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Question Number 102587 by 1549442205 last updated on 10/Jul/20

$$\mathrm{Prove}\mathrm{that}$$$$\cot 7.5°=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$$

Commented by PRITHWISH SEN 2 last updated on 10/Jul/20

$$\mathrm{a}\mathrm{different}\mathrm{approach}$$$$\cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{2\cos 7\theta \cos \theta }{2\cos 7\theta \sin \theta }$$$$=\frac{\cos 8\theta +\cos 6\theta }{\sin 8\theta -\sin 6\theta }$$$$\mathrm{put}\theta =7\cdot 5$$$$\cot 7\cdot 5=\frac{\cos 60+\cos 45}{\sin 60-\sin 45}=\frac{(1+\sqrt{2})}{(\sqrt{3}-\sqrt{2})}=(1+\sqrt{2})(\sqrt{3}+\sqrt{2})$$$$=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$$

Answered by Dwaipayan Shikari last updated on 10/Jul/20

$$Method1)cot7.5°=\frac{cos7.5°}{sin7.5}=\frac{2{cos}^{2}7.5°}{sin15°}=\frac{1+cos15°}{sin15°}=\frac{2cos15°+1+cos30°}{sin30°}$$$$cos15=\frac{\sqrt{3}+1}{2\sqrt{2}}=2(\frac{\sqrt{3}+1}{\sqrt{2}}+1+\sqrt{3}.\frac{1}{2})$$$$=2\left(\frac{\sqrt{6}+\sqrt{2}+\sqrt{4}+\sqrt{3}}{2}\right)$$$$socot7.5°=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\left(\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{d}\right)$$$$$$$$$$$$Method2$$$$cos\frac{\pi }{24}=\sqrt{\frac{1}{2}(1+cos\frac{\pi }{12})}=\sqrt{\frac{1}{2}(1+\frac{\sqrt{3}+1}{2\sqrt{2}})}=\frac{1}{2}\sqrt{\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{2}}}$$$$\frac{1}{2\sqrt{2}}\sqrt{4+\sqrt{6}+\sqrt{2}}$$$$sin\frac{\pi }{24}=\frac{1}{2\sqrt{2}}.\sqrt{4-\sqrt{6}-\sqrt{2}}$$$$cot\frac{\pi }{24}=\sqrt{\frac{4+\sqrt{6}+\sqrt{2}}{4-\sqrt{6}-\sqrt{2}}}=\frac{4+\sqrt{6}+\sqrt{2}}{16-8-2\sqrt{12}}=\frac{1}{2}.\frac{4+\sqrt{6}+\sqrt{2}}{4-\sqrt{12}}=\frac{1}{2}.\frac{(4+\sqrt{6}+\sqrt{2})(4+\sqrt{12})}{4}$$$$$$$$=\frac{1}{8}(16+4\sqrt{6}+4\sqrt{2}+4\sqrt{12}+6\sqrt{2}+2\sqrt{12})$$$$=2+\sqrt{\frac{3}{2}}+\frac{1}{\sqrt{2}}+\sqrt{3}+\frac{3}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}$$$$=\sqrt{4}+\frac{2\sqrt{3}+2+2\sqrt{6}+3+\sqrt{3}}{2\sqrt{2}}=\sqrt{4}+\sqrt{2}+\sqrt{3}+\sqrt{6}$$$$$$

Commented by bemath last updated on 10/Jul/20

$$how\frac{\cos 7.5^o}{\sin 7.5^o}=\frac{2\cos {}^{2}7.5^o}{\sin 7.5^o}?$$

Commented by Dwaipayan Shikari last updated on 10/Jul/20

$$multiplyingby2cos7.5°$$

Commented by Dwaipayan Shikari last updated on 10/Jul/20

$$Itissin15°$$

Commented by bemath last updated on 10/Jul/20

$$oo\sin {15}^o.thank$$

Answered by bobhans last updated on 10/Jul/20

$$\tan {15}^o=\tan (2\times 7.5^o)=\frac{2\tan 7.5^o}{1-\tan {}^2(7.5^o)}$$$$\left(1\right)\tan {15}^o=\tan ({45}^o-{30}^o)=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=$$$$\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$$$$\left(2\right)set\tan 7.5^o=x$$$$(2-\sqrt{3})(1-x^2)=2x$$$$(2-\sqrt{3})x^2+2x-(2-\sqrt{3})=0$$$$x=\frac{-2+\sqrt{4+4{(2-\sqrt{3})}^2}}{2(2-\sqrt{3})}=$$$$\frac{-1+\sqrt{1+7-4\sqrt{3}}}{2-\sqrt{3}}=\frac{-1+\sqrt{8-2\sqrt{12}}}{2-\sqrt{3}}=$$$$\frac{-1+\sqrt{6}-\sqrt{2}}{2-\sqrt{3}}=\frac{(\sqrt{6}-\sqrt{2}-1)(2+\sqrt{3})}{1}=$$$$2\sqrt{6}+3\sqrt{2}-2\sqrt{2}-\sqrt{6}-2-\sqrt{3}=$$$$\sqrt{6}+\sqrt{2}-2-\sqrt{3}=\tan 7.5^o$$$$$$

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