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Question Number 59305 by pete last updated on 07/May/19

$$\mathrm{Witthout}\mathrm{using}\mathrm{mathematical}\mathrm{tables}\mathrm{or}$$$$\mathrm{calculator},\mathrm{find},\mathrm{in}\mathrm{surdform}\left(\mathrm{radicals}\right)$$$$,\mathrm{the}\mathrm{value}\tan 22.5°$$

Answered by ajfour last updated on 07/May/19

$$\tan 2\theta =\frac{2\tan \theta }{1-\tan {}^2\theta }$$$$\mathrm{let}\theta =22.5°,\tan \theta =\tan 22.5°=\mathrm{q}$$$$\Rightarrow 1=\frac{2\mathrm{q}}{1-{\mathrm{q}}^2}$$$$\Rightarrow {\mathrm{q}}^2+2\mathrm{q}+1=2$$$$\Rightarrow \mathrm{q}=\sqrt{2}-1.$$

Commented by pete last updated on 08/May/19

$$\mathrm{thanks}\mathrm{sir}$$

Answered by malwaan last updated on 08/May/19

$$tan\frac{\theta }{2}=\pm \sqrt{\frac{1-cos\theta }{1+sin\theta }}$$$$tan22.5=tan\frac{45}{2}=+\sqrt{\frac{1-cos{45}^{°}}{1+cos{45}^{°}}}$$$$\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}}=\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}$$$$=\sqrt{\frac{{(2-\sqrt{2})}^2}{4-2}}=\frac{2-\sqrt{2}}{\sqrt{2}}$$$$=\frac{2\sqrt{2}-2}{2}=\sqrt{2}-1$$

Commented by pete last updated on 11/May/19

$$\mathrm{Thank}\mathrm{you}\mathrm{boss}$$

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