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Question Number 205429 by mnjuly1970 last updated on 21/Mar/24
If, ϕ = (1/2) (π −cos^( −1) ((1/4) )) ⇒ log_( 2) ( (( 1+ cos(6ϕ ))/(cos^6 (ϕ ))) ) =?
$$ \\ $$$$\:\mathrm{I}{f},\:\:\varphi\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\pi\:−{cos}^{\:−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\:\right)\right) \\ $$$$ \\ $$$$\:\:\:\Rightarrow\:\mathrm{log}_{\:\mathrm{2}} \left(\:\frac{\:\mathrm{1}+\:{cos}\left(\mathrm{6}\varphi\:\right)}{{cos}^{\mathrm{6}} \left(\varphi\:\right)}\:\right)\:=? \\ $$$$ \\ $$
Answered by Ghisom last updated on 21/Mar/24
using trigonometric formulas ⇒ cos 6ϕ =((11)/(16)) cos^6 ϕ =((27)/(512)) log_2 (((27)/(16))/((27)/(512))) =log_2 32 =5
$$\mathrm{using}\:\mathrm{trigonometric}\:\mathrm{formulas}\:\Rightarrow \\ $$$$\mathrm{cos}\:\mathrm{6}\varphi\:=\frac{\mathrm{11}}{\mathrm{16}} \\ $$$$\mathrm{cos}^{\mathrm{6}} \:\varphi\:=\frac{\mathrm{27}}{\mathrm{512}} \\ $$$$\mathrm{log}_{\mathrm{2}} \:\frac{\frac{\mathrm{27}}{\mathrm{16}}}{\frac{\mathrm{27}}{\mathrm{512}}}\:=\mathrm{log}_{\mathrm{2}} \:\mathrm{32}\:=\mathrm{5} \\ $$
Answered by MM42 last updated on 21/Mar/24
cos^(−1) ((1/4))=α⇒cosα=(1/4) cosϕ=sin((1/2)cos^(−1) ((1/4))) (1/2)cos^(−1) ((1/4))=β⇒cos2β=(1/4) sin^2 β=((1−cos2β)/2)=(3/8) ⇒cosϕ=sinβ=((√6)/4) 6ϕ=3π−3cos^(−1) ((1/4)) cos^(−1) ((1/4))=α⇒cosα=(1/4) ⇒cos6ϕ=−cos3α=3cosα−4cos^3 α=((11)/(16)) ⇒cos6ϕ=((11)/(16)) ⇒=log_2 (((1+cos6ϕ)/(cos^6 ϕ)))=log_2 ((((27)/(16))/(6^3 /4^6 )))=5 ✓
$${cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\alpha\Rightarrow{cos}\alpha=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${cos}\varphi={sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\beta\Rightarrow{cos}\mathrm{2}\beta=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${sin}^{\mathrm{2}} \beta=\frac{\mathrm{1}−{cos}\mathrm{2}\beta}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\Rightarrow{cos}\varphi={sin}\beta=\frac{\sqrt{\mathrm{6}}}{\mathrm{4}} \\ $$$$\mathrm{6}\varphi=\mathrm{3}\pi−\mathrm{3}{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\alpha\Rightarrow{cos}\alpha=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{cos}\mathrm{6}\varphi=−{cos}\mathrm{3}\alpha=\mathrm{3}{cos}\alpha−\mathrm{4}{cos}^{\mathrm{3}} \alpha=\frac{\mathrm{11}}{\mathrm{16}} \\ $$$$\Rightarrow{cos}\mathrm{6}\varphi=\frac{\mathrm{11}}{\mathrm{16}} \\ $$$$\Rightarrow={log}_{\mathrm{2}} \left(\frac{\mathrm{1}+{cos}\mathrm{6}\varphi}{{cos}^{\mathrm{6}} \varphi}\right)={log}_{\mathrm{2}} \left(\frac{\frac{\mathrm{27}}{\mathrm{16}}}{\frac{\mathrm{6}^{\mathrm{3}} }{\mathrm{4}^{\mathrm{6}} }}\right)=\mathrm{5}\:\checkmark \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 22/Mar/24
$$\: \\ $$

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