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Question Number 161848 by mnjuly1970 last updated on 23/Dec/21

If tan (α )= 2 find the value of K=(( 1+sin( 8 α)−cos (8α ))/(1+sin( 8α ) + cos (8 α ))) =?

$$ \\ $$$$\:\:\:\:\:\:\mathrm{I}{f}\:\:\:\:{tan}\:\left(\alpha\:\right)=\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}\: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{K}=\frac{\:\:\mathrm{1}+{sin}\left(\:\mathrm{8}\:\alpha\right)−{cos}\:\left(\mathrm{8}\alpha\:\right)}{\mathrm{1}+{sin}\left(\:\mathrm{8}\alpha\:\right)\:+\:{cos}\:\left(\mathrm{8}\:\alpha\:\right)}\:=? \\ $$$$ \\ $$$$ \\ $$

Commented by cortano last updated on 23/Dec/21

K=((2sin^2 4α+sin 8α)/(2cos^2 4α+sin 8α)) =((2sin 4α(sin 4α+cos 4α))/(2cos 4α(cos 4α+sin 4α))) K= tan 4α K=(((2tan 2α)/(1−tan 2α)))=((−(8/3))/(1−((16)/9)))=((−24)/(−7))=((24)/7) R= tan 2α=((2tan α)/(1−tan^2 α)) = (4/(1−4))=−(4/3)

$$\:\:{K}=\frac{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{4}\alpha+\mathrm{sin}\:\mathrm{8}\alpha}{\mathrm{2cos}\:^{\mathrm{2}} \mathrm{4}\alpha+\mathrm{sin}\:\mathrm{8}\alpha}\:=\frac{\mathrm{2sin}\:\mathrm{4}\alpha\left(\mathrm{sin}\:\mathrm{4}\alpha+\mathrm{cos}\:\mathrm{4}\alpha\right)}{\mathrm{2cos}\:\mathrm{4}\alpha\left(\mathrm{cos}\:\mathrm{4}\alpha+\mathrm{sin}\:\mathrm{4}\alpha\right)} \\ $$$$\:\:{K}=\:\mathrm{tan}\:\mathrm{4}\alpha\: \\ $$$$\:{K}=\left(\frac{\mathrm{2tan}\:\mathrm{2}\alpha}{\mathrm{1}−\mathrm{tan}\:\mathrm{2}\alpha}\right)=\frac{−\frac{\mathrm{8}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{16}}{\mathrm{9}}}=\frac{−\mathrm{24}}{−\mathrm{7}}=\frac{\mathrm{24}}{\mathrm{7}} \\ $$$$\:{R}=\:\mathrm{tan}\:\mathrm{2}\alpha=\frac{\mathrm{2tan}\:\alpha}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \alpha}\:=\:\frac{\mathrm{4}}{\mathrm{1}−\mathrm{4}}=−\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Commented by mnjuly1970 last updated on 23/Dec/21

thanks alot ...

$$\:\:\:\:\:{thanks}\:{alot}\:... \\ $$

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