Category: Trigonometry

Question Number 41691 by avishek last updated on 11/Aug/18 $${tan}^{2}20+{tan}^{2}40+{tan}^{2}80=33$$ Answered by ajfour last updated on 12/Aug/18 $$let\tan 20°=m$$$$\mathrm{tan}\:\mathrm{3}\theta=\frac{\mathrm{3tan}\:\theta−\mathrm{tan}\:^{\mathrm{3}}…

Question Number 172608 by Mikenice last updated on 29/Jun/22 Answered by FelipeLz last updated on 03/Jul/22 $$f\left(x\right)=x-\sin \left(x\right)$$$$\int \frac{x\cos \left(x\right)-\sin \left(x\right)}{{[x-\sin \left(x\right)]}^2}dx$$$$\int \frac{x\cos \left(x\right)-x+x-\sin \left(x\right)}{{[x-\sin \left(x\right)]}^2}dx$$$$\int\frac{\left[{x}−\mathrm{sin}\left({x}\right)\right]−{x}\left[\mathrm{1}−\mathrm{cos}\left({x}\right)\right]}{\left[{x}−\mathrm{sin}\left({x}\right)\right]^{\mathrm{2}}…

Question Number 41444 by Rio Michael last updated on 07/Aug/18 $$Giventhatcos(\theta +\frac{\pi }{3})=\frac{1}{2}findthevaluesof\theta intherange0°⩽\theta ⩽360°$$ Commented by maxmathsup by imad last updated on 07/Aug/18 $$cos(\theta +\frac{\pi }{3})=\frac{1}{2}\iff cos(\theta +\frac{\pi }{3})=cos\left(\frac{\pi }{3}\right)\iff \theta +\frac{\pi }{3}=\frac{\pi }{3}+2k\pi or$$$$\theta\:+\frac{\pi}{\mathrm{3}}=−\frac{\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi\:\left({k}\in{Z}\right)\Leftrightarrow\:\:\theta=\mathrm{2}{k}\pi\:\:{or}\:\theta\:=−\frac{\mathrm{2}\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi\:\:{now}\:{let}\:{sove}\:{in}\left[\mathrm{0},\mathrm{2}\pi\right]…

Question Number 106950 by bemath last updated on 08/Aug/20 $${}^{@bemath@}$$$$Given\{\begin{array}{l}2\cos x+7\cos y=5\\ 2\sin x+7\sin y=6\end{array}$$$$\cos (x-y)=?$$ Answered by john santu last updated on 08/Aug/20 $$\:\:\:\:\:\:\:\:\:\blacklozenge\mathrm{JS}\lozenge…

Question Number 41394 by Rio Michael last updated on 07/Aug/18 $$showthata)\frac{1-cos2A}{1+cos2A}\equiv {tan}^{2}A$$$$b)\frac{sin2A}{1+cos2A}\equiv tanA.$$ Commented by mondodotto@gmail.com last updated on 07/Aug/18 $$\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{part}}\:\boldsymbol{\mathrm{B}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{should}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{proved}} \…