If-1-2-pi-cos-1-1-4-log-2-1-cos-6-cos-6-

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

I f, φ = \frac{1}{2} (π - {c o s}^{- 1} (\frac{1}{4}))

\Rightarrow \log_{2} (\frac{1 + c o s (6 φ)}{{c o s}^{6} (φ)}) = ?

Answered by Ghisom last updated on 21/Mar/24

using trigonometric formulas \Rightarrow

\cos 6 φ = \frac{11}{16}

\cos^{6} φ = \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

{c o s}^{- 1} (\frac{1}{4}) = α \Rightarrow c o s α = \frac{1}{4}

c o s φ = s i n (\frac{1}{2} {c o s}^{- 1} (\frac{1}{4}))

\frac{1}{2} {c o s}^{- 1} (\frac{1}{4}) = β \Rightarrow c o s 2 β = \frac{1}{4}

{s i n}^{2} β = \frac{1 - c o s 2 β}{2} = \frac{3}{8}

\Rightarrow c o s φ = s i n β = \frac{\sqrt{6}}{4}

6 φ = 3 π - 3 {c o s}^{- 1} (\frac{1}{4})

{c o s}^{- 1} (\frac{1}{4}) = α \Rightarrow c o s α = \frac{1}{4}

\Rightarrow c o s 6 φ = - c o s 3 α = 3 c o s α - 4 {c o s}^{3} α = \frac{11}{16}

\Rightarrow c o s 6 φ = \frac{11}{16}

\Rightarrow= {l o g}_{2} (\frac{1 + c o s 6 φ}{{c o s}^{6} φ}) = {l o g}_{2} (\frac{\frac{27}{16}}{\frac{6^{3}}{4^{6}}}) = 5 ✓

Commented by mnjuly1970 last updated on 22/Mar/24