Category: Trigonometry

Question Number 93239 by Ar Brandon last updated on 11/May/20 $$\mathrm{Calculate};$$$$i)\cos \left(\arctan x\right)$$$$ii)\cos \left(\arcsin x\right)$$$$iii)\tan \left(\arcsin x\right)$$ Commented by mathmax by abdo last…

Question Number 27605 by ajfour last updated on 10/Jan/18 Answered by mrW2 last updated on 13/Jan/18 $$\frac{DB}{DC}=\frac{AB}{AC}=\frac{c}{b}$$$$\Rightarrow DB=\frac{ac}{b+c}$$$$\Rightarrow DC=\frac{ab}{b+c}$$$$AD=\frac{2\sqrt{s(s-a)bc}}{b+c}$$$$\frac{{AE}}{{AB}}=\frac{{AD}}{{DB}}=\frac{\mathrm{2}\sqrt{{s}\left({s}−{a}\right){bc}}}{{b}+{c}}×\frac{{b}+{c}}{{ac}}=\frac{\mathrm{2}\sqrt{{s}\left({s}−{a}\right){bc}}}{{ac}}…

Question Number 27393 by kumar123 last updated on 06/Jan/18 $$solve$$$${\tan }^{-1}(2x/1-x^2)+{\cot }^{-1}(1-x^2/2x)=\pi /3$$ Answered by $@ty@m last updated on 06/Jan/18…

Question Number 158354 by cortano last updated on 03/Nov/21 $$\sqrt{\sin (\frac{\pi }{4}+x)}+\sqrt{\sin (\frac{\pi }{4}-x)}=\sqrt[4]{2\cos 2x}$$ Commented by tounghoungko last updated on 03/Nov/21 $$\sin (\frac{\pi }{4}+x)+\sin (\frac{\pi }{4}-x)+2\sqrt{\sin (\frac{\pi }{4}+x)\sin (\frac{\pi }{4}-x)}=\sqrt{2\cos 2x}$$$$\left(i\right)\sin (\frac{\pi }{4}+x)+\sin (\frac{\pi }{4}-x)=2\sin \frac{\pi }{4}\cos x=\sqrt{2}\cos x$$$$\left({ii}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+{x}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{2}}−\mathrm{cos}\:\mathrm{2}{x}\right)=\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}} \…

Question Number 92771 by Power last updated on 09/May/20 Terms of Service Privacy Policy Contact: info@tinkutara.com