Can anyone solve this? Prove cos A / sin B - sin A / cos B = 2 cos (A+B) / sin 2B

1 Answer
Apr 30, 2018

Prove cos(A) / sin(B) - sin(A) / cos(B) = (2 cos (A+B) )/ sin(2B)

Multiply the first term by 1 in the form of cos(B)/cos(B):

cos(A) / sin(B)cos(B)/cos(B) - sin(A) / cos(B) = (2 cos (A+B) )/ sin(2B)

Multiply the second term by 1 in the form of sin(B)/sin(B):

cos(A) / sin(B)cos(B)/cos(B) - sin(A) / cos(B)sin(B)/sin(B) = (2 cos (A+B) )/ sin(2B)

Combine the two terms over the common denominator:

(cos(A)cos(B) - sin(A)sin(B))/(sin(B)cos(B)) = (2 cos (A+B) )/ sin(2B)

Use the identity cos (A+B)=cos(A)cos(B) - sin(A)sin(B):

cos (A+B)/(sin(B)cos(B)) = (2 cos (A+B) )/ sin(2B)

Multiply by 1 in the form of 2/2:

(2cos (A+B))/(2sin(B)cos(B)) = (2 cos (A+B) )/ sin(2B)

Use the identity sin(2B) = 2sin(B)cos(B):

(2cos (A+B))/sin(2B) = (2 cos (A+B) )/ sin(2B) Q.E.D.