Category: Trigonometry

Question Number 109715 by bobhans last updated on 25/Aug/20 $$\{\begin{array}{l}\sin 2x+\sin 2y=\frac{5}{4}\\ \cos (x-y)=2\sin (x+y)\end{array}where0<x,y<\frac{\pi }{2}$$$$\cos {}^2(x+y)=?$$$$△\frac{\mathrm{♭}o\mathrm{♭}}{hans}▽$$ Answered by nimnim last updated on 25/Aug/20 $$\Rightarrow\mathrm{2sin}\left({x}+{y}\right)\mathrm{cos}\left({x}−{y}\right)=\frac{\mathrm{5}}{\mathrm{4}}…

Question Number 44159 by rahul 19 last updated on 22/Sep/18 $$If{\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}\mathit{z}=\pi$$$$provethat:$$$$a)x\sqrt{1-x^2}+y\sqrt{1-y^2}+\mathit{z}\sqrt{1-{\mathit{z}}^2}=2xy\mathit{z}$$$$\left.{b}\right)\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +\boldsymbol{{z}}^{\mathrm{4}}…

Question Number 44098 by peter frank last updated on 21/Sep/18 Answered by $@ty@mlastupdatedon21/Sep/18$$\cos {}^{2}C-\cos {}^{2}D$$$\$=(\cos C+\cos D)(\cos C-\cos D)$$$$=\mathrm{2cos}\:\frac{{C}+\mathrm{D}}{\mathrm{2}}\mathrm{cos}\:\frac{{C}−{D}}{\mathrm{2}}.\mathrm{2sin}\:\frac{{C}+{D}}{\mathrm{2}}\mathrm{sin}\:\frac{{D}−{C}}{\mathrm{2}} \…

Question Number 44028 by pieroo last updated on 20/Sep/18 $$\mathrm{Show}\mathrm{that}{\cos }^2\mathrm{x}=\frac{1}{2}(1+\cos 2\mathrm{x})$$ Commented by maxmathsup by imad last updated on 20/Sep/18 $$byscalarproductwecanprovethatcos(a+b)=cosacosb-sinasinb$$$${let}\:{take}\:{a}={b}={x}\:\Rightarrow{cos}\left(\mathrm{2}{x}\right)={cos}^{\mathrm{2}}…

Question Number 109461 by 150505R last updated on 23/Aug/20 Answered by 1549442205PVT last updated on 24/Aug/20 $$\mathrm{Applying}\mathrm{the}\mathrm{inequality}\mathrm{AM}-\mathrm{GM}\mathrm{we}$$$$\mathrm{have}:{\left(2\mathrm{a}\right)}^2={(\mathrm{a}.\tan \alpha +\sqrt{{\mathrm{a}}^2-1}.\tan \beta +\sqrt{{\mathrm{a}}^2+1}.\tan \gamma )}^2$$$$\leqslant\left[\left(\mathrm{a}^{\mathrm{2}} +\left(\sqrt{\mathrm{a}^{\mathrm{2}}…