Question

How do you evaluate `cos (pi/2) cos(pi/7) + sin(pi/2) sin(pi/7)`?

Answer 1

Inserting `cos(pi/2)=0 and sin(pi/2)=1` in the given expression we get

`cos(pi/2)cos(pi/7)+sin(pi/2)sin(pi/7)`

`=0xxcos(pi/7)+1xxsin(pi/7)`

`=sin(pi/7)=0.434`