Question

If cos(A+B) =0, Then sin (A-B) is equal to what?

Answer 1

cos( a+b) = 0 therefore a+b= 90o {cos90 = 0} a=90-b .....1 sin(a-b) = cos( 90- [a-b]) = cos(90 -a+b) ......2 substituting equation 1 in equation 2 cos(90-[90-b]+b) = cos(90-90+b+b) =cos 2b

Answer 2

sin (A-B) = cos 2B = - cos2A  if cos(A+B) =0

Given:

cos(A+B) =0

To Find:

sin (A-B)

Solution:

cos(A+B) =0

cos 90° = 0

=> A + B  = 90°

=> A = 90° - B   or B = 90° - A

Substitute A = 90 - B in sin(A- B)

= sin (90° - B - B)

= sin (90° - 2B)

Using sin(90° - x) = cos x

= cos 2B

Hence sin (A-B) = cos 2B

or

Substitute B = 90° - A in sin(A- B)

= sin( A - (90° - A))

= sin(2A - 90°)

Using sin(-x) = - sinx

= - sin(90° - 2A)

= - cos2A

Hence sin (A-B) = cos 2B = - cos2A