Menu Close

Category: Trigonometry

Question-109910

Question Number 109910 by bemath last updated on 26/Aug/20 Commented by qwerty111 last updated on 26/Aug/20 Commented by qwerty111 last updated on 26/Aug/20 $$\boldsymbol{{Shuni}}\:\:\boldsymbol{{soddalashtirib}}\:\:\boldsymbol{{qo}}'\boldsymbol{{ying}}\:\:\boldsymbol{{chiqadi}}.\boldsymbol{{Siz}}\:\:\boldsymbol{{o}}'\boldsymbol{{zbekmisiz}}\:? \\…

sin-2x-sin-2y-5-4-cos-x-y-2sin-x-y-where-0-lt-x-y-lt-pi-2-cos-2-x-y-o-hans-

Question Number 109715 by bobhans last updated on 25/Aug/20 $$\begin{cases}{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{sin}\:\mathrm{2}{y}=\frac{\mathrm{5}}{\mathrm{4}}}\\{\mathrm{cos}\:\left({x}−{y}\right)=\mathrm{2sin}\:\left({x}+{y}\right)}\end{cases}{where}\:\mathrm{0}<{x},{y}<\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} \left({x}+{y}\right)\:=\:? \\ $$$$\:\:\:\:\:\:\bigtriangleup\frac{\flat{o}\flat}{{hans}}\bigtriangledown \\ $$ 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}}…

The-number-of-solutions-of-the-equation-cos-1-x-2-1-x-2-1-sin-1-2x-x-2-1-tan-1-2x-x-2-1-2pi-3-

Question Number 44172 by rahul 19 last updated on 22/Sep/18 $${The}\:{number}\:{of}\:{solutions}\:{of}\:{the}\:{equation} \\ $$$$\mathrm{cos}^{−\mathrm{1}} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:+\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{2}\pi}{\mathrm{3}}. \\ $$ Commented by tanmay.chaudhury50@gmail.com…

If-sin-1-x-sin-1-y-sin-1-z-pi-prove-that-a-x-1-x-2-y-1-y-2-z-1-z-2-2xyz-b-x-4-y-4-z-4-4x-2-y-2-z-2-2-x-2-y-2-y-2-z-2-z-2-x-2-

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