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Question Number 180459 by mnjuly1970 last updated on 12/Nov/22

Answered by greougoury555 last updated on 12/Nov/22

cos 3α=(1/(2cos α)) 4cos^3 α−3cos α=(1/(2cos α)) cos α= ((√(6+2(√(17))))/4) sin α = ((√(10−2(√(17))))/4) sin 2α = ((√((6+2(√(17)) )(10−2(√(17)) )))/(16)) sin 2α = ((√((3+(√(17)))(5−(√(17)))))/8) =((√(2(√(17))−2))/8)

$$\:\:\mathrm{cos}\:\mathrm{3}\alpha=\frac{\mathrm{1}}{\mathrm{2cos}\:\alpha} \\ $$$$\:\:\mathrm{4cos}\:^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha=\frac{\mathrm{1}}{\mathrm{2cos}\:\alpha} \\ $$$$\:\:\mathrm{cos}\:\alpha=\:\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{17}}}}{\mathrm{4}}\: \\ $$$$\:\mathrm{sin}\:\alpha\:=\:\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{17}}}}{\mathrm{4}} \\ $$$$\:\mathrm{sin}\:\mathrm{2}\alpha\:=\:\frac{\sqrt{\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{17}}\:\right)\left(\mathrm{10}−\mathrm{2}\sqrt{\mathrm{17}}\:\right)}}{\mathrm{16}} \\ $$$$\:\mathrm{sin}\:\mathrm{2}\alpha\:=\:\frac{\sqrt{\left(\mathrm{3}+\sqrt{\mathrm{17}}\right)\left(\mathrm{5}−\sqrt{\mathrm{17}}\right)}}{\mathrm{8}}\:=\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{17}}−\mathrm{2}}}{\mathrm{8}} \\ $$

Commented by mnjuly1970 last updated on 12/Nov/22

bravo sir ..excellent

$$\:\:{bravo}\:{sir}\:..{excellent} \\ $$

Answered by mr W last updated on 12/Nov/22

sin 2α=((sin α)/(cos 3α)) 2 cos α=(1/(cos α(4 cos^2 α−3))) 8 cos^4 α−6 cos^2 α−1=0 cos^2 α=((3+(√(17)))/8) cos 2α=((−1+(√(17)))/4) ⇒sin 2α=((√(2((√(17))−1)))/4)

$$\mathrm{sin}\:\mathrm{2}\alpha=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\mathrm{3}\alpha} \\ $$$$\mathrm{2}\:\mathrm{cos}\:\alpha=\frac{\mathrm{1}}{\mathrm{cos}\:\alpha\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{3}\right)} \\ $$$$\mathrm{8}\:\mathrm{cos}^{\mathrm{4}} \:\alpha−\mathrm{6}\:\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\alpha=\frac{\mathrm{3}+\sqrt{\mathrm{17}}}{\mathrm{8}} \\ $$$$\mathrm{cos}\:\mathrm{2}\alpha=\frac{−\mathrm{1}+\sqrt{\mathrm{17}}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}\alpha=\frac{\sqrt{\mathrm{2}\left(\sqrt{\mathrm{17}}−\mathrm{1}\right)}}{\mathrm{4}} \\ $$

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