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Question Number 205429 by mnjuly1970 last updated on 21/Mar/24

$$$$$$\mathrm{I}f,\phi =\frac{1}{2}(\pi -{cos}^{-1}\left(\frac{1}{4}\right))$$$$$$$$\Rightarrow {\log }_2\left(\frac{1+cos\left(6\phi \right)}{{cos}^6\left(\phi \right)}\right)=?$$$$$$

Answered by Ghisom last updated on 21/Mar/24

$$\mathrm{using}\mathrm{trigonometric}\mathrm{formulas}\Rightarrow$$$$\cos 6\phi =\frac{11}{16}$$$${\cos }^6\phi =\frac{27}{512}$$$${\log }_2\frac{\frac{27}{16}}{\frac{27}{512}}={\log }_{2}32=5$$

Answered by MM42 last updated on 21/Mar/24

$${cos}^{-1}\left(\frac{1}{4}\right)=\alpha \Rightarrow cos\alpha =\frac{1}{4}$$$$cos\phi =sin\left(\frac{1}{2}{cos}^{-1}\left(\frac{1}{4}\right)\right)$$$$\frac{1}{2}{cos}^{-1}\left(\frac{1}{4}\right)=\beta \Rightarrow cos2\beta =\frac{1}{4}$$$${sin}^2\beta =\frac{1-cos2\beta }{2}=\frac{3}{8}$$$$\Rightarrow cos\phi =sin\beta =\frac{\sqrt{6}}{4}$$$$6\phi =3\pi -3{cos}^{-1}\left(\frac{1}{4}\right)$$$${cos}^{-1}\left(\frac{1}{4}\right)=\alpha \Rightarrow cos\alpha =\frac{1}{4}$$$$\Rightarrow cos6\phi =-cos3\alpha =3cos\alpha -4{cos}^3\alpha =\frac{11}{16}$$$$\Rightarrow cos6\phi =\frac{11}{16}$$$$\Rightarrow ={log}_2\left(\frac{1+cos6\phi }{{cos}^6\phi }\right)={log}_2\left(\frac{\frac{27}{16}}{\frac{6^3}{4^6}}\right)=5✓$$$$$$

Commented by mnjuly1970 last updated on 22/Mar/24

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