Question Number 192282 by universe last updated on 14/May/23 $$\:\:\:{find}\:{the}\:{value}\:{of} \\ $$$$\:\:\:\:\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\:+\:\mathrm{4sin}\:\frac{\pi}{\mathrm{9}}\:=\:? \\ $$ Answered by som(math1967) last updated on 14/May/23 $${tan}\mathrm{20}+\mathrm{4}{sin}\mathrm{20} \\ $$$$\frac{{sin}\mathrm{20}+\mathrm{4}{sin}\mathrm{20}{cos}\mathrm{20}}{{cos}\mathrm{20}} \\…
Question Number 192278 by manxsol last updated on 13/May/23 $${S}={arctan}\left(\frac{\mathrm{2}}{\mathrm{1}^{\mathrm{2}} }\right)+{artan}\left(\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }\right)+…….. \\ $$ Commented by Frix last updated on 14/May/23 $$\mathrm{As}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{before}\:\left(\mathrm{but}\:\mathrm{it}\:\mathrm{was}\:\mathrm{deleted}\right): \\ $$$$\mathrm{I}\:\mathrm{approximated}\:\mathrm{it}\:\mathrm{using}\:\mathrm{software}\:\mathrm{and}\:\mathrm{got} \\…
Question Number 192270 by universe last updated on 13/May/23 $$\:{prove}\:{that} \\ $$$$\:\:\mathrm{sin}\:\mathrm{50}°\:+\:\mathrm{sin}\:\mathrm{10}°\:=\:\mathrm{cos}\:\mathrm{20}° \\ $$ Answered by Frix last updated on 13/May/23 $$\mathrm{sin}\:\left(\mathrm{60}°−\mathrm{10}°\right)\:+\mathrm{sin}\:\mathrm{10}°= \\ $$$$=\mathrm{sin}\:\mathrm{60}°\:\mathrm{cos}\:\mathrm{10}°\:−\mathrm{cos60}°\:\mathrm{sin}\:\mathrm{10}°\:+\mathrm{sin}\:\mathrm{10}°\:= \\…
Question Number 61157 by Askash last updated on 29/May/19 $${if} \\ $$$${tan}\:\mathrm{5}\theta\:+\:{tan}\:\mathrm{4}\theta\:=\mathrm{1} \\ $$$${find}\:\mathrm{3}\theta \\ $$$$ \\ $$$$ \\ $$ Commented by mr W last…
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Question Number 126610 by bramlexs22 last updated on 22/Dec/20 $$\:{Given}\:\begin{cases}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}={p}}\\{\mathrm{sec}\:{x}+\mathrm{csc}\:{x}\:=\:{q}\:\:}\end{cases} \\ $$$$\:{p}\neq{q}\:.\:{Find}\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{2}−\mathrm{2sin}\:^{\mathrm{2}} {x}−\mathrm{cs}{c}^{\mathrm{2}} \:{x}\:}\:. \\ $$ Answered by liberty last updated on 22/Dec/20 $$\:{q}\:=\:\left(\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}\right)\Rightarrow{q}^{\mathrm{2}}…
Question Number 126562 by mey3nipaba last updated on 21/Dec/20 $${Prove}\:{that}\:\mathrm{2sin}\:\frac{\theta+\phi}{\mathrm{2}}\mathrm{cos}\:\frac{\theta−\phi}{\mathrm{2}}=\mathrm{sin}\:\theta+\mathrm{sin}\:\emptyset \\ $$$${I}\:{need}\:{help}\:{immediately}\:{please} \\ $$ Commented by mr W last updated on 21/Dec/20 $$\theta=\frac{\theta+\varphi}{\mathrm{2}}+\frac{\theta−\varphi}{\mathrm{2}} \\ $$$$\varphi=\frac{\theta+\varphi}{\mathrm{2}}−\frac{\theta−\varphi}{\mathrm{2}}…
Question Number 192045 by Safiullah_21 last updated on 06/May/23 $$\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{16}}]{\mathrm{2}}\right)}×\left(\frac{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}}}{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}_{} }}\right) \\ $$$$ \\ $$$$\Rightarrow_{} \frac{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)}\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)} \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 126499 by liberty last updated on 21/Dec/20 $$\:\:\:\:\mathrm{4}\:\mathrm{arctan}\:\left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{arctan}\:\left(\frac{\mathrm{1}}{\mathrm{239}}\right)\:=? \\ $$ Answered by benjo_mathlover last updated on 21/Dec/20 $$\:{let}\:{x}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)\:\Rightarrow\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:\mathrm{tan}\:\mathrm{2}{x}=\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}\:=\:\frac{\mathrm{2}/\mathrm{5}}{\mathrm{1}−\mathrm{1}/\mathrm{25}}\: \\…
Question Number 191935 by Rupesh123 last updated on 04/May/23 Answered by som(math1967) last updated on 04/May/23 $${Direction}\:{cosine}\:{of}\:{L} \\ $$$${cos}\alpha,{cos}\beta,{cos}\gamma \\ $$$$\:\therefore{cos}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \beta+{cos}^{\mathrm{2}} \gamma=\mathrm{1} \\…