Question Number 192766 by gopikrishnan last updated on 26/May/23 $${if}\:\pi/\mathrm{2}<{x}<\pi\:{and}\:\sqrt{\mathrm{1}+\mathrm{sin}\:{x}/\mathrm{1}−\mathrm{sin}\:{x}=\:{k}\mathrm{sec}\:{x}\:,{then}\:{k}=} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192763 by TUN last updated on 26/May/23 $${Prove}\:{that}: \\ $$$${sin}\:{A}\:+\:{sin}\:{B}\:+\:{sin}\:{C}\:>\:\mathrm{2} \\ $$$$\left({A},{B},{C}\:\in\:\frac{\pi}{\mathrm{2}}\right) \\ $$ Answered by JDamian last updated on 26/May/23 $${really}? \\…
Question Number 127206 by MathSh last updated on 27/Dec/20 $${Prove}\:{it}: \\ $$$${sin}\frac{\pi}{\mathrm{2}{m}+\mathrm{1}}\centerdot{sin}\frac{\mathrm{2}\pi}{\mathrm{2}{m}+\mathrm{1}}\centerdot…\centerdot{sin}\frac{{m}\pi}{\mathrm{2}{m}+\mathrm{1}}=\frac{\sqrt{\mathrm{2}{m}+\mathrm{1}}}{\mathrm{2}^{{m}} } \\ $$ Answered by Olaf last updated on 28/Dec/20 $$ \\ $$$$\Omega\:=\:\underset{{k}=\mathrm{1}}…
Question Number 192733 by Mingma last updated on 25/May/23 Answered by MM42 last updated on 25/May/23 $$\frac{\mathrm{1}−\frac{\mathrm{2}{tanA}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}}{\mathrm{1}+\frac{\mathrm{1}−{tan}^{\mathrm{2}} {A}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}}={tanA}−\mathrm{1} \\ $$$$\frac{{tan}^{\mathrm{2}} {A}−\mathrm{2}{tanA}+\mathrm{1}}{\mathrm{2}}={tanA}−\mathrm{1} \\ $$$${tan}^{\mathrm{2}}…
Question Number 127169 by bramlexs22 last updated on 27/Dec/20 $${If}\:\left(\mathrm{1}+\mathrm{cos}\:{x}\right)\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:{then}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)\:=? \\ $$$$ \\ $$$${nice}\:{trigonometry}\: \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 127081 by slahadjb last updated on 26/Dec/20 $${Solve}\:{in}\:\mathbb{C} \\ $$$$\left(\frac{\mathrm{1}+{iz}}{\mathrm{1}−{iz}}\right)^{{n}} ={e}_{} ^{{i}\theta_{{n}} } \\ $$$$\theta_{{n}} \:\in\:\mathbb{R} \\ $$$${n}\:\in\:\mathbb{N} \\ $$ Answered by Dwaipayan…
Question Number 127067 by benjo_mathlover last updated on 26/Dec/20 $$\:{If}\:\begin{cases}{\mathrm{sin}\:{x}+\mathrm{sin}\:\mathrm{2}{x}\:=\:{a}}\\{\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{2}{x}\:=\:{b}\:}\end{cases} \\ $$$${show}\:{that}\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{2}{b}. \\ $$ Answered by liberty last updated on 26/Dec/20…
Question Number 127045 by benjo_mathlover last updated on 26/Dec/20 $$\:\:\frac{\mathrm{cos}\:\mathrm{1}°+\mathrm{cos}\:\mathrm{2}°+\mathrm{cos}\:\mathrm{3}°+…+\mathrm{cos}\:\mathrm{44}°}{\mathrm{sin}\:\mathrm{1}°+\mathrm{sin}\:\mathrm{2}°+\mathrm{sin}\:\mathrm{3}°+…+\mathrm{sin}\:\mathrm{44}°}\:=? \\ $$ Answered by liberty last updated on 26/Dec/20 $$\:\frac{\mathrm{cos}\:\left(\mathrm{22}.\mathrm{5}°−\mathrm{21}.\mathrm{5}°\right)+\mathrm{cos}\:\left(\mathrm{22}.\mathrm{5}°−\mathrm{20}.\mathrm{5}°\right)+…+\mathrm{cos}\left(\mathrm{22}.\mathrm{5}°+\mathrm{20}.\mathrm{5}°\right)+\mathrm{cos}\:\left(\mathrm{22}.\mathrm{5}°+\mathrm{21}.\mathrm{5}°\right)\:}{\mathrm{sin}\:\left(\mathrm{22}.\mathrm{5}°−\mathrm{21}.\mathrm{5}°\right)+\mathrm{sin}\:\left(\mathrm{22}.\mathrm{5}°−\mathrm{20}.\mathrm{5}°\right)+…+\mathrm{sin}\:\left(\mathrm{22}.\mathrm{5}°+\mathrm{20}.\mathrm{5}°\right)+\mathrm{sin}\:\left(\mathrm{22}.\mathrm{5}+\mathrm{21}.\mathrm{5}°\right)}\:= \\ $$$$\frac{\mathrm{2cos}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{21}.\mathrm{5}°+\mathrm{2cos}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{20}.\mathrm{5}°+…+\mathrm{2cos}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{0}.\mathrm{5}°}{\mathrm{2sin}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{21}.\mathrm{5}°+\mathrm{2sin}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{20}.\mathrm{5}°+…+\mathrm{2sin}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{0}.\mathrm{5}°}\:= \\ $$$$\frac{\mathrm{cos}\:\mathrm{22}.\mathrm{5}°\left(\mathrm{cos}\:\mathrm{21}.\mathrm{5}°+\mathrm{cos}\:\mathrm{20}.\mathrm{5}°+…+\mathrm{cos}\:\mathrm{0}.\mathrm{5}°\right)}{\mathrm{sin}\:\mathrm{22}.\mathrm{5}°\left(\mathrm{cos}\:\mathrm{21}.\mathrm{5}°+\mathrm{cos}\:\mathrm{20}.\mathrm{5}°+…+\mathrm{cos}\:\mathrm{0}.\mathrm{5}°\right)}\:= \\…
Question Number 192536 by cortano12 last updated on 20/May/23 $$\:\:\mathrm{R}=\frac{\mathrm{3sin}\:\mathrm{5}°+\mathrm{4cos}\:\mathrm{5}°−\mathrm{5cos}\:\mathrm{58}°+\mathrm{35}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{13}°}{\mathrm{cos}\:\mathrm{5}°}=? \\ $$ Answered by Tomal last updated on 20/May/23 $$\:\mathrm{R}=\frac{\mathrm{3sin}\:\mathrm{5}°+\mathrm{4cos}\:\mathrm{5}°−\mathrm{5cos}\:\mathrm{58}°+\mathrm{35}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{13}°}{\mathrm{cos}\:\mathrm{5}°}=? \\ $$$$\left.{R}=\frac{\left(\mathrm{3}×\mathrm{0}.\mathrm{0872}\right)+\left(\mathrm{4}×\mathrm{0}.\mathrm{996}\right)−\left(\mathrm{5}×\mathrm{0}.\mathrm{53}\right)+\left(\mathrm{35}\:\underbrace{\frown}\right.}{}\mathrm{2}×\mathrm{0}.\mathrm{974}\right) \\ $$ Terms…
Question Number 192537 by cortano12 last updated on 20/May/23 $$\:\begin{cases}{\mathrm{tan}\:\left(\alpha+\mathrm{2}\beta\right)=\sqrt{\mathrm{1}+\mathrm{2k}}}\\{\mathrm{tan}\:^{\mathrm{2}} \left(\alpha+\beta\right)\left\{\mathrm{1}+\mathrm{k}\:\mathrm{tan}\:^{\mathrm{2}} \beta\right\}=\mathrm{k}+\mathrm{tan}\:^{\mathrm{2}} \beta}\end{cases} \\ $$$$\:\mathrm{Find}\:\mathrm{cot}\:\mathrm{2}\beta\:. \\ $$ Answered by a.lgnaoui last updated on 20/May/23 $$\:\:\:\:\mathrm{tan}\:\left[\left(\alpha+\beta\right)+\beta\right]=\sqrt{\mathrm{1}+\mathrm{2k}}\:\:\:\:\:\:\:\:\:\:\:…