Question

If , sinA+sinB=a, and cosA+cosB=b, then how do we prove that:


cos(A+B)=(b2- a2)/ (b2+a2)

sin(A+B)=2ab/(b2+a2)

Answer 1

Step-by-step explanation:

Let SinA + SinB = a ------------(1)

and CosA + CosB = b ------------(2)

Sq the two eqns, and then subtract them

You will get, Cos2A+Cos2B+2Cos(A+B) = b2-a2.

So, 2Cos(A+B)Cos(A-B)+2Cos(A+B) = b2-a2.

or Cos(A+B){2Cos(A-B)+2} = b2-a2 ----------------(3)

Now Sq the two eqns and add, ie (2)^2+(1)^2

You will get 2+2Cos(A-B) = a2+b2 -----------[Use this in (3)]

You'd get Cos(A+B){a2+b2) = (b2-a2).

or Cos(A+B) = (b2-a2)/{a2+b2)

Answer 2

Step-by-step explanation:

cos(a+b)=b2-a2/a2+b2

√1-sin^2(a+b)=b2-a2 /b2+a2. (since sin2a +cos2a=1)

1-sin^2(a+b)=(b2-a2/a2+b2)^2

=b4+a4-2a^2b^2/b4+a4+2a^2b^2

sin2(a+b)=1-(b4+a4-2a2b2/b4+a4+2a2b2)

sin2(a+b)=4a^2b^2/a2+b2

sin(a+b)=√4a^2b^2/a2+b2

sin(a+b)=2ab/a2+b2

Hence,proved