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Given sinA=12/13 and cosB=4/5, how do i solve cos2A

 May 12, 2020
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\(cos(2A)=2cos^2(A)-1\)

Note \(cos^2(A)=1-sin^2(A)\)     (Pythagoras identity, \(sin^2(x)+cos^2(x)=1\) rearrange and solve for \(cos^2(x)\))

\(2(1-sin^2(A))-1\)

\(cos(2A)=2-2sin^2(A)-1\)

Given \(sin(A)=\frac{12}{13}\) , then \(sin^2(A)=\frac{144}{169}\)

Substitute 

\(cos(2A)=2-2(\frac{144}{169})-1\)

\(cos(2A)=-\frac{119}{169}=-0.704142\)

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 May 12, 2020
edited by Guest  May 12, 2020
edited by Guest  May 12, 2020

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