A Maths Trigo Proving

Prove that (cosA - sinB)/ [cos(A - B)] = [cos(A + B)] / (cosA + sinB)

Cross multiply, simplify left side to show equal to right side.

Cross multiply, simplify left side to show equal to right side.

We cannot use cross multiply in the proofing ie we need to keep to the rule

using the LHS as a function and work until it is equal to the RHS in terms of a

function or using the RHS as a function and work until is equal to the LHS in terms

of a function.

Hi,

Thanks for sharing a nice question :)

1. Start with LHS.

2. Multiply it by (cos A + sin B) / (cos A + sin B), because we want to see (cos A + sin B) in the denominator eventually.

3. Numerator then becomes (cos A)^2 - (sin B)^2. Keep the denominator as it is.

4. Use the cosine double-angle identity twice to show

(cos A)^2 - (sin B)^2 = cos (A + B) cos (A - B).

5. Cancel cos (A - B), which is common in both numerator and denominator.

The original question may be rephrased in two parts, since Cambridge examiners are more inclined to guide candidates in systematic problem-solving:

Show that cos (A + B) cos (A - B) = (cos A)^2 - (sin B)^2.

Hence prove that (cos A - sin B) / [cos (A - B)] = [cos (A + B)] / (cos A + sin B).

Thanks!

Cheers,
Wen Shih