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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}°} \\ $$$$\Rightarrow\mathrm{2sin}\:{x}\mathrm{cos}\:\mathrm{55}°=\mathrm{sin}\:\left({x}+\mathrm{5}°\right) \\ $$$$\mathrm{cos}\:\mathrm{55}° \\…
Question Number 104904 by bemath last updated on 24/Jul/20 $$\mathrm{4cos}\:^{\mathrm{2}} {x}\:\mathrm{sin}\:{x}\:−\mathrm{2sin}\:^{\mathrm{2}} {x}\:=\:\mathrm{3sin}\:{x} \\ $$$${where}\:−\frac{\pi}{\mathrm{2}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$ Answered by bramlex last updated on 24/Jul/20 $$\mathrm{sin}\:{x}\:\left(\mathrm{4cos}\:^{\mathrm{2}} {x}−\mathrm{2sin}\:{x}−\mathrm{3}\right)\:=\:\mathrm{0}…
Question Number 104894 by bramlex last updated on 24/Jul/20 $${prove}\:{that}\:\mathrm{cos}\:^{\mathrm{6}} {a}\:−\mathrm{sin}\:^{\mathrm{6}} {a}\:=\: \\ $$$$\mathrm{cos}\:\mathrm{2}{a}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{a}\right)\: \\ $$ Answered by john santu last updated on 24/Jul/20…
Question Number 170371 by cortano1 last updated on 22/May/22 $$\:{solve}\:\frac{\mathrm{sin}\:\left({x}+\mathrm{18}°\right)}{\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{sin}\:\mathrm{18}°}{\mathrm{sin}\:\mathrm{12}°} \\ $$ Answered by greougoury555 last updated on 22/May/22 $$\:\mathrm{sin}\:\mathrm{18}°=\mathrm{2sin}\:\mathrm{12}°\mathrm{sin}\:\mathrm{48}° \\ $$$$\:\frac{\mathrm{sin}\:\left({x}+\mathrm{18}°\right)}{\mathrm{sin}\:{x}}\:=\:\frac{\mathrm{2sin}\:\mathrm{12}°\:\mathrm{sin}\:\mathrm{48}°}{\mathrm{sin}\:\mathrm{12}°} \\ $$$$\mathrm{sin}\:\left({x}+\mathrm{18}°\right)=\mathrm{2sin}\:{x}\:\mathrm{sin}\:\mathrm{48}° \\…