#cos((5pi)/7 "rad")# is the equivalent of #cos(900/7)^o#, and since you are working with a denominator of #7#, I would not expect you to have to evaluate the exact values (with radicals and all) for this on a test. But here goes!
This is not obvious, but you have to start from here so that you can use sum and difference identities:
#cos((5pi)/7) = cos((10pi)/14)#
You could just say:
#cos(u+v) = cosucosv - sinusinv#
#=> cos(2x) = cos^2x - sin^2x#
#color(blue)(cos((5pi)/7)) = cos((5pi)/14 + (5pi)/14) color(blue)(= cos^2((5pi)/14) - sin^2((5pi)/14))#
Or, you could get these down to first-degree trig functions by noting that:
#cos(x) = sin(x+pi/2) = sin(x+(7pi)/14)#
Also:
#sinx = -sin(-x)#
#cosx = cos(-x)#
#sinx = sin(xpm2pi)#
#cosx = cos(xpm2pi)#
Therefore, you can do this:
#cos((10pi)/14) = cos(-(10pi)/14) = cos((18pi)/14)#
#cos((18pi)/14) = sin((25pi)/14) = -sin(-(25pi)/14) = -sin((3pi)/14)#
Overall, you get:
#cos((5pi)/7) = -sin((3pi)/14)#
Now we can use the sum identity for #sin#.
#sin(upmv) = sinucosv pm cosusinv#
#=> sin(2x) = 2sincosx#
Finally, we have the result:
#color(blue)(cos((5pi)/7)) = -sin((3pi)/14) = -sin((3pi)/28 + (3pi)/28) color(blue)(= -2sin((3pi)/28)cosx((3pi)/28))#
