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Find 2(theta) given cos(theta) = 5/8 and 0 < theta < pi/2
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use x for theta
cos(x)=5/8
cos^2(x)=25/64
sin^2(x)=1-cos^2(x)=1-25/64=64/64-25/64=39/64
Identity:cos(2x)=cos^2(x)-sin^2(x)
=25/64-39/64=-14/64=-7/32
2x=arccos(-7/32)
..
check:(w/calculator)
x=arccos(5/8)=0.89566..
2x=(2*.89566)=1.7913..
arccos(-7/32)=1.7913..