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`cos^2(3pi/5)*cos^2(4pi/5)`

Answer» Here, we will use ,
`cos 36^@ = (sqrt5+1)/4 and cos72^@ = (sqrt5-1)/4`
Now, `cos^2((3pi)/5) = (cos(108^@))^2 = (cos(pi - 72^@))^2 = (-cos72^@)^2`
`=((sqrt5-1)/4)^2 = (5+1-2sqrt5)/16 = (3-sqrt5)/8`
Now, `cos^2((4pi)/5) = (cos(144^@))^2 = (cos(pi - 36^@))^2 = (-cos36^@)^2`
`=((sqrt5+1)/4)^2 = (5+1+2sqrt5)/16 = (3+sqrt5)/8`
`:. cos^2((3pi)/5)*cos^2((4pi)/5) = (3-sqrt5)/8 ** (3+sqrt5)/8 = (9-5)/64 = 4/64 = 1/16`
`:. cos^2((3pi)/5)*cos^2((4pi)/5) = 1/16`


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