Question Number 135604 by liberty last updated on 14/Mar/21 $$\:\frac{\mathrm{sin}\:\left(\pi/\mathrm{7}\right).\mathrm{cos}\:\left(\pi/\mathrm{14}\right)}{\mathrm{tan}\:\left(\mathrm{3}\pi/\mathrm{14}\right)\left(\mathrm{2cos}\:\left(\pi/\mathrm{7}\right)−\mathrm{1}\right)}\:=? \\ $$ Answered by EDWIN88 last updated on 14/Mar/21 $$\:\mathrm{Remark}\:\begin{cases}{\mathrm{sin}\:\mathrm{3x}=\mathrm{3sin}\:\mathrm{x}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{x}}\\{\mathrm{cos}\:\mathrm{3x}=\mathrm{4cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{3cos}\:\mathrm{x}}\end{cases} \\ $$$$\mathrm{let}\:\mathrm{x}\:=\:\frac{\pi}{\mathrm{14}}\:\Rightarrow\frac{\mathrm{sin}\:\left(\pi/\mathrm{7}\right)\mathrm{cos}\:\left(\pi/\mathrm{14}\right)}{\mathrm{tan}\:\left(\mathrm{3}\pi/\mathrm{14}\right)\left(\mathrm{2cos}\:\left(\pi/\mathrm{7}\right)−\mathrm{1}\right)}\: \\…
Question Number 135603 by liberty last updated on 14/Mar/21 $$\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{17}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{17}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{17}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{17}}\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 70052 by ahmadshahhimat775@gmail.com last updated on 30/Sep/19 $$ \\ $$$$\mathrm{sin}\:{A}+\mathrm{sin}\:{B}={n}\:\:\:\:\:\:\:\:\:{and}\:\:\:\:\:\:\:\:\mathrm{cos}\:{A}+\mathrm{cos}\:{B}={m} \\ $$$$\mathrm{sin}\:\left({A}+{B}\right)=? \\ $$ Answered by $@ty@m123 last updated on 30/Sep/19 $$\mathrm{sin}\:{A}+\mathrm{sin}\:{B}={n} \\…
Question Number 135594 by liberty last updated on 14/Mar/21 $${Trigonometry} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{sin}\:\mathrm{2}{x}+\mathrm{5}\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:{find}\:{the}\:{solution} \\ $$ Answered by EDWIN88 last updated on 14/Mar/21…
Question Number 4429 by alib last updated on 24/Jan/16 $$\left({sin}\:{x}\:+\sqrt{\mathrm{3}}\:{cos}\:{x}\right)\:{sin}\:\mathrm{3}{x}\:=\:\mathrm{2} \\ $$ Answered by Yozzii last updated on 24/Jan/16 $${sinx}+\sqrt{\mathrm{3}}{cosx}=\sqrt{\mathrm{1}+\mathrm{3}}{sin}\left({x}+\pi/\mathrm{3}\right) \\ $$$$=\mathrm{2}{sin}\left({x}+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\left({sinx}+\sqrt{\mathrm{3}}{cosx}\right){sin}\mathrm{3}{x}=\mathrm{2}………\left(\Upsilon\right) \\…
Question Number 135497 by EDWIN88 last updated on 13/Mar/21 $$\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\:=\:\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)+\:\sqrt{\mathrm{7}}\:\mathrm{cos}\:\mathrm{x} \\ $$ Answered by john_santu last updated on 13/Mar/21 $$\Leftrightarrow\:\mathrm{sin}\:^{\mathrm{2}} \left({x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{sin}\:^{\mathrm{2}} \left({x}−\frac{\pi}{\mathrm{4}}\right)=\sqrt{\mathrm{7}}\:\mathrm{cos}\:{x} \\…
Question Number 69953 by Shamim last updated on 29/Sep/19 $$\mathrm{if}\:\mathrm{tan}\:\theta+\mathrm{sec}\:\theta=\:\sqrt{\mathrm{3}}\:\mathrm{than}\:\mathrm{find}\:\mathrm{out}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\underset{} {\:}\theta\:\mathrm{where}\:\mathrm{0}^{\mathrm{o}} \leqslant\theta\leqslant\mathrm{2}\pi. \\ $$ Commented by Shamim last updated on 29/Sep/19 $$\mathrm{ans}\:\mathrm{ki}\:\mathrm{only}\:\mathrm{x}=\frac{\pi}{\mathrm{6}}??? \\…
Question Number 135435 by physicstutes last updated on 13/Mar/21 $$\mathrm{solve}\:\mathrm{in}\:\mathrm{0}\:\leqslant\:{x}\:\leqslant\:\mathrm{180}°\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\mathrm{sin}\:\mathrm{3}{x}\:+\:\mathrm{cos}\:{x}\:=\:\mathrm{0}\: \\ $$ Answered by mr W last updated on 13/Mar/21 $$\mathrm{cos}\:{x}=−\mathrm{sin}\:\mathrm{3}{x}=\mathrm{sin}\:\left(−\mathrm{3}{x}\right)=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{3}{x}\right) \\ $$$$\frac{\pi}{\mathrm{2}}+\mathrm{3}{x}=\mathrm{2}{k}\pi\pm{x}…
Question Number 135419 by liberty last updated on 13/Mar/21 $$\sqrt{\mathrm{3}}\:\mathrm{tan}\:{x}.\mathrm{cot}\:{x}\:+\sqrt{\mathrm{3}}\:\mathrm{tan}\:{x}−\mathrm{cot}\:{x}−\mathrm{1}\:=\:\mathrm{0} \\ $$ Answered by benjo_mathlover last updated on 13/Mar/21 $$\Rightarrow\sqrt{\mathrm{3}}\:\mathrm{tan}\:{x}\left(\mathrm{cot}\:{x}+\mathrm{1}\right)−\left(\mathrm{cot}\:{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{3}}\:\mathrm{tan}\:{x}−\mathrm{1}\right)\left(\mathrm{cot}\:{x}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\\{\mathrm{cot}\:{x}=−\mathrm{1}\Rightarrow\mathrm{tan}\:{x}=−\mathrm{1}}\end{cases} \\…
Question Number 69871 by Shamim last updated on 28/Sep/19 $$\mathrm{Here},\:\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}\:\:} =\:\mathrm{4}\sqrt{\mathrm{mn}\:}\mathrm{and}\:\mathrm{tan}\theta+\mathrm{sin}\theta=\:\mathrm{m} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\:\mathrm{tan}\theta−\mathrm{sin}\theta=\:\mathrm{n}. \\ $$ Answered by MJS last updated on 29/Sep/19 $$\mathrm{first}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve} \\…