Question

cos[2cos^-1(4/5)]

Answer 1

let \[\large A=\cos^{-1}\frac{4}{5} \]\[\large \cos (2\cos^{-1}\frac{ 4 }{ 5 })=\cos (2A) = 2\cos^2(A)-1=2\left( \cos A \right)^2 -1\]

Answer 2

@arianarose can you do the rest?

Answer 3

I have as an identity \[\cos (2A) = \cos ^{2}A - \sin ^{2}A\]
Which starts out to look like yours, but I don't know how you got -1? And even then, I don't know where to go next.

Answer 4

also, \[\large \sin^2A=1-\cos^2A\] so \[\cos 2A=\cos^2A - \sin^2 A = \cos^2A-(1-\cos^2 A)=2\cos^2A-1\]

Answer 5

since we chose \[\large A=\cos^{-1}\frac{ 4 }{ 5 }\] then \[\large \cos A = \frac{ 4 }{ 5 }\]

Answer 6

I'm so confused, I don't know how to write this out. . .
\[\cos2(4/5) =\cos ^{2}(4/5)-\sin ^{2}(3/5) ?\]
\[\cos ^{2}(4/5)-(1-\cos ^{2}(4/5) = 2\cos(4/5) -1 ?\]

Is that how you wrote it?

Answer 7

\[\large \cos A =\frac{ 4 }{ 5 }\]\[\large 2\cos^2A-1 = 2(\frac{ 4 }{ 5 })^2-1=2(16/25)-1=32/25-1=7/25\]

Answer 8

Okay. I think I mostly understand it now. Thank you for your help.