Consider the equation 3cos2A+2cos2B=4 It is solved as 3(1−sin2A)+2(1−sin2B)‌‌=4 5−3sin2A−2sin2B‌‌=4 3sin2A‌‌=1−2sin2B 3sin2B‌‌=cos‌2‌B Now ‌
3‌sin‌A
sin‌B
=‌
2‌cos‌B
cos‌A
3‌sin‌A‌cos‌A=2‌cos‌B‌sin‌B 3‌sin‌2‌A=2‌sin‌2‌B Now cos(A+2B)=cos‌A‌cos‌2‌B−sin‌A‌sin‌2‌B =cos‌A(3sin2A)−sin‌A(‌