Question 963370
{{{sin^2(theta)+cos^2(theta)=1}}}
{{{25/49+cos^2(theta)=1}}}
{{{cos^2(theta)=24/49}}}
{{{cos(theta)=0 +- (2/7)sqrt(6)}}}
Since it's in the first quadrant,
{{{cos(theta)=(2/7)sqrt(6)}}}
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{{{cos(2*theta)=cos^2(x)-sin^2(x)}}}
{{{cos(2*theta)=24/49-25/49}}}
{{{cos(2*theta)=-1/49}}}