Question Number 42175 by lucha116 last updated on 19/Aug/18 $${sin}\mathrm{3}{x}+{sin}\mathrm{5}{x}=\mathrm{2}\left({cos}^{\mathrm{2}} \mathrm{2}{x}−{sin}^{\mathrm{2}} \mathrm{3}{x}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 42174 by lucha116 last updated on 19/Aug/18 $${cos}^{\mathrm{2}} \mathrm{3}{xcos}\mathrm{2}{x}−{cos}^{\mathrm{2}} {x}=\mathrm{0} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 19/Aug/18 $$\left(\frac{\mathrm{1}+{cos}\mathrm{6}{x}}{\mathrm{2}}\right){cos}\mathrm{2}{x}−\left(\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${cos}\mathrm{2}{x}+{cos}\mathrm{2}{x}.{cos}\mathrm{6}{x}−\mathrm{1}−{cos}\mathrm{2}{x}=\mathrm{0} \\…
Question Number 107684 by bemath last updated on 12/Aug/20 $$\:\:\:\:\:“\mathcal{B}{e}\mathcal{M}{ath}“ \\ $$$$\:\mathrm{ln}\:\mathrm{tan}\:\mathrm{1}°+\mathrm{ln}\:\mathrm{tan}\:\mathrm{2}°+\mathrm{ln}\:\mathrm{tan}\:\mathrm{3}°+…+\mathrm{ln}\:\mathrm{tan}\:\mathrm{89}°=? \\ $$ Answered by Her_Majesty last updated on 12/Aug/20 $${ln}\:{tan}\:{x}°\:=−{ln}\:{tan}\:\left(\mathrm{90}−{x}\right)° \\ $$$$\Rightarrow\:{answer}\:{is}\:{ln}\:{tan}\:\mathrm{45}°\:=\mathrm{0} \\…
Question Number 107672 by bemath last updated on 12/Aug/20 $$\:\:\:\:\:\spadesuit\mathcal{B}{e}\mathcal{M}{ath}\spadesuit \\ $$$$\:\:\:\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)}\:? \\ $$ Commented by Her_Majesty last updated on 12/Aug/20 $$\mathrm{2} \\ $$ Commented…
Question Number 107555 by anonymous last updated on 11/Aug/20 Answered by bshahid010 last updated on 11/Aug/20 $${if}\:{cos}\alpha=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow{sin}\alpha=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${and}\:{sin}\beta=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow{cos}\beta=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${Now}\:{cos}\left(\alpha+\beta\right)={cos}\alpha.{cos}\beta−{sin}\alpha.{sin}\beta \\ $$$$\Rightarrow−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow−\left(\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}\right)…
Question Number 107534 by bemath last updated on 11/Aug/20 $$\:\:\:\vDash\mathcal{B}{e}\mathcal{M}{ath}\vDash \\ $$$$\:\mathrm{sin}\:^{\mathrm{6}} {x}\:+\:\mathrm{cos}\:^{\mathrm{6}} {x}\:=\:\frac{\mathrm{7}}{\mathrm{16}}\:;\:{x}\:\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$ Commented by hgrocks last updated on 11/Aug/20 7 and 16 both my favorite numbers ���� Commented…
Question Number 107470 by bemath last updated on 11/Aug/20 $$\:\:\:\:\:\:\:\:\circlearrowleft\mathcal{B}{e}\mathcal{M}{ath}\circlearrowright \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{3}°\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{4}°\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{41}°\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{42}°\right)=? \\ $$$$\left(\mathrm{2}\right){f}\left({x}\right)={g}\left({h}\left({x}\right)\right);\:{h}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}. \\ $$$${If}\:{f}\:'\left(−\mathrm{1}\right)=\mathrm{14}\:{then}\:{g}\:'\left(\mathrm{5}\right)=? \\ $$ Answered by john santu last updated…
Question Number 41902 by upadhyayrakhi20@gmail.com last updated on 15/Aug/18 $$\mathrm{tan3}\theta\:\mathrm{tan2}\theta\:=\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:^{\mathrm{solution}……} \\ $$ Answered by MJS last updated on 15/Aug/18 $$\mathrm{tan}\:\mathrm{3}\theta\:\mathrm{tan}\:\mathrm{2}\theta\:−\mathrm{1}=\mathrm{0} \\ $$$$−\mathrm{2}\frac{\mathrm{cos}\:\mathrm{5}\theta}{\mathrm{cos}\:\mathrm{5}\theta\:+\mathrm{cos}\:\theta}=\mathrm{0} \\…
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Question Number 107364 by bemath last updated on 10/Aug/20 $$\:\:\:\:\:\:\circledcirc\mathscr{BEMATH}\circledcirc\: \\ $$$${If}\:\mathrm{tan}\:{x}+\mathrm{sec}\:{x}\:=\:{b}\:\Rightarrow\:\mathrm{cos}\:{x}\:=\:? \\ $$ Answered by bobhans last updated on 10/Aug/20 $$\:\:\:\:\:\:\Rrightarrow\mathbb{BOB}\mathcal{HANS}\Lleftarrow \\ $$$$\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{sec}\:\mathrm{x}}{\mathrm{tan}\:\mathrm{x}−\mathrm{sec}\:\mathrm{x}}\:×\:\mathrm{tan}\:\mathrm{x}+\mathrm{sec}\:\mathrm{x}\:=\:\flat \\…