This is like the preceding problem except that you need to g

This is like the preceding problem, except that you need to get all answers correctly to get credit.
Suppose

cosu = 3/5

sinu is positive ==> u lies in the first quadrant

since both sinu and cosu are positive only in Quadrant I

cosu = adjacent side / hypotenuse = 3/5

==> hypotenuse = 5 , adjacent side = 3

We have from pythagorous theorem hypotenuse² = opposite side² + adjacent side²

==> 5² = opposite side² + 3²

==> opposite side² = 25 -9

==> opposite side² = 16

==> opposite side = 4

==> sinu = opposite side / hypotenuse = 4/5

sin(u - ) = sin[ - ( - u)]

==> -sin( - u) since sin(-x) = -sinx

==> -sinu since sin( - x) = sinx

==> - 4/5

==> sin(u - ) = -4/5

cos(u - ) = cos[-( -u)]

==> cos( -u) since cos(-x) = cosx

==> -cosu since cos( -x) = -cosx

==> -3/5

Hence cos(u - ) = -3/5

sin(u - /2) = sin(-(/2 -u))

==> -sin(/2 -u) since sin(- x) = -sinx

==> -cos(u) since sin(/2 - x) = cosx

==> -3/5

Hence sin(u - /2) = -3/5

cos(u - /2) = cos(-(/2 -u))

==> cos(/2 -u) since cos(- x) = cosx

==> sin(u) since cos(/2 - x) = sinx

==> 4/5

Hence cos(u - /2) = 4/5