 x \\ \\ \text{When } & y = 0, \\ y & = 6\sin x \cos x + 2\cos^2 x \\ 0 & = 6\sin x \cos x + 2\cos^2 x \\ 0 & = (2\cos x)(3\sin x + \cos x) \\ \\ 2\cos x = 0 \phantom{00} & \text{or} \phantom{000} 3\sin x + \cos x = 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} 3\sin x + \cos x & = 0 \\ 3\sin x & = -\cos x \\ {3\sin x \over \cos x} & = -1 \\ {\sin x \over \cos x} & = -{1 \over 3} \\ \tan x & = -{1 \over 3} \phantom{000000} [\text{2nd & 4th quadrants since } \tan x < 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left( {1 \over 3} \right) \\ & = 18.43^\circ \end{align}

\begin{align} x & = 180^\circ - 18.43^\circ, 360^\circ - 18.43^\circ \\ & = 161.57^\circ, 341.57^\circ \\ & \approx 161.6^\circ, 341.6^\circ \\ \\ \\ x & = 90^\circ, 161.6^\circ, 270^\circ, 341.6^\circ \end{align}

(ii)

\begin{align} \text{When } & y = 2, \\ 2 & = 6 \sin x \cos x + 2\cos^2 x \\ 1 & = 3 \sin x \cos x + \cos^2 x \\ 1 & = 3\sin x \cos x + (1 - \sin^2 x) \phantom{000000} [\text{Since } \sin^2 A + \cos^2 A = 1 \rightarrow \cos^2 A = 1 - \sin^2 A] \\ 1 & = 3\sin x \cos x + 1 - \sin^2 x \\ 0 & = 3\sin x \cos x - \sin^2 x \\ 0 & = (\sin x)(3\cos x - \sin x) \\ \\ \sin x = 0 \phantom{00} & \text{or} \phantom{000} 3\cos x - \sin x = 0 \end{align}

$$ \sin x = 0 $$

\begin{align} x & = 0^\circ \text{ (Reject)}, 180^\circ, 360^\circ \text{ (Reject)} \end{align}

\begin{align} 3\cos x - \sin x & = 0 \\ -\sin x & = -3\cos x \\ \sin x & = 3\cos x \\ {\sin x \over \cos x} & = 3 \\ \tan x & = 3 \phantom{00000000} [\text{1st & 3rd quadrants since } \tan x > 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (3) \\ & = 71.57^\circ \end{align}

\begin{align} x & = 71.57^\circ, 180^\circ + 71.57^\circ \\ & = 71.57^\circ, 251.57^\circ \\ & \approx 71.6^\circ, 251.6^\circ \\ \\ \\ \therefore x & = 71.6^\circ, 180^\circ, 251.6^\circ \end{align}

(i)

\begin{align} \cos 2x & = \cos^2 x - \sin^2 x \\ & = (\cos x + \sin x)(\cos x - \sin x) \\ & = (a)(b) \\ & = ab \end{align}
\begin{align} \sin 2x & = 2\sin x \cos x \\ & = 2\sin x \cos x + 1 - 1 \\ & = 2\sin x \cos x + (\sin^2 x + \cos^2 x) - 1 \phantom{000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = \sin^2 x + 2\sin x \cos x + \cos^2 x - 1 \\ & = (\sin x + \cos x)^2 - 1 \\ & = (a)^2 - 1 \\ & = a^2 -1 \end{align}
\begin{align} \text{Alternatively, } \sin 2x & = 2\sin x \cos x \\ -\sin 2x & = -2\sin x \cos x \\ -\sin 2x & = -2\sin x \cos x + 1 - 1 \\ -\sin 2x & = -2\sin x \cos x + (\sin^2 x + \cos^2 x) - 1 \\ -\sin 2x & = -2\sin x \cos x + \sin^2 x + \cos^2 x - 1 \\ -\sin 2x & = \sin^2 x -2\sin x \cos x + \cos^2 x - 1 \\ -\sin 2x & = (\sin x - \cos x)^2 - 1 \\ -\sin 2x & = (b)^2 - 1 \\ -\sin 2x & = b^2 - 1 \\ \sin 2x & = -(b^2 - 1) \\ \sin 2x & = 1 - b^2 \end{align}

(ii)

\begin{align} a^2 + b^2 & = (\cos x + \sin x)^2 + (\cos x - \sin x)^2 \\ & = \cos^2 x + 2\sin x \cos x + \sin^2 x + \cos^2 x - 2\sin x \cos x + \sin^2 x \\ & = 2\cos^2 x + 2\sin^2 x \\ & = 2(\cos^2 x + \sin^2 x) \\ & = 2(1) \phantom{000000000000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = 2 \end{align}

Show question

\begin{align} \text{L.H.S } & = \cos 3x \\ & = \cos (2x + x) \\ & = \cos 2x \cos x - \sin 2x \sin x \phantom{0000000000000000000} [\text{Addition formula: } \cos (A + B)] \\ & = (2\cos^2 x - 1)(\cos x) - (2\sin x \cos x)(\sin x) \phantom{0000} [\text{Double angle formula}] \\ & = 2\cos^3 x - \cos x - 2\sin^2 x \cos x \\ & = 2\cos^3 x - \cos x - 2(1 - \cos^2 x) \cos x \phantom{000000000.} [\text{From } \sin^2 A + \cos^2 A = 1 \rightarrow \sin^2 A = 1 - \cos^2 A] \\ & = 2\cos^3 x - \cos x - 2\cos x + 2 \cos^3 x \\ & = 4\cos^3 x - 3\cos x \\ & = \text{R.H.S} \end{align}

(i)

\begin{align} \cos 3x & = 4\cos^3 x - 3\cos x \phantom{000000} [\text{Proved earlier}] \\ \\ \therefore \cos [3(10)]^\circ & = 4\cos^3 10^\circ - 3\cos 10^\circ \\ \cos 30^\circ & = 4\cos^3 10^\circ - 3\cos 10^\circ \\ \\ \therefore 8\cos^3 10^\circ - 6\cos 10^\circ & = 2(4\cos^3 10^\circ - 3\cos 10^\circ) \\ & = 2(\cos 30^\circ) \\ & = 2\left( {\sqrt{3} \over 2} \right) \phantom{0000000} [\text{Special value: } \cos 30^\circ] \\ & = \sqrt{3} \end{align}

(ii)

\begin{align} 8\cos^3 x & = 6\cos x - 1 \\ 8\cos^3 x - 6\cos x & = - 1 \\ 4\cos^3 x - 3\cos x & = -0.5 \\ \\ \therefore \cos 3x & = -0.5 \phantom{000000} [\text{2nd & 3rd quadrants since } \cos 3x < 0 ] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.5) \\ & = 60^\circ \end{align}

\begin{align} 0^\circ < \phantom{.} & x < 180^\circ \\ 0^\circ < \phantom{.} & 3x < 540^\circ \\ \\ 3x & = 180^\circ - 60^\circ, 180^\circ + 60^\circ \\ & = 120^\circ, 240^\circ, 120^\circ + 360^\circ \\ & = 120^\circ, 240^\circ, 480^\circ \\ \\ x & = 40^\circ, 80^\circ, 160^\circ \end{align}

Prove question

\begin{align} \require{cancel} \text{R.H.S } & = \cot x - 2\cot 2x \\ \\ & = {1 \over \tan x} - 2\left( {1 \over \tan 2x} \right) \\ \\ & = {1 \over \tan x} - {2 \over \tan 2x} \\ \\ & = {1 \over \tan x} - {2 \over {2\tan x \over 1 - \tan^2 x}} \phantom{00000000} [\text{Double angle formula}] \\ \\ & = {1 \over \tan x} - \left( 2 \div {2\tan x \over 1 - \tan^2 x} \right) \\ \\ & = {1 \over \tan x} - \left( \cancel{2} \times {1 - \tan^2 x \over \cancel{2}\tan x} \right) \\ \\ & = {1 \over \tan x} - {1 - \tan^2 x \over \tan x} \\ \\ & = {1 - (1 - \tan^2 x) \over \tan x} \\ \\ & = {1 - 1 + \tan^2 x \over \tan x} \\ \\ & = {\tan^\cancel{2} x \over \cancel{\tan x}} \\ \\ & = \tan x \\ \\ & = \text{L.H.S} \end{align}

Hence question

\begin{align} \tan x & = \cot x - 2\cot 2x \\ \\ \tan 20^\circ & = \cot 20^\circ - 2\cot 40^\circ \\ 2\tan 20^\circ & = 2\cot 20^\circ - 4\cot 40^\circ \\ \\ 4\tan 40^\circ & = 4(\tan 40^\circ) \\ & = 4(\cot 40^\circ - 2\cot 80^\circ) \\ & = 4\cot 40^\circ - 8\cot 80^\circ \\ \\ 8\tan 80^\circ & = 8(\tan 80^\circ) \\ & = 8(\cot 80^\circ - 2\cot 160^\circ) \\ & = 8\cot 80^\circ - 16\cot 160^\circ \\ \\ \\ \text{L.H.S } & = 2\tan 20^\circ + 4\tan 40^\circ + 8 \tan 80^\circ \\ & = (2\cot 20^\circ - 4\cot 40^\circ) + (4\cot 40^\circ - 8\cot 80^\circ) + (8\cot 80^\circ - 16\cot 160^\circ) \\ & = 2\cot 20^\circ - 16\cot 160^\circ \\ & = 2\left( {1 \over \tan 20^\circ} \right) - 16 \left( {1 \over \tan 160^\circ} \right) \\ & = {2 \over \tan 20^\circ} - {16 \over \tan 160^\circ} \\ & = {2 \over \tan 20^\circ} - {16 \over \tan (180^\circ - 20^\circ)} \\ & = {2 \over \tan 20^\circ} - {16 \over -\tan 20^\circ} \phantom{000000000000} [\tan (180^\circ - \theta) = -\tan \theta] \\ & = {2 \over \tan 20^\circ} + {16 \over \tan 20^\circ} \\ & = {18 \over \tan 20^\circ} \\ & = 18 \left( {1 \over \tan 20^\circ} \right) \\ & = 18 \cot 20^\circ \\ & = 9(2 \cot 20^\circ) \\ & = 9(\cot 10^\circ - \tan 10^\circ) \phantom{0000000000000} [\tan x = \cot x - 2\cot 2x \rightarrow 2\cot 2x = \cot x - \tan x]\\ & = \text{R.H.S} \end{align}

\begin{align} \cos 2A & = 2\cos^2 A - 1 \\ \cos 2A + 1 & = 2\cos^2 A \\ \therefore \cos^2 A & = {\cos 2A + 1 \over 2} \phantom{000} \text{ --- (1)} \\ \\ \cos 2A & = 1 - 2\sin^2 A \\ 2\sin^2 A & = 1 - \cos 2A \\ \sin^2 A & = {1 - \cos 2A \over 2} \phantom{000} \text{ --- (2)} \\ \\ \text{From } & (1) \text{, let } A = 2A, \\ \\ \cos^2 2A & = {\cos 4A + 1 \over 2} \phantom{000} \text{ --- (3)} \end{align}
\begin{align} \text{L.H.S } & = \cos^6 \theta + \sin^6 \theta \\ & = (\cos^2 \theta)^3 + (\sin^2 \theta)^3 \phantom{0000000000000000.} [\text{Apply: } a^3 + b^3 = (a + b)(a^2 - ab + b^2)] \\ & = (\cos^2 \theta + \sin^2 \theta)[(\cos^2 \theta)^2 - (\cos^2 \theta)(\sin^2 \theta) + (\sin^2 \theta)^2] \\ & = (1)[(\cos^2 \theta)^2 - (\cos^2 \theta)(\sin^2 \theta) + (\sin^2 \theta)^2] \phantom{000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = (\cos^2 \theta)^2 - (\cos^2 \theta)(\sin^2 \theta) + (\sin^2 \theta)^2 \\ & = \left( {\cos 2\theta + 1 \over 2} \right)^2 - \left( {\cos 2\theta + 1 \over 2} \right) (\sin^2 \theta) + (\sin^2 \theta)^2 \phantom{000000000000000} [\text{Apply (1)}] \\ & = \left( {\cos 2\theta + 1 \over 2} \right)^2 - \left( {\cos 2\theta + 1 \over 2} \right) \left( {1 - \cos 2\theta \over 2} \right) + \left( {1 - \cos 2\theta \over 2} \right)^2 \phantom{000} [\text{Apply (2)}] \\ & = {(\cos 2\theta + 1)^2 \over 4} - {1 - \cos^2 2\theta \over 4} + {(1 - \cos 2\theta)^2 \over 4} \\ & = {(\cos^2 2\theta + 2\cos 2\theta + 1) - (1 - \cos^2 2\theta) + (1 - 2\cos 2\theta + \cos^2 2\theta) \over 4} \\ & = {\cos^2 2\theta + 2\cos 2\theta + 1 - 1 + \cos^2 2\theta + 1 - 2\cos 2\theta + \cos^2 2\theta \over 4} \\ & = {3\cos^2 2\theta + 1 \over 4} \\ & = {3\left( {\cos 4\theta + 1 \over 2} \right) + 1 \over 4} \phantom{0000000000} [\text{Apply (3)}] \\ & = {{3\cos 4\theta + 3 \over 2} + {2 \over 2} \over 4} \\ & = {{3\cos 4\theta + 5 \over 2} \over 4} \\ & = {3\cos 4\theta + 5 \over 2} \div 4 \\ & = {3\cos 4\theta + 5 \over 2} \times {1 \over 4} \\ & = {3\cos 4\theta + 5 \over 8} \\ & = \text{R.H.S} \end{align}

(i)

\begin{align} \cos 2x & = 2\cos^2 x - 1 \\ \cos 2x + 1 & = 2\cos^2 x \\ \therefore \cos^2 x & = {1 \over 2}(\cos 2x + 1) \\ \\ 3 + 4\cos^2 x & = 3 + 4\left( {\cos 2x + 1 \over 2} \right) \\ & = 3 + 2(\cos 2x + 1) \\ & = 3 + 2\cos 2x + 2 \\ & = 2\cos 2x + 5 \\ \\ & \therefore p = 2, q = 5 \end{align}

(ii)

\begin{align} y = 3 + 4\cos^2 x & = 2\cos 2x + 5 \\ \\ \text{Comparing with } & y = a\cos bx + c, \\ \text{Amplitude, } a & = 2 \\ \\ \text{Period } & = {2\pi \over b} \\ & = {2\pi \over 2} \\ & = \pi \\ \\ \text{Center line: } & y = 5 \\ \\ \text{Maximum value} & = 5 + 2 \\ & = 7 \\ \\ \text{Minimum value} & = 5 - 2 \\ & = 3 \end{align}

(iii)

\begin{align} \sin 2x & = 2\sin x \cos x \\ {1 \over 2} \sin 2x & = \sin x \cos x \\ \\ y & = 4 - 3(\sin x \cos x) \\ y & = 4 - 3\left( {1 \over 2}\sin 2x \right) \\ y & = 4 - 1.5 \sin 2x \\ y & = -1.5 \sin 2x + 4 \\ \\ \text{Comparing with } & y = a\sin bx + c, \\ a & = -1.5 \text{ (Shape is inverted)} \\ \\ \text{Amplitude} & = 1.5 \\ \\ \text{Period } & = {2\pi \over b} \\ & = {2\pi \over 2} \\ & = \pi \\ \\ \text{Center line: } & y = 4 \\ \\ \text{Maximum value} & = 4 + 1.5 \\ & = 5.5 \\ \\ \text{Minimum value} & = 4 - 1.5 \\ & = 2.5 \end{align}

(iv)

\begin{align} 4\cos^2 x + 3\sin x \cos x - 1 & = 0 \\ 4\cos^2 x - 1 & = -3\sin x \cos x \\ 4\cos^2 x - 1 + 4 & = -3\sin x \cos x + 4 \\ \\ \underbrace{3 + 4\cos^2 x}_{y = 2\cos x + 5} & = \underbrace{4 - 3\sin x \cos x}_{y = -1.5\sin x + 4} \\ \\ \text{Since graphs meet 4 times} & \text{, there are } 4 \text{ solutions} \end{align}

\begin{align} \cos 2A & = 2 \cos^2 A - 1 \\ \\ \text{Let } & A = {A \over 2}, \\ \cos \left[2 \left(A \over 2\right) \right] & = 2 \cos^2 {A \over 2} - 1 \\ \cos A & = 2 \cos^2 {A \over 2} - 1 \\ \\ \\ \text{Consider } \sqrt{2 + 2 \cos x} & = \sqrt{2 + 2 \left( 2 \cos^2 {x \over 2} - 1 \right)} \\ & = \sqrt{2 + 4 \cos^2 {x \over 2} - 2 } \\ & = \sqrt{4 \cos^2 {x \over 2} } \\ & = -2 \cos {x \over 2}, \text{ since } 135^\circ < {x \over 2} < 180^\circ \text{ and } \cos {x \over 2} < 0 \\ \\ \\ \cos 2A & = 1 - 2\sin^2 A \\ \\ \text{Let } & A = {A \over 4}, \\ \cos \left[2 \left(A \over 4\right) \right] & = 1 - 2 \sin^2 {A \over 4} \\ \cos {A \over 2} & = 1 - 2 \sin^2 {A \over 4} \\ \\ \\ \therefore \sqrt{ 2 + \sqrt{2 + 2 \cos x} } & = \sqrt{ 2 - 2 \cos {x \over 2} } \\ & = \sqrt{ 2 - 2 \left( 1 - 2 \sin^2 {x \over 4} \right) } \\ & = \sqrt{ 2 - 2 + 4 \sin^2 {x \over 4} } \\ & = \sqrt{ 4 \sin^2 {x \over 4} } \\ & = 2 \sin {x \over 4}, \text{ since } 67.5^\circ < {x \over 4} < 45^\circ \text{ and } \sin {x \over 4} > 0 \end{align}