Question

please prove the aforementioned question....​

Answer 1

Step-by-step explanation:

Representing, angle Theta as: A

and, angle Phi as: ∅

So, x cosA - y sinA = x cos∅ - y sin∅

★ x cosA - x cos∅ = y sinA - y sin∅

★ x (cosA - cos∅) = y (sinA - sin∅)

★ x [ 2 sin(A + ∅)/2 sin(∅ - A)/2 ] = y [ 2 cos(A + ∅)/2 sin(A - ∅)/2 ]

★ x [ sin(A + ∅)/2 (-sin(A - ∅)/2) ] = y [ cos(A + ∅)/2 sin(A - ∅)/2 ]

★ -x [ sin(A + ∅)/2 ] = y [ cos(A + ∅)/2 ]

★ -x = y [ cot(A + ∅)/2 ]

★ x + y [ cot(A - ∅)/2 ] = 0

Hence, Proved.

• cos C - cos D = 2 sin(C + D)/2 sin(D - C)/2
• sin C - sin D = 2 cos(C + D)/2 sin(C - D)/2

Thanks! ✨