Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 69034 by ahmadshah last updated on 18/Sep/19

Answered by $@ty@m123 last updated on 18/Sep/19

$$4\cos x-3\sec x-2\tan x$$$$4\cos x-\frac{3}{\cos x}-\frac{2\sin x}{\cos x}$$$$=\frac{4\cos {}^{2}x-3-2\sin x}{\cos x}$$$$=\frac{4\cos {}^{3}x-3\cos x-2\sin x\cos x}{{\cos }^{2}x}$$$$=\frac{\cos 3x-\sin 2x}{{\cos }^{2}x}$$$$=\frac{\cos 54-\sin 36}{{\cos }^{2}18}$$$$=\frac{\cos 54-\cos 54}{{\cos }^{2}18}$$$$=0$$

Answered by MJS last updated on 18/Sep/19

$$\sin 18°=-\frac{1}{4}+\frac{\sqrt{5}}{4}$$$$\cos 18°=\frac{\sqrt{10+2\sqrt{5}}}{4}$$$$\Rightarrow \sec 18°=\frac{2\sqrt{2}}{\sqrt{5+\sqrt{5}}};\tan 18°=\frac{(-1+\sqrt{5})\sqrt{2}}{2\sqrt{5+\sqrt{5}}}$$$$4\frac{\sqrt{10+2\sqrt{5}}}{4}-3\frac{2\sqrt{2}}{\sqrt{5+\sqrt{5}}}-2\frac{(\sqrt{5}-1)\sqrt{2}}{2\sqrt{5+\sqrt{5}}}=$$$$=\sqrt{10+2\sqrt{5}}-\frac{6\sqrt{2}+(1-\sqrt{5})\sqrt{2}}{\sqrt{5+\sqrt{5}}}=$$$$=\sqrt{10+2\sqrt{5}}-\frac{(5+\sqrt{5})\sqrt{2}}{\sqrt{5+\sqrt{5}}}=$$$$=\sqrt{10+2\sqrt{5}}-\sqrt{5+\sqrt{5}}\sqrt{2}=$$$$=\sqrt{10+2\sqrt{5}}-\sqrt{10+2\sqrt{5}}=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com