How do you evaluate $cos(2Arctan(3/4))$ ?

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Answer 1 · 2016-07-12T18:32:21

$cos(2arctan(3/4))=7/25.$

Let $arctan(3/4)=theta$ , sothat, by defn. of $arctan$ function, $tantheta=3/4$ , and, $theta in (-pi/2,pi/2)=(-pi/2,0)uu{0}uu(0,pi/2).$

Note that, $tantheta=3/4 >0, theta !in (-pi/2,0]$ , but, $theta in (0,pi/2).$

Now, reqd. value $=cos(2arctan(3/4))=cos2theta=(1-tan^2theta)/(1+tan^2theta)={1-(3/4)^2}/{1+(3/4)^2}=(16-9)/(16+9)=7/25.$

How do you evaluate cos(2Arctan(3/4))cos(2Arctan(3/4))?