Question 1191002

Prove that : {{{cos(72) =(1/2)sqrt(4sin^2(36)-1)}}}

solve left side:

{{{cos (72)}}} .....use identity {{{cos( 72)  =cos (90 - 18)}}}°

use identity

{{{cos(a-b)=sin(a) sin(b) + cos(a) cos(b)}}}........where {{{a=90}}}° and {{{b=18}}}°


 {{{cos (90 - 18) =sin(90) sin(18) + cos(90) cos(18)}}}

{{{ cos (90 - 18) =1* sin(18) + 0*cos(18)}}}

{{{cos (90 - 18) = sin(18) }}}


{{{cos (90 - 18) =(1/4) (sqrt(5) - 1)}}}


 

solve right side:

{{{(1/2)sqrt(4sin^2(36)-1)}}}.....{{{sin^2(36)=5/8 - sqrt(5)/8}}}

={{{(1/2)sqrt(4 (5/8 - sqrt(5)/8) - 1)}}}

={{{(1/2)sqrt((1/2) (3 - sqrt(5)))}}}

={{{(1/2)(sqrt(5)/2 - 1/2)}}}

={{{(1/4)(sqrt(5) - 1)}}}


so, both sides have same answer

then

{{{cos(72) =(1/2)sqrt(4sin^2(36)-1)}}}

{{{(1/4)(sqrt(5) - 1)=(1/4)(sqrt(5) - 1)}}} ->proven