document.write( "Question 250835: find cos θ given that cos2θ = 5/7 and 0≤θ<π/2 \n" ); document.write( "
Algebra.Com's Answer #184443 by drk(1908)\"\" \"About 
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Cos2x = cos(x + x) = cosx*cosx -sinx*sinx = cos^2(x) - sin^2(x)
\n" ); document.write( "We have the following identity: cos^2(x) + sin^2(x) = 1 ; sin^2(x) = 1 - cos^2(x)
\n" ); document.write( "By substitution,
\n" ); document.write( "2cos^2(x) - 1 = 5/7
\n" ); document.write( "cos^2(x) = 6/7
\n" ); document.write( "cos(x) = sqrt(6/7)
\n" ); document.write( "x = cos^(-1) (6/7)
\n" ); document.write( "at this point you put it into your calculator.
\n" ); document.write( "x ~ 22.5 degrees, or ~ pi/8.\r
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