Ifa +b+c=3pi/2 then cos 2a + cos 2b + cos 2c is ?

Here is the solution

As A+B+C=3π/2, we have A+B=(3π/2−C) and cos(A+B)=cos(3π/2−C)=−sinC Hence cos2A+cos2B+cos2C =2cos(2A+2B/2)cos(2A−2B/2)+cos2C =2cos(A+B)cos(A−B)+cos2C =2cos(3π/2−C)cos(A−B)+1−2...