Question

prove that acos2B + bcosC = ccosB - 2bcosAcosB

Answer 1

Step-by-step explanation:

Given in the question

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

now,

putting A in place of B in (A-B) ,we get

cos(A-A)= cosA cosA + sinA sinA

=> cos0 = cos^2A + sin^2A

Since , cos 0 = 1 , therefore

cos^2A + sin^2 A = 1 ………(1)

Again ,

putting B in place of A in (A-B), we get

cos(B-B) = cos B cos B + sin B sinB

=> cos0 =cos^2B + sin^2 B

=> cos^2B + sin^2B = 1 ……..(2)

Adding equation (1) and (2) , we get

=> cos^2A + sin^2A + cos^2B + sin^2B = 1+1

=> cos^2A + sin^2B = (1-sin^2A)+(1-cos^2B)

=> cos^2A + sin^2B = cos^2A + sin^2B ………(3)

As the question is changed now ,earlier it was different,but still we can prove that .

Let's put A=B=x on the right side of the equation (3) we get

cos²A+sin²B= cos²x + sin²x =1