Question Number 170752 by balirampatel last updated on 30/May/22 $${Solve}:−\:\:\frac{{sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}}{{sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}}\:=\:\frac{\sqrt{\mathrm{3}\:}\:−\:\mathrm{1}}{\:\sqrt{\mathrm{3}\:}\:+\:\mathrm{1}}\: \\ $$ Answered by Rasheed.Sindhi last updated on 30/May/22 $$\:\:\frac{{sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}}{{sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}}\:=\:\frac{\sqrt{\mathrm{3}\:}\:−\:\mathrm{1}}{\:\sqrt{\mathrm{3}\:}\:+\:\mathrm{1}}\: \\ $$$$\begin{array}{|c|}{\frac{{a}}{{b}}=\frac{{c}}{{d}}\Rightarrow\frac{{a}+{b}}{{a}−{b}}=\frac{{c}+{d}}{{c}−{d}}}\\\hline\end{array} \\ $$$$\:\:\frac{\left({sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}\right)+\left({sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}\right)}{\left({sin}\boldsymbol{{x}}\:+\:{cos}\boldsymbol{{x}}\right)−\left({sin}\boldsymbol{{x}}\:−\:{cos}\boldsymbol{{x}}\right)} \\…
Question Number 170662 by MikeH last updated on 28/May/22 $$\mathrm{Given}\:\mathrm{that}\: \\ $$$${I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{Show}\:\mathrm{that} \\ $$$$\:{I}_{{n}+\mathrm{2}} \:=\:\left(\frac{{I}_{{n}+\mathrm{1}} }{{I}_{{n}+\mathrm{4}} }\right){I}_{{n}} \\…
Question Number 105112 by bemath last updated on 26/Jul/20 $$\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:^{\mathrm{2}} \left({x}+\frac{\mathrm{3}\pi}{\mathrm{2}}\right)\:+\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{0}\: \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi\: \\ $$ Answered by john santu last updated on 28/Jul/20 $$\Leftrightarrow\:\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{cos}\:^{\mathrm{2}} \left({x}\right)+\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)\:=\:\mathrm{0}…
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Question Number 39486 by ajfour last updated on 06/Jul/18 $$\mathrm{sin}\:\theta=\mathrm{sin}\:\alpha\mathrm{sin}\:\left(\frac{\theta+\alpha}{\mathrm{2}}\right) \\ $$$${Express}\:\theta\:{explicitly}\:{in}\:{terms}\:{of}\:\alpha. \\ $$ Commented by math khazana by abdo last updated on 07/Jul/18 $$\Rightarrow\mathrm{2}\:{sin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\theta}{\mathrm{2}}\right)={sin}\alpha\left({sin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\alpha}{\mathrm{2}}\right)+\right.…
Question Number 170503 by libaolin last updated on 25/May/22 $$\mathrm{tan90}°=? \\ $$ Commented by mr W last updated on 25/May/22 $${you}\:{repeat}\:{the}\:{same}\:{question}\:{till} \\ $$$${you}\:{have}\:{got}\:{a}\:{wrong}\:{answer}? \\ $$$${when}\:{you}\:{look}\:{at}\:{the}\:{definition}\:{of}…
Question Number 39388 by maxmathsup by imad last updated on 05/Jul/18 $${calculate}\:{A}\:={tan}\left(\frac{\pi}{\mathrm{5}}\right).{tan}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right).{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{5}}\right).{tan}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jul/18 $${tan}\mathrm{36}.{tan}\mathrm{72}.{tan}\mathrm{108}.{tan}\mathrm{144} \\ $$$${tan}\mathrm{108}={tan}\left(\mathrm{180}−\mathrm{72}\right)=−{tan}\mathrm{72} \\…
Question Number 170449 by cortano1 last updated on 24/May/22 $$\:\:\:{Solve}\:\frac{\mathrm{sin}\:\mathrm{12}°}{\mathrm{sin}\:\mathrm{24}°\:\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{sin}\:\mathrm{72}°}{\mathrm{sin}\:\left(\mathrm{36}°+{x}\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170448 by libaolin last updated on 24/May/22 $$\mathrm{tan90}°= \\ $$ Commented by mokys last updated on 24/May/22 $${undefind} \\ $$ Terms of Service…
Question Number 170445 by cortano1 last updated on 24/May/22 $$\:\:\:{Solve}\:\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\left({x}+\mathrm{5}°\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2cos}\:\mathrm{55}°} \\ $$ Answered by thfchristopher last updated on 24/May/22 $$\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\left({x}+\mathrm{5}°\right)}=\frac{\mathrm{1}}{\mathrm{2cos}\:\mathrm{55}°} \\ $$$$\Rightarrow\mathrm{2sin}\:{x}\mathrm{cos}\:\mathrm{55}°=\mathrm{sin}\:\left({x}+\mathrm{5}°\right) \\ $$$$\mathrm{cos}\:\mathrm{55}° \\…