How to find the exact value of cos(2tan^-1(9/40))?

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Answer 1 · 2018-02-05T09:26:58

#cos(2tan^(-1)(9/40))=1519/1681#

Let #2tan^(-1)(9/40)=A#

i.e. #tan^(-1)(9/40)=A/2# and #tan(A/2)=9/40#

then as #cos2theta=(1-tan^2theta)/(1+tan^2theta)#

we have #cosA=cos(2tan^(-1)(9/40))#

= #(1-tan^2(A/2))/(1+tan^2(A/2))#

= #(1-(9/40)^2)/(1+(9/40)^2)#

= #(1-81/1600)/(1+81/1600)#

= #(1519/1600)/(1681/1600)#

= #1519/1681#