Prove that: cos(30-A)-cos(30+A)=sinA

please solve it with details.

cos (a-b) = cosa*cos b + sin b*sin a

cos (a+b) = cosa*cos b - sin b*sin a

We'll put a = 30 and b = A.

cos (30+A) = cos 30*cos A - sin A*sin 30

cos (30-A) = cos 30*cos A + sin A*sin30

Now, we'll replace these in the original equation:

cos (30-A) - cos (30+A)

= cos 30*cos A + sin A*sin30 - (cos 30*cos A - sin A*sin 30)

= cos 30*cos A + sin A*sin30 - cos 30*cos A + sin A*sin 30

the cos30*cosA is subtracted so we are left with 2sinAsin30

sin 30 = 1/2, so 2sinAsin30 = 2sinA*(1/2) = sinA as 2*(1/2) = 1

Do you know this identity: cos(a - b) = cos(a).cos(b) + sin(a).sin(b)

Do you know this identity: cos(a + b) = cos(a).cos(b) - sin(a).sin(b)



= cos(a - b) - cos(a + b)

= [cos(a).cos(b) + sin(a).sin(b)] - [cos(a).cos(b) - sin(a).sin(b)]

= cos(a).cos(b) + sin(a).sin(b) - cos(a).cos(b) + sin(a).sin(b)

= sin(a).sin(b) + sin(a).sin(b)

= 2.sin(a).sin(b)



= cos(a - b) - cos(a + b) → adapt it to your case where: b = 30

= cos(a - 30) - cos(a + 30)

= 2.sin(a).sin(30) → you know that: sin(30) = 1/2

= 2.sin(a) * (1/2)

= sin(a)

cos(30-A)-cos(30+A)=sinA
L.H.S.
= cos(30-A)-cos(30+A)
= -2 sin[(30-A+30+A)/2] sin[(30-A-30-A)/2] <-- cos A - cos B = -2 sin [(A+B)/2] sin [(A-B)/2]
= -2 sin30 sin (-A)
= -2 (1/2) (-sinA)
= sinA
= R.H.S.

cos 30 cos A + sin 30 sin A
-
cos 30 cos A - sin 30 sin A

2 sin 30 sin A = 2 (1/2) sin A = sin A