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TRIGONOMETRIC RATIOS OF SUBMULTIPLE ANGLES (Part-8)

TRIGONOMETRIC RATIOS OF SUBMULTIPLE ANGLES (Part-8)

 

$\,10.\,$ If $\,\,\tan\frac{\beta}{2}=4\tan\frac{\alpha}{2},\,$ prove that $\,\,\tan\frac{\beta-\alpha}{2}=\frac{3\sin\alpha}{5-3\cos\alpha}$

Sol. We have, $\,\,\tan\frac{\beta}{2}=4\tan\frac{\alpha}{2}\cdots(1) $

Now, $\,\,\tan\frac{\beta-\alpha}{2}\\=\frac{\tan\frac{\beta}{2}-\tan\frac{\alpha}{2}}{1+\tan\frac{\beta}{2}\tan\frac{\alpha}{2}}\\=\frac{4\tan\frac{\alpha}{2}-\tan\frac{\alpha}{2}}{1+4\tan\frac{\alpha}{2}.\tan\frac{\alpha}{2}}\,\,[\text{By (1)}]\\=\frac{3\tan\frac{\alpha}{2}}{1+4\tan^2\frac{\alpha}{2}}\\=\frac{3\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}} \times \frac{\cos^2\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}+4\sin^2\frac{\alpha}{2}}\\=\frac{3.\left(2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}\right)}{2\cos^2\frac{\alpha}{2}+4.2\sin^2\frac{\alpha}{2}}\\=\frac{3\sin\alpha}{1+\cos\alpha+4(1-\cos\alpha)}\\=\frac{3\sin\alpha}{5-3\cos\alpha}\,\,\text{(proved)}$

$\,11.\,$ Prove that , $\,\,4\sin 27^{\circ}=\sqrt{5+\sqrt5}-\sqrt{3-\sqrt5}$

Sol. $\,\,(\sin27^{\circ}+\cos27^{\circ})^2\\=\sin^227^{\circ}+\cos^227^{\circ}+2\sin 27^{\circ}\cos27^{\circ}\\=1+\sin54^{\circ}\\ =1+\frac{\sqrt5+1}{4}\\=\frac 14(5+\sqrt5) \\ \Rightarrow (\sin 27^{\circ}+\cos27^{\circ})=\frac 12\sqrt{5+\sqrt5}\cdots(1) \\ \text{Similarly,}\\(\sin 27^{\circ}-\cos27^{\circ})^2\\=1-\sin 54^{\circ}\\=1-\frac{\sqrt5+1}{4}\\=\frac{3-\sqrt5}{4} \\ \Rightarrow (\sin 27^{\circ}-\cos 27^{\circ})=\frac 12\sqrt{3-\sqrt5}\cdots(2)$

Adding (1) and (2) , we get $\,\,2\sin 27^{\circ}=\frac 12(\sqrt{5+\sqrt5}-\sqrt{3-\sqrt5}) \\ \Rightarrow 4\sin 27^{\circ}=\sqrt{5+\sqrt5}-\sqrt{3-\sqrt5}$

$\,12.\,$ Prove that, $\,(\sin 59^{\circ}-\sin 13^{\circ})+(\sin49^{\circ}-\sin23^{\circ})=\cos5^{\circ}$

Sol.$\,\,\,(\sin 59^{\circ}-\sin 13^{\circ})+(\sin49^{\circ}-\sin23^{\circ})\\=2\cos\frac{59^{\circ}+13^{\circ}}{2}\sin\frac{59^{\circ}-13^{\circ}}{2}+2\cos\frac{49^{\circ}+23^{\circ}}{2}\sin\frac{49^{\circ}-23^{\circ}}{2}\\=2\cos36^{\circ}\sin23^{\circ}+2\cos36^{\circ}\sin13^{\circ}\\=2\cos36^{\circ}(\sin23^{\circ}+\sin13^{\circ})\\=2\cos36^{\circ} \times 2\sin\frac{23^{\circ}+13^{\circ}}{2}\cos\frac{23^{\circ}-13^{\circ}}{2}\\=2\cos36^{\circ} \times 2\sin18^{\circ}\cos5^{\circ}\\=2.\frac{\sqrt5+1}{4}.2.\frac{\sqrt5-1}{4}\cos5^{\circ}\\=\cos5^{\circ} \,\,\text{(proved)}$

$\,13.\,$ If $\,\,\sin x+\sin y=2\sin(x+y)\,\,\text{and if}\,\, x+y \neq 0,\,$ then show that $\,\,\tan\frac x2 \tan\frac y2=\frac 13.$

Sol. $\,\,\sin x+\sin y=2\sin(x+y) \\ \Rightarrow 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}=2.2\sin\frac{x+y}{2}\cos\frac{x+y}{2} \\ \Rightarrow \frac{\cos\frac{x-y}{2}}{\cos\frac{x+y}{2}}=\frac 21  \\ \Rightarrow \frac{\cos\frac{x-y}{2} -\cos\frac{x+y}{2} }{\cos\frac{x-y}{2} +\cos\frac{x+y}{2} }=\frac{2-1}{2+1} \\ \Rightarrow \frac{2\sin\frac x2\sin\frac y2}{2\cos\frac x2\cos\frac y2}=\frac 13 \\ \Rightarrow \tan\frac x2\tan\frac y2=\frac 13\,\,\text{(proved)}$

$\,14.\,$ If $\,\,\cos\theta=\frac{2\cos\psi-1}{2-\cos\psi},\,$ show that $\,\,\tan\frac{\theta}{2}=\pm\sqrt3\tan\frac{\psi}{2}\,\,$ and hence prove that $\,\,\sin\psi=\pm\frac{\sqrt3\sin\theta}{2+\cos\theta}$

Sol. $\,\,\cos\theta=\frac{2\cos\psi-1}{2-\cos\psi}\cdots(1)$

Now, $\,\,\tan^2\frac{\theta}{2}\\=\frac{1-\cos\theta}{1+\cos\theta}\\=\frac{1-\frac{2\cos\psi-1}{2-\cos\psi}}{1+\frac{2\cos\psi-1}{2-\cos\psi}}\,\,[\text{By (1)}]\\=\frac{2-\cos\psi-2\cos\psi+1}{2-\cos\psi+2\cos\psi-1}\\=\frac{3(1-\cos\psi)}{1+\cos\psi}\\=\frac{3 \times 2\sin^2\frac{\psi}{2}}{2\cos^2\frac{\psi}{2}}\\=3\tan^2\frac{\psi}{2} \\ \therefore \,\, \tan\frac{\theta}{2}=\pm\sqrt3\tan\frac{\psi}{2}\,\,\,\text{(proved)}$

Next, $\,\sin\psi\\=\frac{2\tan\frac{\psi}{2}}{1+\tan^2\frac{\psi}{2}}\\=\frac{2.\left(\pm\frac{1}{\sqrt3}\tan\frac{\theta}{2}\right)}{1+\frac 13\tan^2\frac{\theta}{2}}\\=\pm \frac{2}{\sqrt3} \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}.\frac{3\cos^2\frac{\theta}{2}}{3\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}}\\=\pm \frac{\sqrt3 \left(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)}{2\cos^2\frac{\theta}{2}+1}\\=\pm \sqrt3 \frac{3\sin\theta}{1+\cos\theta+1} \\ \therefore \sin\psi=\pm \frac{\sqrt3\sin\theta}{2+\cos\theta}\,\,\text{(proved)}$

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