Exercise 5. 2 [ Page 167 ][br][br]1. (a) Define submultiple angle with an example.[br]Solution: [br]If [math] A [/math] be any angle, then [math] \frac{A}{2}, \frac{A}{3}... [/math] etc. are called sub-multiple angles of A.[br]1. (b) Write [math] \sin A, \cos A \text{ and } \tan A [/math] in terms of [math] \frac{A}{ 2} . [/math][br]Solution: [br][math] \displaystyle \sin A = 2\sin \frac{A}{2} \cos \frac{A}{2} [/math] [br][br][math]\displaystyle \sin A = \frac{2\tan \frac{A}{2}}{1+\tan^2 \frac{A}{2}} [/math][br][br][math] \cos A = \cos^2 \frac{A}{2} - \sin^2 \frac{A}{2} [/math][br][br][math] \cos A = 2\cos^2 \frac{A}{2} - 1 [/math][br][br][math] \cos A = 1- 2\sin^2 \frac{A}{2} [/math][br][br][math] \displaystyle \cos A = \frac{1-\tan^2 \frac{A}{2}}{1+\tan^2 \frac{A}{2}} [/math][br][br][math] \displaystyle \tan A = \frac {2\tan \frac{A}{2} }{1-\tan^2 \frac{A}{2} } [/math][br][br]1. (c) Write [math] \sin A [/math] in terms of [math] \sin \frac{A}{3} [/math] and [math] \cos A [/math] in terms of [math] \cos \frac{A}{3} [/math].[br]Solution:[br][math] \sin A = 3\sin \frac{A}{3} - 4\sin^3 \frac{A}{3} [/math][br][br][math] \cos A = 4\cos^3 \frac{A}{3} - 3\cos \frac{A}{3} [/math] [br][br]2. (a) If [math] \sin \frac{\beta}{2} = \frac{1}{7} [/math] and [math] \cos \frac{\beta}{2} =\frac{7}{9} [/math], find the value of [math] \sin \beta [/math].[br]Solution:[br]Given, [math] \sin \frac{\beta}{2} = \frac{1}{7} [/math] and [math] \cos \frac{\beta}{2} =\frac{7}{9} [/math][br]Now,[br][math] \begin{align} \sin \beta \ & = 2\sin \frac{\beta}{2}\cos\frac{\beta}{2}\\ \ \ & = 2\times \frac{1}{7} \times \frac{7}{9} \\ \ \ & = \frac{2}{9}\\ \therefore \sin \beta \ & = \frac{2}{9}\end{align} [/math][br]2 (b) If [math] \sin \frac{\theta}{2} = \frac{1}{\sqrt{2}}, [/math] find the value of [math] \cos \theta [/math].[br]Solution:[br][math] \begin{align} \cos \theta \ & = 1-2\sin^2 \frac{\theta }{2}\\ \ \ & = 1 - 2\left( \frac{1}{\sqrt{2}} \right)^2\\ \ \ & = 1- 2\times \frac{1}{2} \\ \ \ & = 1 - 1 \\ \ \ & = 0 \\ \therefore \cos \theta \ & = 0 \end{align} [/math] [br][br]2. (c) If [math] \cos \frac{\theta}{3} = \frac{1}{2} [/math], find the value of [math] \cos \theta [/math].[br]Solution:[br][math] \begin{align} \cos \theta \ & = 4\cos^3 \frac{\theta}{3}-3\cos\frac{\theta}{3}\\ \ \ & = 4 \left(\frac{1}{2} \right)^3-3\times \left(\frac{1}{2} \right)\\ \ \ & = 4 \times \frac{1}{8} -\frac{3}{2} \\ \ \ & = \frac{1}{2} -\frac{3}{2} \\ \ \ & = \frac{1-3}{2} \\ \ \ & = \frac{-2}{2} \\ \ \ & = -1 \\ \end{align} [/math] [br]3. (a) If [math] \sin \frac{A}{2} = \frac{1}{2} [/math], find the value of [math] \sin A, \cos A [/math] and [math] \tan A [/math].[br]Solution:[br]Given, [math] \sin \frac{A}{2} = \frac{1}{2}[/math][br]Now,[br][math] \begin{align} \cos A \ & = 1-2\sin^2 \frac{A}{2}\\ \ \ & = 1-2\left( \frac{1}{2}\right)^2\\ \ \ & = 1 -2\times \frac{1}{4} \\ \ \ & = 1 - \frac{1}{2}\\ \ \ & = \frac{2-1}{2} \\ \ \ & = \frac{1}{2} \\ \therefore \cos A \ & = \frac{1}{2} \\ \end{align} [/math][br]Also,[br][math] \begin{align} \sin A \ & = \sqrt{1-\cos^2 A }\\ \ \ & = \sqrt{1-\left(\frac{1}{2} \right)^2}\\ \ \ & = \sqrt{1-\frac{1}{4}}\\ \ \ & = \sqrt{\frac{4-1}{4}} \\ \ \ & = \sqrt{\frac{3}{4}}\\ \ \ & = \frac{\sqrt{3}}{2}\\ \therefore \sin A \ & = \frac{\sqrt{3}}{2}\\ \end{align} [/math] [br]Also,[br][math] \begin{align} \tan A \ & = \frac{\sin A }{ \cos A } \\ \ \ & = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \\ \ \ & = \frac{\sqrt{3}}{1}\\ \therefore \tan A \ \ & = \sqrt{3}\\ \end{align} [/math] [br][br]3. (b) If [math] \cos \frac{\theta}{2} = \frac{4}{5}[/math], find the value of [math] \sin \theta , \cos \theta [/math] and [math] \tan \theta [/math]. [br]Solution:[br] [math]\begin{align} \cos \theta \ & = 2\cos^2 \frac{\theta}{2}-1\\ \ \ & = 2 \left( \frac{4}{5}\right)^2-1 \\ \ \ & = 2\times \frac{16}{25} -1\\ \ \ & =\frac{32-25}{25}\\ \ \ & = \frac{7}{25}\\ \therefore \cos \theta \ & = \frac{7}{25} \end{align} [/math][br]Also,[br][math] \begin{align} \sin \theta \ & = \sqrt{1-\cos^2 \theta }\\ \ \ & = \sqrt{1-\left(\frac{7}{25} \right)^2} \\ \ \ & = \sqrt{1-\frac{49}{625}} \\ \ \ & = \sqrt{\frac{625-49}{625}}\\ \ \ & = \sqrt{\frac{576}{625}}\\ \ \ & = \frac{24}{25}\\ \therefore \sin \theta \ & = \frac{24}{25}\\ \end{align}[/math] [br]Also,[br][math]\begin{align} \tan \theta \ & = \frac{\sin \theta}{\cos \theta}\\ \ \ & = \frac{\frac{24}{25}}{\frac{7}{25}}\\ \ \ & = \frac{24}{7}\\ \therefore \tan \theta \ & = \frac{24}{7} \\ \end{align} [/math][br][br]3. (c) If [math] \tan \frac{\theta}{2} = \frac{4}{3} [/math], find the value of [math] \sin \theta, \cos \theta [/math] and [math] \tan \theta [/math].[br]Solution:[br]Given, [math] \tan \frac{\theta}{2} = \frac{4}{3} [/math][br][math]\begin{align} \sin \theta \ & = \frac{2\tan \frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}} \\ \ \ & = \frac{2\times\frac{4}{3}}{1+\left( \frac{4}{3} \right)^2}\\ \ \ & =\frac{\frac{8}{3}}{1+\frac{16}{9}} \\ \ \ & =\frac{\frac{8}{3}}{\frac{9+16}{9}} \\ \ \ & =\frac{\frac{8}{3}}{\frac{25}{9}} \\ \ \ & = \frac{8}{3} \times \frac{9}{25} \\ \ \ & = \frac{24}{25}\\ \therefore \sin \theta \ & = \frac{24}{25}\\ \end{align} [/math] [br][br][math] \begin{align} \cos \theta \ & = \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}} \\ \ \ & = \frac{1-\left( \frac{4}{3} \right)^2}{1+\left( \frac{4}{3} \right)^2}\\ \ \ & =\frac{1-\frac{16}{9}}{1+\frac{16}{9}} \\ \ \ & =\frac{\frac{9-16}{9}}{\frac{9+16}{9}} \\ \ \ & =\frac{\frac{-7}{9}}{\frac{25}{9}} \\ \ \ & = \frac{-7}{9} \times \frac{9}{25} \\ \ \ & = \frac{-7}{25}\\ \therefore \cos \theta \ & = \frac{-7}{25}\\ \end{align} [/math] [br][br][math] \begin{align} \tan \theta \ & = \frac{2\tan \frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}} \\ \ \ & = \frac{2\times\frac{4}{3}}{1-\left( \frac{4}{3} \right)^2}\\ \ \ & =\frac{\frac{8}{3}}{1-\frac{16}{9}} \\ \ \ & =\frac{\frac{8}{3}}{\frac{9-16}{9}} \\ \ \ & =\frac{\frac{8}{3}}{\frac{-7}{9}} \\ \ \ & = \frac{8}{3} \times \frac{9}{-7} \\ \ \ & = \frac{-24}{7}\\ \therefore \tan \theta \ & = \frac{-24}{7}\\ \end{align}[/math][br][br]4. (a) If [math] \cos \frac{\gamma}{3} = \frac{1}{2} [/math], find the value of [math] \cos \gamma [/math] and [math] \sin \gamma [/math] .[br]Solution:[br]Given, [math] \cos \frac{\gamma}{3} = \frac{1}{2} [/math][br][math]\begin{align} \cos \gamma \ & =4\cos^3 \frac{\gamma}{3}-3\cos \frac{\gamma}{3}\\ \ \ & = 4 \left( \frac{1}{2}\right)^3-3\times \frac{1}{2}\\ \ \ & = 4\times \frac{1}{8}- \frac{3}{2}\\ \ \ & =\frac{1}{2} - \frac{3}{2} \\ \ \ & = \frac{1-3}{2}\\ \ \ & = \frac{-2}{2}\\ \ \ & = -1 \\ \therefore \cos \gamma\ & = -1 \end{align} [/math][br]Again,[br][math] \begin{align} \sin \gamma \ & = \sqrt{1-\cos^2 \gamma}\\ \ \ & = \sqrt{1-(-1)^2 } \\ \ \ & = \sqrt {1-1 }\\ \ \ & = \sqrt{0} \\ \ \ & = 0 \\ \therefore \sin \gamma \ & = 0 \end{align}[/math][br][br]4. (b) If [math] \sin \frac{B}{3} = \frac{1}{2} [/math], find the value of [math] \sin B [/math].[br]Solution:[br]Given, [math] \sin \frac{B}{3} = \frac{1}{2} [/math][br]Now,[br] [math]\begin{align} \sin B \ & = 3\sin \frac{B}{3} -4\sin^3 \frac{B}{3} \\ \ \ & = 3 \times \frac{1}{2} - 4 \left(\frac{1}{2} \right)^3 \\ \ \ & = \frac{3}{2}-4\times \frac{1}{8} \\ \ \ & = \frac{3}{2}- \frac{1}{2} \\ \ \ & = \frac{3-1}{2}\\ \ \ & = \frac{2}{2} \\ \ \ & = 1 \\ \therefore \sin B \ & = 1 \end{align} [/math][br][br]4. (c) If [math] \tan \frac{A}{3} = 1 [/math], find the value of [math] \tan A [/math].[br]Solution:[br]Given, [math] \tan \frac{A}{3} = 1 [/math][br]Now,[br][math]\begin{align} \tan A \ & = \frac{3\tan \frac{A}{3} - \tan^3 \frac{A}{3} }{1-3\tan^2 \frac{A}{3} } \\ \ \ & = \frac{3\times 1-(1)^3}{1-3(1)^2}\\ \ \ & =\frac{3-1}{1-3}\\ \ \ & = \frac{2}{-2}\\ \ \ & = -1 \\ \therefore \tan A \ & =-1 \end{align} [/math][br][br]5. (a) If [math] \sin\frac{\alpha}{2}=\frac{1}{2}\left(a+\frac{1}{a}\right) [/math], show that, [math] \cos\alpha=-\frac{1}{2}\left(a^2+\frac{1}{a^2}\right) [/math].[br]Solution:[br][math] \begin{align} \text{LHS } \ \ & =\cos \alpha \\  \ \ & =1-2\sin^2\frac{\alpha}{2} \\  \ \ & =1-2\left(\sin\frac{\alpha}{2}\right)^{^2} \\  \ \ & =1-2\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)\right\}^2 \\ \ \ & =1-2\times\frac{1}{4}\left(a+\frac{1}{a}\right)^2 \\  \ \ & =1-\frac{1}{2}\left(a^2+2\times a\times\frac{1}{a}+\frac{1}{a^2}\right) \\  \ \ & =1-\frac{1}{2}\left(a^2+2+\frac{1}{a^2}\right) \\  \ \ & =1-\frac{1}{2}\left(a^2+\frac{1}{a^2}+2\right) \\  \ \ & =1-\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)-\frac{1}{2}\times2 \\  \ \ & =1-\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)-1 \\ \ \ & =-\frac{1}{2}\left(a^2+\frac{1}{a^2}\right) \\  \ \ & = \text{ RHS} \end{align} [/math] [br][br]5. (b) If [math] \cos\frac{\theta}{2}=\frac{1}{2}\left(a+\frac{1}{a}\right) [/math], prove that [math] \cos\theta =\frac{1}{2}\left(a^2+\frac{1}{a^2}\right) [/math].[br]Solution:[br][math] \begin{align} \text{LHS } \ \ & =\cos\theta \\ \ \ & =2\cos^2\frac{\theta}{2}-1 \\  \ \ & =2\left(\cos\frac{\theta}{2}\right)^2-1 \\ \ \ & =2\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)\right\}^2-1 \\ \ \ & =2\times\frac{1}{4}\left(a+\frac{1}{a}\right)^2-1 \\ \ \ & =\frac{1}{2}\left(a+\frac{1}{a}\right)^2-1 \\ \ \ & =\frac{1}{2}\left(a^2+2\times a\times\frac{1}{a}+\frac{1}{a^2}\right)-1 \\ \ \ & =\frac{1}{2}\left(a^2+\frac{1}{a^2}+2\right)-1 \\ \ \ & =\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)+\frac{1}{2}\times2-1 \\ \ \ & =\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)+1-1 \\ \ \ & =\frac{1}{2}\left(a^2+\frac{1}{a^2}\right) \\ \ \ & = \text{ RHS} \end{align} [/math] [br][br]5. (c) If [math] \sin\frac{\alpha}{3}=\frac{1}{2}\left(a+\frac{1}{a}\right) [/math] show that: [math] \sin\alpha =-\frac{1}{2}\left(a^3+\frac{1}{a^3}\right) [/math][br]Solution:[br][math]\begin{align} \text{LHS } \ & =\sin \alpha \\ \ & =3\sin\frac{\alpha}{3} -4\sin^3\frac{\alpha}{3} \\  \ & =3\times\frac{1}{2}\left(a+\frac{1}{a}\right)-4\times\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)\right\}^3\\  \ & =\frac{3}{2}\left(a+\frac{1}{a}\right)-4\times\frac{1}{8}\left(a+\frac{1}{a}\right)^3\\ [br]\ & =\frac{3}{2} \left( a + \frac{1}{a} \right) - \frac{1}{2} \left(a+\frac{1}{a}\right)^3\\ [br]\ & =\frac{3}{2} \left( a + \frac{1}{a} \right) - \frac{1}{2} \left\{a^3+ \frac{1}{a^3}+3\times a\times \frac{1}{a} \left( 1+ \frac{1}{a} \right) \right\}\\ [br]\ & =\frac{3}{2} \left( a + \frac{1}{a} \right) - \frac{1}{2} \left\{a^3+ \frac{1}{a^3}+3 \left( a + \frac{1}{a} \right) \right\}\\ [br]\ & =\frac{3}{2} \left( a + \frac{1}{a} \right) - \frac{1}{2} \left(a^3+ \frac{1}{a^3}\right) - \frac{3}{2} \left( a+ \frac{1}{a} \right) \\ [br]\ & = - \frac{1}{2} \left(a^3+ \frac{1}{a^3}\right) \\ [br] \ & =\text{ RHS}\\ \end{align} [/math][br]Alternative[br][math] \begin{align} \text{LHS } \ \ & =\sin\alpha \\ \ \ & =3\sin\frac{\alpha}{3}-4\sin^3\frac{\alpha}{3} \\  \ \ & =3\times\frac{1}{2}\left(a+\frac{1}{a}\right)-4\times\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)\right\}^3 \\ \ \ & =\frac{3}{2}\left(a+\frac{1}{a}\right)-4\times\frac{1}{8}\left(a+\frac{1}{a}\right)^3 \\  \ \ & =\frac{1}{2}\left\{3\left(a+\frac{1}{a}\right)-\left(a+\frac{1}{a}\right)^3\right\} \\   \ \ & =-\frac{1}{2}\left\{\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)\right\} \\ \ \ & =-\frac{1}{2}\left\{\left(a+\frac{1}{a}\right)^3-3\times a\times\frac{1}{a}\left(a+\frac{1}{a}\right)\right\} \\ \ \ & =-\frac{1}{2}\left(a^3+\frac{1}{a^3}\right) \left[\because a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\right] \\ \ \ & = \text{ RHS } \end{align} [/math][br][br]5. (d) If [math] \cos\frac{\phi}{3}=\frac{1}{2}\left(a+\frac{1}{a}\right)[/math], prove that: [math] \cos\phi =\frac{1}{2}\left(a^3+\frac{1}{a^3}\right) [/math][br]Solution:[br][math] \begin{align} \cos \phi & =4\cos^3\frac{\phi}{3} - 3\cos \frac{\phi}{3} \\  \ & =4\times\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)\right\}^3 - 3\times\frac{1}{2}\left(a+\frac{1}{a}\right) \\  [br]\ & =4\times\frac{1}{8}\left(a+\frac{1}{a}\right)^3-\frac{3}{2}\left(a+\frac{1}{a}\right)\\ [br]\ & = \frac{1}{2} \left(a+\frac{1}{a}\right)^3 - \frac{3}{2} \left( a + \frac{1}{a} \right) \\ [br]\ & =\frac{1}{2} \left\{a^3+ \frac{1}{a^3}+3\times a\times \frac{1}{a} \left( 1+ \frac{1}{a} \right) \right\}- \frac{3}{2} \left( a + \frac{1}{a} \right) \\ [br]\ & =\frac{1}{2} \left\{a^3+ \frac{1}{a^3}+3 \left( a + \frac{1}{a} \right) \right\}-\frac{3}{2} \left( a + \frac{1}{a} \right) \\ [br]\ & =\frac{1}{2} \left(a^3+ \frac{1}{a^3}\right) - \frac{3}{2} \left( a+ \frac{1}{a} \right) -\frac{3}{2} \left( a + \frac{1}{a} \right) \\ [br]\text{or,} \cos \phi & = \frac{1}{2} \left(a^3+ \frac{1}{a^3}\right) \\ [br]\therefore \cos \phi & - \frac{1}{2} \left(a^3+ \frac{1}{a^3}\right) = 0 \\ [br]\end{align} [/math][br]Alternative[br][math] \begin{align} \cos\phi & =4\cos^3\frac{\phi}{3}-3\cos\frac{\phi}{3} \\  \ \ & =4\left(\cos\frac{\phi}{3}\right)^3-3\cos\frac{\phi}{3} \\  \ \ & =4\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)^3\right\}-3\times\frac{1}{2}\left(a+\frac{1}{a}\right)^2 \\  \ \ & =4\times\frac{1}{8}\left(a+\frac{1}{a}\right)^3-\frac{3}{2}\left(a+\frac{1}{a}\right) \\  \ \ & =\frac{1}{2}\left(a+\frac{1}{a}\right)^3-\frac{3}{2}\left(a+\frac{1}{a}\right) \\  \ \ & =\frac{1}{2}\left\{\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)\right\} \\  \ \ & =\frac{1}{2}\left\{\left(a+\frac{1}{a}\right)^3-3\times a\times\frac{1}{a}\left(a+\frac{1}{a}\right)\right\} \\ [br][ \because & x^3+y^3 =\left(x+y\right)^3-3xy\left(x+y\right) ] \\  [br]\ \text{or,} \cos \phi & =\frac{1}{2}\left(a^3+\frac{1}{a^3}\right) \\[br]\therefore \cos \phi & - \frac{1}{2}\left(a^3+\frac{1}{a^3}\right) = 0 \\ \end{align} [/math] [br][br]6. (a) If [math] \cos 30^{\circ} = \frac{\sqrt{3}}{2} [/math], prove that:[br][math] (i.) \cos 15^{\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}} [/math] [math] (ii) \sin 15^{\circ} = \frac{\sqrt{3}-1}{2\sqrt{2}} [/math] and [math] (iii) \tan 15^{\circ} = 2 -\sqrt{3} [/math][br]Solution:[br] [math] \begin{align}(i) \text{ We know },\ & \\ \cos^2 A \ & =\frac{1+\cos2A}{2}\\ \cos^2 15^{\circ} \ & = \frac{1+\cos30^{\circ}}{2}\\ \ \ & = \frac{1+\frac{\sqrt{3}}{2}}{2}\\ \ \ & = \frac{\frac{2+\sqrt{3}}{2}}{2}\\ \ \ & = \frac{2+\sqrt{3}}{4}\\ \ \ & = \frac{2+\sqrt{3}}{4}\times \frac{2}{2}\\ \ \ & = \frac{ 4 +2 \sqrt{3}}{8}\\ \ \ & = \frac{ 3 +2 \sqrt{3}+1}{8}\\ \ \ & = \frac{ (\sqrt{3})^2 +2\times \sqrt{3}\times 1+1^2}{(2\sqrt{2})^2}\\ \ \ & = \frac{(\sqrt{3}+1)^2}{(2\sqrt{2})^2}\\ \ \ & = \left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)^2\\ \text{or,} \cos^2 15^{\circ} \ & = \left(\frac{\sqrt{3}+1}{2} \right)^2\\ \therefore \cos 15^{\circ} \ & = \frac{\sqrt{3}+1}{2} \\ \end{align} [/math] [br][br][math] \begin{align} (ii) \text{ We know },\ & \\ \sin^2 A \ & =\frac{1-\cos2A}{2}\\ \sin^2 15^{\circ} \ & = \frac{1-\cos30^{\circ}}{2}\\ \ \ & = \frac{1-\frac{\sqrt{3}}{2}}{2}\\ \ \ & = \frac{\frac{2-\sqrt{3}}{2}}{2}\\ \ \ & = \frac{2-\sqrt{3}}{4}\\ \ \ & = \frac{2-\sqrt{3}}{4}\times \frac{2}{2}\\ \ \ & = \frac{ 4 -2 \sqrt{3}}{8}\\ \ \ & = \frac{ 3 -2 \sqrt{3}+1}{8}\\ \ \ & = \frac{ (\sqrt{3})^2 -2\times \sqrt{3}\times 1+1^2}{(2\sqrt{2})^2}\\ \ \ & = \frac{(\sqrt{3}-1)^2}{(2\sqrt{2})^2}\\ \ \ & = \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2\\ \text{or,} \sin^2 15^{\circ} \ & = \left(\frac{\sqrt{3}-1}{2\sqrt{2}} \right)^2\\ \therefore \sin 15^{\circ} \ & = \frac{\sqrt{3}-1}{2\sqrt{2}} \\ \end{align} [/math][br][br][math] \begin{align} (iii) \text{ We know },\ & \\ \tan^2 A \ & =\frac{1-\cos2A}{1+\cos2A}\\ \tan^2 15^{\circ} \ & = \frac{1-\cos30^{\circ}}{1+\cos30^{\circ}}\\ \ \ & = \frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}}\\ \ \ & = \frac{\frac{2-\sqrt{3}}{2}}{\frac{2+\sqrt{3}}{2}}\\ \ \ & = \frac{2-\sqrt{3}}{2+\sqrt{3}}\\ \ \ & = \frac{2-\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}\\ \ \ & = \frac{\left(2-\sqrt{3}\right)^2}{4-3}\\ \ \ & = \frac{(2-\sqrt{3})^2}{1}\\ \ \ & = (2-\sqrt{3})^2 \\ \text{or,} \tan^2 15^{\circ} \ & = (2-\sqrt{3})^2 \\ \text{or,} \tan 15^{\circ} \ & = (2-\sqrt{3})\\ \therefore \tan 15^{\circ} \ & = 2-\sqrt{3}\\ \end{align}[/math][br][br]6(b) If [math] \cos 45^{circ} = \frac{1}{\sqrt{2}}, [/math] show that:[br](i) [math] \sin 22 \frac{1}{2} ^{\circ} = \frac{1}{2} \sqrt{ 2-\sqrt{2}} [/math][br](ii) [math] \cos 22\frac{1}{2}^{\circ} = \frac{1}{2}\sqrt{2+\sqrt{2}}[/math][br](iii) [math] \tan 22\frac{1}{2}^{\circ} = \sqrt{3-2\sqrt{2}} [/math][br]Solution:[br] [math] \begin{align} (i) \text{We know,}\ & \\ \sin^2 A\ & =\frac{1-\cos2A}{2}\\ \text{or, }\sin^2 22\frac{1}{2}^{\circ}\ & =\frac{1-\cos45^{\circ}}{2}\\ \ & = \frac{1-\frac{1}{\sqrt{2}}}{2}\\ \ & = \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{2}\\ \ & = \frac{\sqrt{2}-1}{2\sqrt{2}}\\ \ & = \frac{\sqrt{2}-1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\\ & = \frac{2-\sqrt{2}}{4}\\ \text{or, }\sin^2 22\frac{1}{2}^{\circ}\ & = \frac{2-\sqrt{2}}{4}\\ \text{or, }\sin^2 22\frac{1}{2}^{\circ}\ & =\sqrt{\frac{2-\sqrt{2}}{4}}\\ \therefore \sin 22\frac{1}{2}^{\circ}\ & =\frac{1}{2}\sqrt{2-\sqrt{2}}\\ \end{align} [/math][br][math] \begin{align} (ii) \text{We know,}\ & \\ \cos^2 A\ & =\frac{1+\cos2A}{2}\\ \text{or, }\cos^2 22\frac{1}{2}^{\circ}\ & =\frac{1+\cos45^{\circ}}{2}\\ \ & = \frac{1+\frac{1}{\sqrt{2}}}{2}\\\ & = \frac{\frac{\sqrt{2}+1}{\sqrt{2}}}{2}\\ \ & = \frac{\sqrt{2}+1}{2\sqrt{2}}\\ \ & = \frac{\sqrt{2}+1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ \ & = \frac{2+\sqrt{2}}{4}\\ \text{or, }\cos^2 22\frac{1}{2}^{\circ}\ & = \frac{2+\sqrt{2}}{4}\\ \text{or, }\cos^2 22\frac{1}{2}^{\circ}\ & =\sqrt{\frac{2+\sqrt{2}}{4}}\\ \therefore \cos 22\frac{1}{2}^{\circ}\ & =\frac{1}{2}\sqrt{2+\sqrt{2}}\\ \end{align} [/math][br][br][math] \begin{align} (ii) \text{We know,}\ & \\ \tan^2 A\ & =\frac{1-\cos2A}{1+\cos2A}\\ \text{or, }\tan^2 22\frac{1}{2}^{\circ}\ & =\frac{1-\cos45^{\circ}}{1+\cos45^{\circ}}\\ \ & = \frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}\\ \ & = \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{\frac{\sqrt{2}+1}{\sqrt{2}}}\\ \ & = \frac{\sqrt{2}-1}{\sqrt{2}+1}\\ \ & = \frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}\\ \ & = \frac{\left(\sqrt{2}-1\right)^2}{(\sqrt{2})^2-1^2}\\ \ & = \frac{(\sqrt{2})^2 -2\times \sqrt{2}\times 1 +1^2 }{2-1}\\ \ & = \frac{2-2\sqrt{2}+1}{1}\\ \ & = 3-2\sqrt{2}\\ \text{or, }\tan^2 22\frac{1}{2}^{\circ}\ & = 3-2\sqrt{2}\\ \therefore \tan 22\frac{1}{2}^{\circ}\ & = \sqrt{ 3-2\sqrt{2}}\\ \end{align}[/math][br][br]7. (a) Prove that: [math] \frac{\sin \theta }{1+ \cos \theta} = \tan \frac{\theta}{2} [/math][br]Solution:[br] [math] \begin{align} \text{ LHS } \ & = \frac{\sin \theta }{1+\cos \theta}\\ \ \ & = \frac{2\sin \frac{\theta}{2} \cos \frac{\theta}{2} }{1+ 2\cos^2 \frac{\theta}{2} -1 }\\ \ \ & = \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} }{2\cos^2 \frac{\theta}{2} } \\ \ \ & = \frac{ \sin \frac{\theta}{2} }{ \cos \frac{\theta}{2} } \\ \ \ & = \tan \frac{\theta}{2} \\ \ \ & = \text { RHS } \end{align} [/math][br][br]7. (b) Prove that: [math] \frac{\sin \theta}{ 1-\cos \theta} = \cot \frac{\theta}{2} [/math][br]Solution:[br][math]\begin{align} \text{ LHS } \ & = \frac{\sin \theta}{ 1-\cos \theta}\\ \ \ & = \frac{2\sin \frac{\theta}{2} \cos \frac{\theta}{2} } {1-(1-2\sin^2 \frac{\theta}{2})} \\ \ \ & = \frac{ 2\sin \frac{\theta}{2} \cos \frac{\theta}{2}}{1-1+2\sin^2 \frac{\theta}{2}}\\ \ \ & = \frac{ 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} } {2\sin^2 \frac{\theta}{2} } \\ \ \ & =\frac{ \cos \frac{\theta}{2}}{\sin \frac{\theta}{2} } \\ \ \ & = \cot \frac{\theta}{2} \\ \ \ & = \text{ RHS } \\ \end{align}[/math][br][br]7. (c) [math] \frac{ 1-\cos \alpha }{1+\cos \alpha} = \tan^2 \frac{\alpha}{2} [/math] [br]Solution:[br][math] \begin{align} \text{ LHS } \ & = \frac{ 1-\cos \alpha }{1+\cos \alpha} \\ \ \ & = \frac{ 1 - (1-2\sin^2 \frac{\alpha}{2} ) }{ 1+ ( 2\cos^2 \frac{\alpha}{2} - 1) } \\ \ \ & = \frac{ 1-1+2\sin^2 \frac{\alpha}{2} }{1+2\cos^2 \frac{\alpha}{2}-1} \\ \ \ & = \frac{ 2\sin^2 \frac{\alpha}{2}} { 2\cos^2 \frac{\alpha}{2} } \\ \ \ & = \frac{ \sin^2 \frac{\alpha}{2}} { \cos^2 \frac{\alpha}{2}} \\ \ \ & = \tan^2 \frac{\alpha}{2} \\ \ \ & = \text{ RHS } \\ \end{align} [/math][br][br]7. (d) [math] \frac{ 1-\cos \beta }{1+\cos \beta} = \tan^2 \frac{\beta}{2} [/math] [br]Solution:[br][math] \begin{align} \text{ LHS } \ & = \frac{ 1+\cos \beta }{1-\cos \beta} \\ \ \ & = \frac{ 1 + (2\cos^2 \frac{\beta}{2}-1 ) }{ 1- (1- 2\sin^2 \frac{\beta}{2}) } \\ \ \ & = \frac{ 1+2\cos^2 \frac{\beta}{2} -1 }{1-1+2\sin^2 \frac{\beta}{2}} \\ \ \ & = \frac{ 2\cos^2 \frac{\beta}{2}} { 2\sin^2 \frac{\beta}{2} } \\ \ \ & = \frac{ \cos^2 \frac{\beta}{2}} { \sin^2 \frac{\beta}{2}} \\ \ \ & = \tan^2 \frac{\beta}{2} \\ \ \ & = \text{ RHS } \\ \end{align} [/math][br][br]7 (e) [math] 1+\sin \theta = \left(\sin \frac{\theta}{2} + \cos \frac{\theta}{2} \right)^2[/math][br]Solution:[br][math] \begin{align} \text{ LHS} \ & = 1+\sin \theta \\ \ & = \sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2} + 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} \\ \ & = \left(\sin \frac{\theta}{2} + \cos \frac{\theta}{2} \right)^2 \\ \ & =\text{ RHS}\\ \end{align} [/math][br][br]7 (f) [math] 1 - \sin A = \left(\sin \frac{A}{2} - \cos \frac{A}{2} \right)^2[/math][br]Solution:[br][math] \begin{align} \text{ LHS} \ & = 1 - \sin A \\ \ & = \sin^2 \frac{A}{2} + \cos^2 \frac{A}{2} - 2\sin \frac{A}{2} \cos \frac{A}{2} \\ \ & = \left(\sin \frac{A}{2} - \cos \frac{A}{2} \right)^2 \\ \ & =\text{ RHS}\\ \end{align} [/math][br][br]7. (g) Prove that: [math] \frac{1+\cos \theta + \sin \theta }{1-\cos \theta + \sin \theta}=\cot \frac{\theta}{2} [/math][br]Solution:[br][math]\begin{align} \text{LHS } \ & = \frac{1+\cos \theta + \sin \theta }{1-\cos \theta + \sin \theta}\\ \ & = \frac{1+(2\cos^2 \frac{\theta}{2} - 1) + 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} }{1-(1-2\sin^2 \frac{\theta}{2}) + 2\sin \frac{\theta}{2} \cos\frac{\theta}{2}}\\ \ & = \frac{2\cos^2 \frac{\theta}{2} + 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} }{2\sin^2 \frac{\theta}{2} +2\sin \frac{\theta}{2} \cos \frac{\theta}{2} }\\ \ & = \frac{2\cos \frac{\theta}{2} ( \cos \frac{\theta}{2} + \sin \frac{\theta}{2} )}{2\sin \frac{\theta}{2} ( \sin \frac{\theta}{2} + \cos \frac{\theta}{2} ) }\\ \ & = \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2} }\\ \ & = \cot \frac{\theta}{2}\\ \ & = \text{ RHS} \end{align}[/math][br][br]7. (i) Prove that: [math] \frac{\sin^3 \frac{\theta}{2} + \cos^3 \frac{\theta}{2}}{\sin \frac{\theta}{2} + \cos \frac{\theta}{2}} = 1 - \frac{1}{2}\sin \theta [/math][br]Solution:[br][math] \begin{align} \text{LHS } \ & = \frac{\sin^3 \frac{\theta}{2} + \cos^3 \frac{\theta}{2}}{\sin \frac{\theta}{2} + \cos \frac{\theta}{2}} \\ \ & = \frac{\left(\sin \frac{\theta}{2} + \cos \frac{\theta}{2}\right) \left(\sin^2 \frac{\theta}{2} - 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} + \cos^2 \frac{\theta}{2}\right) }{\sin \frac{\theta}{2} + \cos \frac{\theta}{2} }\\ \ & = \sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2} - 2\sin \frac{\theta}{2} \cos \frac{\theta}{2}\\ \ & = 1 - \sin \theta \\ \ & = \text{ RHS} \end{align}[/math][br][br]7. (j) [math] \frac{2\sin A + \sin 2A}{2\sin A - \sin 2A}=\cot^2 \frac{A}{2}[/math][br]Solution:[br][math]\begin{align} \text{LHS } \ & = \frac{2\sin A + \sin 2A}{2\sin A - \sin 2A}\\ \ & = \frac{2\sin A + 2\sin A \cos A}{2\sin A - 2\sin A \cos A}\\ \ & = \frac{2\sin A( 1+ \cos A)}{2\sin A ( 1- \cos A )}\\ \ & = \frac{1+ \cos A }{1-\cos A}\\ \ & = \frac{1+ (2\cos^2 \frac{A}{2} -1) }{1- (1- 2\sin^2 \frac{A}{2})}\\ \ & = \frac{1+ 2\cos^2 \frac{A}{2} -1 }{1- 1 + 2\sin^2 \frac{A}{2}}\\ \ & = \frac{2\cos^2 \frac{A}{2}}{2\sin^ \frac{A}{2}}\\ \ & = \frac{\cos^2 \frac{A}{2}}{\sin^2 \frac{A}{2}}\\ \ & = \cot^2 \frac{A}{2} \\ \ & = \text{ RHS} \end{align} [/math][br][br]7. (k) Prove that: [math] \frac{\sin 2\alpha}{1+\cos 2\alpha}. \frac{\cos \alpha}{1+\cos \alpha} = \tan \frac{\alpha}{2} [/math][br]Solution:[br][math] \begin{align} \text{LHS}\ & = \frac{\sin 2\alpha}{1+\cos 2\alpha}. \frac{\cos \alpha}{1+\cos \alpha}\\ \ & = \frac{2\sin \alpha \cos \alpha}{1+(2\cos^2 \alpha - 1)}. \frac{\cos \alpha }{1+(2\cos^2 \frac{\alpha}{2})-1}\\ \ & = \frac{2\sin \alpha \cos \alpha }{2\cos^2 \alpha}. \frac{\cos \alpha}{2\cos^2 \frac{\alpha}{2}}\\ \ & = \frac{\sin \alpha}{2\cos^2 \frac{\alpha}{2}}\\ \ & = \frac{2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}}{2\cos^2 \frac{\alpha}{2}}\\ \ & = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\\ \ & = \tan \frac{\alpha}{2}\\ \ & = \text{ RHS} \end{align} [/math][br][br]7. (l) Prove that: [math]\frac{ \sin \beta + \sin \frac{\beta}{2}}{1+\cos \beta + \cos \frac{\beta}{2}} = \tan \frac{\beta}{2} [/math][br]Solution:[br][math] \begin{align} \text{LHS } \ & = \frac{ \sin \beta + \sin \frac{\beta}{2}}{1+\cos \beta + \cos \frac{\beta}{2}}\\ \ & = \frac{2\sin \frac{\beta}{2} \cos \frac{\beta}{2} + \sin \frac{\beta}{2}}{1+(2\cos^2 \frac{\beta}{2} - 1 ) + \cos \frac{\beta}{2}}\\ \ & = \frac{\sin \frac{\beta}{2} ( 2\cos\frac{\beta}{2}+ 1)}{2\cos^2 \frac{\beta}{2} + \cos \frac{\beta}{2}}\\ \ & = \frac{\sin \frac{\beta}{2} ( 2\cos \frac{\beta}{2} + 1 )}{\cos \frac{\beta}{2} ( 2\cos \frac{\beta}{2} + 1)}\\ \ & = \frac{\sin \frac{\beta}{2}}{\cos \frac{\beta}{2}}\\ \ & = \tan \frac{\beta}{2}\\ \ & = \text { RHS}\\ \end{align} [/math][br][br]7. (m) Prove that: [math] \frac{\cos \frac{\theta}{2} -\sqrt{1+\sin \theta}}{ \sin \frac{\theta}{2} -\sqrt{1+\sin \theta}} = \tan \frac{\theta}{2}[/math][br]Solution:[br][math] \begin{align} \text{LHS} \ & = \frac{\cos \frac{\theta}{2} -\sqrt{1+\sin \theta}}{ \sin \frac{\theta}{2} -\sqrt{1+\sin \theta}}\\ \ & = \frac{\cos \frac{\theta}{2} - \sqrt{\sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2} + 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} }}{\sin \frac{\theta}{2} - \sqrt{\sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2} + 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} }}\\ \ & = \frac{\cos \frac{\theta}{2} -\sqrt{(\sin \frac{\theta}{2} + \cos \frac{\theta}{2} )^2 }}{\sin \frac{\theta}{2} -\sqrt{(\sin \frac{\theta}{2} + \cos \frac{\theta}{2} )^2 }}\\ \ & = \frac{\cos \frac{\theta}{2} - ( \sin \frac{\theta}{2} + \cos \frac{\theta}{2} ) }{\sin \frac{\theta}{2} - ( \cos \frac{\theta}{2} + \cos \frac{\theta}{2} )}\\ \ & = \frac{\cos \frac{\theta}{2} - \sin \frac{\theta}{2} - \cos \frac{\theta}{2} }{\sin \frac{\theta}{2} - \sin \frac{\theta}{2} - \cos \frac{\theta}{2} }\\ \ & = \frac{-\sin \frac{\theta}{2} }{- \cos \frac{\theta}{2}}\\ \ & = \frac{\sin \frac{\theta}{2} }{ \cos \frac{\theta}{2} }\\ \ & = \tan \frac{\theta}{2} \\ \ & = \text{ RHS} \end{align}[/math][br][br]8. (a) Prove that: [math] \frac{2\tan\left(\frac{\pi^c}{4}-\frac{\theta}{2}\right)}{1+\tan^2\left(\frac{\pi^c}{4}-\frac{\theta}{2}\right)}= \cos \theta [/math][br]Solution:[br][br][math]\begin{align} \text{LHS}\ & = \frac{2\tan\left(\frac{\pi^c}{4}-\frac{\theta}{2}\right)}{1+\tan^2\left(\frac{\pi^c}{4}-\frac{\theta}{2}\right)}\\ \ & = \sin 2\left(\frac{\pi^c}{4}-\frac{\theta}{2}\right) \\ \ & = \sin \left( \frac{\pi^c}{2}-\theta \right)\\ \ & = \cos \theta \\ \ & = \text{ RHS }\\ \end{align}[/math][br][br][br][br]8. (b) Prove that: [math] 1-2\sin^2 \left(\frac{\pi^c}{4}-\frac{A}{2}\right)= \sin A[/math][br]Solution:[br][math] \begin{align} \text{LHS}\ & = 1-2\sin^2 \left(\frac{\pi^c}{4}-\frac{A}{2}\right)\\ \ & = \cos 2 \left(\frac{\pi^c}{4}-\frac{A}{2}\right) \\ \ & = \cos \left( \frac{\pi^c}{2}-A \right) \\ \ & = \sin A \\ \ & = \text{ RHS }\\ \end{align} [/math][br][br]8. (c) Prove that: [math] \frac{1-\tan^2\left( \frac{\pi^c}{4}-\frac{\theta}{4}\right)}{1+\tan^2\left( \frac{\pi^c}{4}-\frac{\theta}{4}\right)}=\sin \frac{\theta}{2}[/math][br]Solution:[br][math] \begin{align} \text{LHS}\ & = \frac{1-\tan^2\left( \frac{\pi^c}{4}-\frac{\theta}{4}\right)}{1+\tan^2\left( \frac{\pi^c}{4}-\frac{\theta}{4}\right)} \\ \ & = \cos 2 \left( \frac{\pi^c}{4}-\frac{\theta}{4} \right) \\ \ & = \cos \left( \frac{\pi^c}{2}-\frac{\theta}{2} \right) \\ \ & = \sin \frac{\theta}{2} \\ \ & = \text{ RHS }\\ \end{align}[/math][br][br]8. (d) Prove that: [math] \cos^2 \left(\frac{\pi^c}{4}-\frac{\theta}{4}\right)-\sin^2 \left(\frac{\pi^c}{4}-\frac{\theta}{4}\right) = \sin \frac{\theta}{2} [/math][br]Solution:[br][math] \begin{align} \text{LHS}\ & = \cos^2 \left(\frac{\pi^c}{4}-\frac{\theta}{4}\right)-\sin^2 \left(\frac{\pi^c}{4}-\frac{\theta}{4}\right) \\ \ & = \cos 2 \left( \frac{\pi^c}{4}-\frac{\theta}{4} \right) \\ \ & = \cos \left( \frac{\pi^c}{2}-\frac{\theta}{2} \right) \\ \ & = \sin \frac{\theta}{2} \\ \ & = \text{ RHS }\\ \end{align}[/math][br]9. (a) Prove that: [math] \left(\cos A - \cos B \right)^2 + \left(\sin A - \sin B\right)^2 = 4 \sin^2 \frac{A-B}{2} [/math][br]Solution:[br][math] \begin{align} \text{LHS}\ & = \left(\cos A - \cos B \right)^2 + \left(\sin A - \sin B\right)^2 \\ \ & = \cos^2 A -2\cos A \cos B + \cos^2 B + \sin^2 A -2\sin A \sin B +\sin^2 B \\ \ & = \sin^2 A + \cos^2 A + \cos^2 B +\sin^2 B - 2( \cos A \cos B + \sin A \sin B) \\ \ & = 1+1-2\cos (A-B) \\ \ & = 2 -2 \left( 1 - 2\sin^2 \frac{A-B}{2} \right)\\ \ & = 2 - 2 + 4\sin^2 \frac{A-B}{2} \\ \ & = 4\sin^2 \frac{A-B}{2} \\ \ & = \text{ RHS }\\ \end{align} [/math][br][br]9. (b) Prove that: [math] \left(\cos A + \cos B \right)^2 + \left(\sin A + \sin B\right)^2 = 4 \cos^2 \frac{A-B}{2} [/math][br]Solution:[br][math] \begin{align} \text{LHS}\ & = \left(\cos A + \cos B \right)^2 + \left(\sin A + \sin B\right)^2 \\ \ & = \cos^2 A +2\cos A \cos B + \cos^2 B + \sin^2 A +2\sin A \sin B +\sin^2 B \\ \ & = \sin^2 A + \cos^2 A + \cos^2 B +\sin^2 B + 2( \cos A \cos B + \sin A \sin B) \\ \ & = 1+1+ 2\cos (A-B) \\ \ & = 2 + 2 \left(2\cos^2 \frac{A-B}{2} -1 \right)\\ \ & = 2 + 4\cos^2 \frac{A-B}{2} - 2 \\ \ & = 4\cos^2 \frac{A-B}{2} \\ \ & = \text{ RHS }\\ \end{align} [/math][br][br]10. (a) Prove that: [math] \tan \left( 45^{\circ}-\frac{A}{2}\right)=\sqrt{\frac{1-\sin A}{1+\sin A}}=\frac{\cos A}{1+\sin A} [/math][br]Solution:[br][math] \begin{align} \text{LHS}\ & = \tan \left( 45^{\circ}-\frac{A}{2}\right) \\ \ & = \frac{\tan 45^{\circ}-\tan \frac{A}{2}}{1+\tan 45^{\circ}\tan \frac{A}{2}}\\ \ & = \frac{1-\tan \frac{A}{2}}{1+1\times \tan \frac{A}{2}}\\ \ & = \frac{1-\tan \frac{A}{2}}{1+\tan \frac{A}{2}}\\ \ & = \frac{1-\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}}{1+\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}}\\ \ & = \frac{\frac{\cos \frac{A}{2}-\sin \frac{A}{2}}{\cos\frac{A}{2}}}{\frac{\cos \frac{A}{2}+\sin \frac{A}{2}}{\cos\frac{A}{2}}}\\ \ & = \frac{\cos \frac{A}{2}- \sin \frac{A}{2}}{\cos \frac{A}{2}+ \sin \frac{A}{2}} ......(i)\\ \ & = \sqrt{\left(\frac{\cos \frac{A}{2}- \sin \frac{A}{2}}{\cos \frac{A}{2}+ \sin \frac{A}{2}} \right)^2} \\ \ & = \sqrt{\frac{\left( \cos \frac{A}{2}- \sin \frac{A}{2} \right)^2}{\left( \cos \frac{A}{2}+ \sin \frac{A}{2} \right)^2}}\\ \ & = \sqrt{\frac{\cos^2 \frac{A}{2}+\sin^2 \frac{A}{2}-2\cos\frac{A}{2}\sin \frac{A}{2}}{\cos^2 \frac{A}{2}+\sin^2 \frac{A}{2}+ 2\cos\frac{A}{2}\sin \frac{A}{2}}}\\ \ & = \sqrt{\frac{1-\sin A}{1+\sin A}}\\ \ & = \text{MID TERM}\\ \end{align} [/math][br] [math] \begin{align} \text{Again}\ & \\ \text{LHS}\ & = \tan \left( 45^{\circ}-\frac{A}{2}\right) \\ \ & = \frac{\tan 45^{\circ}-\tan \frac{A}{2}}{1+\tan 45^{\circ}\tan \frac{A}{2}}\\ \ & = \frac{1-\tan \frac{A}{2}}{1+1\times \tan \frac{A}{2}}\\ \ & = \frac{1-\tan \frac{A}{2}}{1+\tan \frac{A}{2}}\\ \ & = \frac{1-\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}}{1+\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}}\\ \ & = \frac{\frac{\cos \frac{A}{2}-\sin \frac{A}{2}}{\cos\frac{A}{2}}}{\frac{\cos \frac{A}{2}+\sin \frac{A}{2}}{\cos\frac{A}{2}}}\\ \ & = \frac{\cos \frac{A}{2}- \sin \frac{A}{2}}{\cos \frac{A}{2}+ \sin \frac{A}{2}}\\ \ & = \frac{\cos \frac{A}{2}- \sin \frac{A}{2}}{\cos \frac{A}{2}+ \sin \frac{A}{2}}\times \frac{\cos \frac{A}{2}+ \sin \frac{A}{2}}{\cos \frac{A}{2}+ \sin \frac{A}{2}} \\ \ & = \frac{\cos^2 \frac{A}{2}- \sin^2 \frac{A}{2}}{\left(\cos \frac{A}{2}+ \sin \frac{A}{2}\right)^2}\\ \ & = \frac{\cos A}{\sin^2 \frac{A}{2}+\cos^2 \frac{A}{2}+2\sin \frac{A}{2}\cos \frac{A}{2}}\\ \ & = \frac{\cos A}{1+\sin A} \\ \ & = \textbf{RHS}\\ \end{align} [/math][br][br]10. (b) Prove that: [math] \tan \left( 45^{\circ}+\frac{A}{2}\right)= \sec A + \tan A [/math][br]Solution: [br][math] \begin{align} \text{Again}\ & \\ \text{LHS}\ & = \tan \left( 45^{\circ}+ \frac{A}{2}\right) \\ \ & = \frac{\tan 45^{\circ}+ \tan \frac{A}{2}}{1 -\tan 45^{\circ}\tan \frac{A}{2}}\\ \ & = \frac{1+\tan \frac{A}{2}}{1-1\times \tan \frac{A}{2}}\\ \ & = \frac{1+\tan \frac{A}{2}}{1-\tan \frac{A}{2}}\\ \ & = \frac{1+\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}}{1-\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}}\\ \ & = \frac{\frac{\cos \frac{A}{2}+\sin \frac{A}{2}}{\cos\frac{A}{2}}}{\frac{\cos \frac{A}{2}-\sin \frac{A}{2}}{\cos\frac{A}{2}}}\\ \ & = \frac{\cos \frac{A}{2}+ \sin \frac{A}{2}}{\cos \frac{A}{2}- \sin \frac{A}{2}}\\ \ & = \frac{\cos \frac{A}{2}+ \sin \frac{A}{2}}{\cos \frac{A}{2} - \sin \frac{A}{2}}\times \frac{\cos \frac{A}{2}+ \sin \frac{A}{2}}{\cos \frac{A}{2}+ \sin \frac{A}{2}} \\ \ & = \frac{\left(\cos \frac{A}{2}+ \sin \frac{A}{2}\right)^2}{\cos^2 \frac{A}{2} - \sin^2 \frac{A}{2}}\\ \ & = \frac{\sin^2 \frac{A}{2}+\cos^2 \frac{A}{2}+2\sin \frac{A}{2}\cos \frac{A}{2}}{\cos A}\\ \ & = \frac{1+\sin A}{\cos A} \\ \ & = \frac{1}{\cos A}+ \frac{\sin A}{\cos A}\\ \ & = \sec A + \tan A \\ \ & = \textbf{RHS}\\ \end{align} [/math][br][br]10. (c) Prove that: [math] \tan \left( \frac{\pi^c}{4}+\frac{\theta}{2}\right) + \tan \left( \frac{\pi^c}{4}-\frac{\theta}{2}\right)= 2\sec \theta [/math] [br] Solution:[br] [math] \begin{align} \text{LHS}\ & = \tan \left( \frac{\pi^c}{4}+\frac{\theta}{2}\right) + \tan \left( \frac{\pi^c}{4}-\frac{\theta}{2}\right) \\ \ & = \frac{\tan \frac{\pi^c}{4}+ \tan \frac{\theta}{2}}{1 -\tan \frac{\pi^c}{4}\tan \frac{\theta}{2}} + \frac{\tan \frac{\pi^c}{4}-\tan \frac{\theta}{2}}{1+\tan \frac{\pi^c}{4}\tan \frac{\theta}{2}} \\ \ & = \frac{1+\tan \frac{\theta}{2}}{1-1\times \tan \frac{\theta}{2}}+\frac{1-\tan \frac{\theta}{2}}{1+1\times \tan \frac{\theta}{2}}\\ \ & = \frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}+\frac{1-\tan \frac{\theta}{2}}{1+ \tan \frac{\theta}{2}}\\ \ & = \frac{1+\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}}{1- \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}}+\frac{1-\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}}{1+\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}}\\ \ & = \frac{\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos\frac{\theta}{2}}}{\frac{\cos \frac{\theta}{2}- \sin \frac{\theta}{2}}{\cos\frac{\theta}{2}}}+\frac{\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos\frac{\theta}{2}}}{\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos\frac{\theta}{2}}}\\ \ & = \frac{\cos \frac{\theta}{2}+ \sin \frac{\theta}{2}}{\cos \frac{\theta}{2} - \sin \frac{\theta}{2}}+ \frac{\cos \frac{\theta}{2}- \sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+ \sin \frac{\theta}{2}}\\ \ & = \frac{\left(\cos \frac{\theta}{2}+ \sin \frac{\theta}{2}\right)^2 + \left(\cos \frac{\theta}{2} -\sin \frac{\theta}{2} \right)^2}{\cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2}} \\ \ & = \frac{\cos^2 \frac{\theta}{2}+2\cos \frac{\theta}{2} \sin \frac{\theta}{2}+\sin^2 \frac{\theta}{2}+\cos^2 \frac{\theta}{2}-2\cos \frac{\theta}{2} \sin \frac{\theta}{2}+\sin^2 \frac{\theta}{2}}{\cos \theta}\\ \ & = \frac{\sin^2 \frac{\theta}{2}+ \cos^2 \frac{\theta}{2}+\sin^2 \frac{\theta}{2}+ \cos^2 \frac{\theta}{2}}{\cos \theta}\\ \ & = \frac{1+1}{\cos \theta}\\ \ & = \frac{2}{\cos \theta}\\ \ & = 2\sec \theta \\ \ & = \textbf{RHS}\\ \end{align} [/math] [br]10. (d) [math] \cot \left( \frac{\theta}{2} + 45^{\circ}\right) -\tan \left( \frac{\theta}{2} - 45^{\circ}\right) = \frac{2\cos \theta}{1+\sin \theta } [/math][br]Solution:[br] [math] \begin{align} \text{LHS } \ & = \cot \left( \frac{\theta}{2} + 45^{\circ}\right) -\tan \left( \frac{\theta}{2} - 45^{\circ}\right) \\ \ & = \frac{\cot \frac{\theta}{2} \cot 45^{\circ}-1}{\cot \frac{\theta}{2}+\cot 45^{\circ}} - \frac{\tan \frac{\theta}{2}-\tan 45^{\circ}}{1+\tan \frac{\theta}{2}\tan 45^{\circ}}\\ \ & = \frac{\cot \frac{\theta}{2} \times 1 -1}{\cot \frac{\theta}{2}+1} - \frac{\tan \frac{\theta}{2}-1}{1+\tan \frac{\theta}{2}\times 1 }\\ \ & = \frac{\cot \frac{\theta}{2}-1}{\cot \frac{\theta}{2}+1} - \frac{\tan \frac{\theta}{2}-1}{1+\tan \frac{\theta}{2} }\\ \ & = \frac{\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}-1}{\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}+1} - \frac{\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}-1}{1+\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}}\\ \ & = \frac{\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\sin \frac{\theta}{2}}}{\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\sin \frac{\theta}{2}}} - \frac{\frac{\sin \frac{\theta}{2}-\cos \frac{\theta}{2}}{\cos \frac{\theta}{2}}}{\frac{\cos \frac{\theta}{2}+ \sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}}\\ \ & = \frac{\cos \frac{\theta}{2}-\sin \frac{\theta }{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta }{2}} - \frac{\sin \frac{\theta}{2}-\cos \frac{\theta }{2}}{\cos \frac{\theta}{2}+ \sin \frac{\theta }{2}} \\ \ & = \frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}-\left(\sin \frac{\theta}{2}-\cos \frac{\theta}{2}\right)}{\cos \frac{\theta}{2}+ \sin \frac{\theta }{2}}\\ \ & = \frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}-\sin \frac{\theta}{2}+ \cos \frac{\theta}{2}}{\cos \frac{\theta}{2}+ \sin \frac{\theta }{2}}\\ \ & = \frac{2\cos \frac{\theta}{2}-2\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+ \sin \frac{\theta }{2}}\\ \ & = \frac{2\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)}{\cos \frac{\theta}{2}+ \sin \frac{\theta }{2}}\times \frac{\cos \frac{\theta}{2}+ \sin \frac{\theta }{2}}{\cos \frac{\theta}{2}+ \sin \frac{\theta }{2}}\\ \ & = \frac{2\left(\cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2}\right)}{\left(\cos \frac{\theta}{2}+ \sin \frac{\theta }{2}\right)^2}\\ \ & = \frac{2\cos \theta}{\cos^2 \frac{\theta}{2}+2\cos \frac{\theta}{2} \sin \frac{\theta}{2}+\sin^2 \frac{\theta}{2}}\\ \ & = \frac{2\cos \theta}{\sin^2 \frac{\theta}{2}+\cos^2 \frac{\theta}{2}+2\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\\ \ & = \frac{2\cos \theta}{1+\sin \theta}\\ \end{align} [/math][br][br] 11. (a) Prove that: [math] \frac{\sin 2A}{1+ \cos 2A} . \frac{\cos A}{1+ \cos A}=\tan \frac{A}{2} [/math] Solution:[br][math] \begin{align} \text{LHS } \ & = \frac{\sin 2A}{1+ \cos 2A} . \frac{\cos A}{1+ \cos A} \\ \ & = \frac{2\sin A \cos A}{1+2\cos^2 A - 1} . \frac{\cos A}{1+\cos A}\\ \ & = \frac{2\sin A \cos A}{2\cos^2 A} . \frac{\cos A}{1+\cos A}\\ \ & = \frac{\sin A}{1+\cos A}\\ \ & = \frac{2\sin \frac{A}{2} \cos \frac{A}{2}}{1+2\cos^2 \frac{A}{2}-1}\\ \ & = \frac{2\sin \frac{A}{2} \cos \frac{A}{2}}{2\cos^2 \frac{A}{2}}\\ \ & = \frac{ \sin \frac{A}{2}}{\cos \frac{A}{2}}\\ \ & = \tan \frac{A}{2}\\ \ & = \text{ RHS} \end{align} [/math][br][br] 11. (b) Prove that: [math] \frac{1+\cos 2\theta}{\sin 2\theta} . \frac{1+\cos \theta }{\cos \theta }= \cot \frac{\theta}{2} [/math][br] Solution: [br][math] \begin{align} \text{LHS} \ & = \frac{1+\cos 2\theta}{\sin 2\theta} . \frac{1+\cos \theta }{\cos \theta } \\ \ & = \frac{1+2\cos^2 \theta -1 }{2\sin \theta \cos \theta }. \frac{1+\cos \theta }{\cos \theta}\\ \ & = \frac{2\cos^2 \theta }{2\sin \theta \cos \theta }. \frac{1+\cos \theta }{\cos \theta}\\ \ & = \frac{1+\cos \theta}{\sin \theta}\\ \ & = \frac{1+2\cos^2\frac{ \theta}{2} -1 }{2\sin \frac{\theta}{2} \cos \frac{\theta}{2} }\\ \ & = \frac{2\cos^2\frac{ \theta}{2} }{2\sin \frac{\theta}{2} \cos \frac{\theta}{2} }\\ \ & = \frac{\cos\frac{ \theta}{2} }{\sin \frac{\theta}{2} }\\ \ & = \cot \frac{\theta}{2}\\ \ & = \text{ RHS}\\ \end{align} [/math][br][br] 11. (c) Prove that: [math]\left( 1+ \sin \frac{\pi^c}{8} \right)\left( 1+ \sin \frac{\pi^c}{8} \right)\left( 1+ \sin \frac{\pi^c}{8} \right)\left( 1+ \sin \frac{\pi^c}{8} \right)\left( 1+ \sin \frac{\pi^c}{8} \right) = \frac{1}{8} [/math][br]Solution:[br][math] \begin{align} \text{LHS } \ & = \left( 1+ \sin \frac{\pi^c}{8} \right)\left( 1+ \sin \frac{3\pi^c}{8} \right)\left( 1 - \sin \frac{5\pi^c}{8} \right)\left( 1- \sin \frac{7\pi^c}{8} \right) \\ \ & = \left( 1+ \sin \frac{\pi^c}{8} \right)\left( 1+ \sin \frac{3\pi^c}{8} \right)\left( 1 - \sin \frac{8\pi^c - 3\pi^c}{8} \right)\left( 1- \sin \frac{8\pi^c - \pi^c}{8} \right) \\ \ & = \left( 1+ \sin \frac{\pi^c}{8} \right)\left( 1+ \sin \frac{3\pi^c}{8} \right)\left[ 1-\sin \left( \pi^c -\frac{3\pi^c}{8}\right) \right]\left[ 1-\sin \left( \pi^c -\frac{\pi^c}{8}\right) \right]\\ \ & = \left( 1+ \sin \frac{\pi^c}{8} \right)\left( 1 + \sin \frac{3\pi^c}{8} \right)\left( 1 - \sin \frac{3\pi^c}{8} \right)\left( 1 -\sin \frac{\pi^c}{8} \right)\\ \ & = \left( 1- \sin^2 \frac{\pi^c}{8} \right)\left( 1 - \sin^2 \frac{3\pi^c}{8} \right)\\ \ & = \cos^2 \frac{\pi^c}{8} \times \cos^2 \frac{3\pi^c}{8}\\ \ & = \frac{1+\cos 2 \left( \frac{\pi^c}{8}\right)}{2} \times \frac{1+\cos 2 \left( \frac{3\pi^c}{8}\right)}{2}\\ \ & = \frac{1+\cos \left( \frac{\pi^c}{4}\right)}{2} \times \frac{1+\cos \left( \frac{3\pi^c}{4}\right)}{2}\\ \ & = \frac{1+\cos 45^{\circ}}{2}\times \frac{1+\cos 135^{\circ}}{2}\\ \ & = \frac{1}{4}(1+\cos 45^{\circ})(1+\cos 135^{\circ})\\ \ & = \frac{1}{4}\left( 1+\frac{1}{\sqrt{2}} \right) \left( 1 -\frac{1}{\sqrt{2}} \right)\\ \ & = \frac{1}{4}\left[ 1^2 - \left(\frac{1}{\sqrt{2}} \right) \right]\\ \ & = \frac{1}{4}\left( 1^2 - \frac{1}{2}\right)\\ \ & = \frac{1}{4}\left( \frac{1}{2} \right)\\ \ & = \frac{1}{8}\\ \ & = \text{RHS} \end{align} [/math][br][br]11. d) Prove that: [math]\cos^6\frac{\theta}{4}-\sin^6\frac{\theta}{4}=\cos\frac{\theta}{2}\left(1-\frac{1}{4}\sin^2\frac{\theta}{2}\right) [/math][br]Solution:[br] [math] \begin{align} \text{LHS } \ \ & =\cos^6\frac{\theta}{4}-\sin^6\frac{\theta}{4}\\ \ \ & =\left(\cos^2\frac{\theta}{4}\right)^3-\left(\sin^2\frac{\theta}{4}\right)^3\\ \ \ & =\left(\cos^2\frac{\theta}{4}-\sin^2\frac{\theta}{4}\right)\left\{\left(\cos^2\frac{\theta}{4}\right)^2+\cos^2\frac{\theta}{4}\sin^2\frac{\theta}{4}+\left(\sin^2\frac{\theta}{4}\right)^2\right\}\\\ \ & =\cos\frac{\theta}{2}\left\{\left(\cos^2\frac{\theta}{4}\right)^2+2\cos^2\frac{\theta}{4}\sin^2\frac{\theta}{4}+\left(\sin^2\frac{\theta}{4}\right)^2-\cos^2\frac{\theta}{4}\sin^2\frac{\theta}{4}\right\}\\ \ \ & =\cos\frac{\theta}{2}\left\{\left(\cos^2\frac{\theta}{4}+\sin^2\frac{\theta}{4}\right)^2-\frac{1}{4}\left(2\sin\frac{\theta}{4}\cos\frac{\theta}{4}\right)^2\right\}\\ \ \ & =\cos\frac{\theta}{2}\left\{\left(1\right)^2-\frac{1}{4}\left(\sin\frac{\theta}{2}\right)^2\right\}\\ \ \ & =\cos\frac{\theta}{2}\left(1-\frac{1}{4}\sin^2\frac{\theta}{2}\right)\\ \ & = \text{ RHS }\\ \end{align} [/math][br][br] 12. If [math] 2\tan \frac{A}{2} = 3\tan \frac{B}{2} [/math] then prove that: [math] \cos A = \frac{13\cos B -5}{13-5\cos B }[/math][br]Solution:[br] Given,[br][math] \begin{align} 2\tan \frac{A}{2} \ & = 3\tan \frac{B}{2}\\ \text{or, } \tan \frac{A}{2} \ & =\frac{3\tan \frac{B}{2}}{2} ......(i) \end{align} [/math][br][math] \begin{align} \textbf{Now,} \ & \\ \text{LHS} \ & = \cos A \\ \ & = \frac{1-\tan^2 \frac{A}{2}}{1+\tan^2 \frac{A}{2}}\\ \ & = \frac{1-\left(\tan \frac{A}{2} \right)^2}{1+\left(\tan \frac{A}{2} \right)^2 }\\ \ & = \frac{1-\left( \frac{3\tan \frac{B}{2}}{2} \right)^2}{1+\left( \frac{3\tan \frac{B}{2}}{2} \right)^2}\\ \ & = \frac{1-\frac{9\tan^2 \frac{B}{2}}{4}}{1+\frac{9\tan^2 \frac{B}{2}}{4}}\\ \ & = \frac{\frac{4-9\tan^2 \frac{B}{2}}{4}}{\frac{4+9\tan^2 \frac{B}{2}}{4}}\\ \ & = \frac{4-9\tan^2 \frac{B}{2}}{4+9\tan^2 \frac{B}{2}}\\ \ & = \frac{4-9\times \frac{1-\cos B}{1+\cos B}}{4+9\times \frac{1-\cos B}{1+\cos B}}\\ \ & = \frac{\frac{4(1+\cos B)-9(1-\cos B)}{1+\cos B}}{\frac{4(1+\cos B)+ 9(1-\cos B)}{1+\cos B}}\\ \ & = \frac{4+4\cos B - 9 + 9 \cos B}{4+4\cos B + 9 - 9 \cos B}\\ \ & = \frac{13\cos B -5}{13 - 5\cos B }\\ \ & = \text{RHS} \\ \end{align} [/math]