Verified answer

Make use of the identity cos(2x) = 2 cos²(x) - 1

cos(2*π/8) = 2 cos²(π/8) - 1

cos(π/4) = 2 cos²(π/8) - 1

1/√2 = 2 cos²(π/8) - 1

2 cos²(π/8) = 1 + 1/√2

2 cos²(π/8) = (√2 + 1)/√2

cos²(π/8) = (√2 + 1)/(2√2)

cos(π/8) = √[(√2 + 1)/(2√2)]

Hello grey

its very possible to find cos(TT/8) if you know cos(TT/4) = 1/(√2)

I will show you how to do this

Do you know the trignametrical identity about cos2a = 2cos^2a - 1

so cosTT/4 = 2cos^2(TT/8) - 1

2cos^2(TT/8) = 1/(√2) + 1

cos^2 (TT/8) = 1/2 [ 1 + (√2)] /(√2)

cos(tt/8) = 1/(√2)[ { 1 + (√2) /(√2)} ]^ 1/2

cos pi/4 = 2cos^2 pi/8 -1

2cos^2 pi/8 -1 = 1/(√2)

cos^2 pi/8 = 1/(√2)+1

cos pi/8 = √(1/(√2)+1 )