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Question Number 12572 by tawa last updated on 25/Apr/17

$$\mathrm{prove}\mathrm{that}\mathrm{if},$$$$\sin \left(\theta \right)=\frac{1-\mathrm{x}}{1+\mathrm{x}}\mathrm{then}\tan (\frac{\mathrm{x}}{4}-\frac{\theta }{2})=\sqrt{\mathrm{x}}$$

Commented by mrW1 last updated on 26/Apr/17

$$Ithinkyoumean\mathrm{t}a\mathrm{n}(\frac{\pi }{4}-\frac{\theta }{2})=\sqrt{\mathrm{x}}$$

Answered by mrW1 last updated on 26/Apr/17

$$\sin \theta =\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=\frac{1-x}{1+x}$$$$lett=\tan \frac{\theta }{2}$$$$\Rightarrow \frac{2t}{1+t^2}=\frac{1-x}{1+x}$$$$(1-x)t^2-2(1+x)t+(1-x)=0$$$$t=\frac{2(1+x)\pm \sqrt{4{(1+x)}^2-4{(1-x)}^2}}{2(1-x)}$$$$=\frac{(1+x)\pm 2\sqrt{x}}{(1-x)}=\frac{{(1\pm \sqrt{x})}^2}{(1+\sqrt{x})(1-\sqrt{x})}$$$$\Rightarrow t=\frac{1+\sqrt{x}}{1-\sqrt{x}}or\frac{1-\sqrt{x}}{1+\sqrt{x}}$$$$$$$$\tan (\frac{\pi }{4}-\frac{\theta }{2})=\frac{\tan \frac{\pi }{4}-\tan \frac{\theta }{2}}{1+\tan \frac{\pi }{4}\times \tan \frac{\theta }{2}}$$$$=\frac{1-t}{1+t}=\frac{1-\frac{1-\sqrt{x}}{1+\sqrt{x}}}{1+\frac{1-\sqrt{x}}{1+\sqrt{x}}}=\sqrt{x}$$$$or$$$$=\frac{1-t}{1+t}=\frac{1-\frac{1+\sqrt{x}}{1-\sqrt{x}}}{1+\frac{1+\sqrt{x}}{1-\sqrt{x}}}=-\sqrt{x}$$

Commented by tawa last updated on 26/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{really}\mathrm{appreciate}\mathrm{your}\mathrm{effort}$$

Answered by ajfour last updated on 26/Apr/17

$$\tan (\frac{\pi }{4}-\frac{\theta }{2})=\frac{1-\tan \left(\theta /2\right)}{1+\tan \left(\theta /2\right)}$$$$\sin \theta =\frac{2\tan \left(\theta /2\right)}{1+\tan {}^2\left(\theta /2\right)}=\frac{1-x}{1+x}=\frac{p}{q}\left(say\right)$$$$\frac{q-p}{q+p}={\left[\frac{1-\tan \left(\theta /2\right)}{1+\tan \left(\theta /2\right)}\right]}^2=\frac{(1+x)-(1-x)}{(1+x)+(1-x)}$$$$\Rightarrow {\tan }^2(\frac{\pi }{4}-\frac{\theta }{2})=\frac{2x}{2}=x$$$$\Rightarrow \tan (\frac{\pi }{4}-\frac{\theta }{2})=\pm \sqrt{\mathit{x}}.$$

Commented by mrW1 last updated on 26/Apr/17

$$Canyoupleasecheck:$$$$InmywayIget\tan (\frac{\pi }{4}-\frac{\theta }{2})=\pm \sqrt{x}$$$$butinyourwayyougetonly\sqrt{x.}$$$$$$$$Forexample:$$$$x=3$$$$\sin \theta =\frac{1-3}{1+3}=-\frac{2}{4}=-\frac{1}{2}$$$$\Rightarrow \theta =\frac{7\pi }{6}or-\frac{\pi }{6}$$$$\frac{\pi }{4}-\frac{\theta }{2}=\frac{\pi }{4}-\frac{7\pi }{12}=-\frac{\pi }{3}or=\frac{\pi }{4}+\frac{\pi }{12}=\frac{\pi }{3}$$$$\tan (\frac{\pi }{4}-\frac{\theta }{2})=\tan (-\frac{\pi }{3})=-\sqrt{3}$$$$or$$$$\tan (\frac{\pi }{4}-\frac{\theta }{2})=\tan \left(\frac{\pi }{3}\right)=\sqrt{3}$$$$$$$$Issomethingwronghere?$$

Commented by ajfour last updated on 26/Apr/17

$$ihadassumed\sin \theta =\frac{p}{q}>0,$$$${shouldn}^{\prime }thave.$$

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