Question Number 43268 by Rauny last updated on 09/Sep/18 $$\mathrm{probably},\:\mathrm{cos}\:{nx}=\mathrm{2}^{{n}−\mathrm{1}} \mathrm{cos}^{{n}} \:{x}−{n}\mathrm{2}^{{n}−\mathrm{3}} \mathrm{cos}^{{n}−\mathrm{2}} \: \\ $$$$+\frac{\left({n}−\mathrm{3}\right){n}}{\mathrm{2}}\mathrm{2}^{{n}−\mathrm{5}} \mathrm{cos}^{{n}−\mathrm{4}} \:{x}\ldots \\ $$$$\mathrm{wow} \\ $$ Commented by Rauny…
Question Number 43267 by Rauny last updated on 09/Sep/18 $$\mathrm{cos}\:\mathrm{2}{x}=\mathrm{2cos}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{3}{x}=\mathrm{4cos}^{\mathrm{3}} \:{x}−\mathrm{3cos}\:{x} \\ $$$$\mathrm{cos}\:\mathrm{4}{x}=\mathrm{8cos}^{\mathrm{4}} \:{x}−\mathrm{8cos}^{\mathrm{2}} \:{x}+\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{5}{x}=\mathrm{16cos}^{\mathrm{5}} \:{x}−\mathrm{20cos}^{\mathrm{3}} +\mathrm{5cos}\:{x}\: \\ $$$$\mathrm{cos}\:\mathrm{6}{x}=\mathrm{32cos}^{\mathrm{6}} \:{x}−\mathrm{48cos}^{\mathrm{4}}…
Question Number 43266 by gunawan last updated on 09/Sep/18 $$\mathrm{cos}^{\mathrm{3}} {x}+\mathrm{cos}^{−\mathrm{3}} {x}=\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{2}{x}=… \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18 $${t}={cosx}+{isinx}={e}^{{ix}} \\…
Question Number 108780 by 150505R last updated on 19/Aug/20 Answered by bobhans last updated on 19/Aug/20 $$\mathrm{cot}\:^{\mathrm{2}} \:\mathrm{81}°\:=\:\mathrm{tan}\:^{\mathrm{2}} \:\mathrm{9}\: \\ $$$$\mathrm{cot}\:^{\mathrm{2}} \:\mathrm{63}°\:=\:\mathrm{tan}\:^{\mathrm{2}} \:\mathrm{27}\: \\ $$$$\left(\ast\right)\:\frac{\mathrm{1}}{\mathrm{2}−\mathrm{cot}\:^{\mathrm{2}}…
Question Number 43192 by Raj Singh last updated on 08/Sep/18 Commented by Rauny last updated on 09/Sep/18 $$\mathrm{cosec}\:{x}+\mathrm{sec}\:{x}=\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}=\mathrm{2} \\ $$$$\mathrm{cos}\:{x}+\mathrm{sin}\:{x}=\mathrm{2sin}\:{x}\mathrm{cos}\:{x} \\ $$$${A}={B}\Rightarrow{A}^{\mathrm{2}} ={B}^{\mathrm{2}} , \\…
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Question Number 108644 by bemath last updated on 18/Aug/20 $$\:\:\:\frac{{bemath}}{\bigstar} \\ $$$${find}\:{the}\:{value}\:{of}\: \\ $$$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{9}}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{9}}\right)\mathrm{sin}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)?\: \\ $$ Commented by bemath last updated on 18/Aug/20 $${great}\:{answer}\:{all}\:{master} \\…
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