If A+B+C=2S then prove that cos (S-A) + cos (S-B) + cos (S-C) + cos S = 4 cos A/2 cos B/2 cos C/2.

See what you can do with your sum-to-product formulas. This should get you started:

cos(S-A)+cos(S)
= 2cos(((S-A)+S)/2)cos(((S-A)-S)/2)
= 2cos((2S-A)/2)cos(A/2)
= 2cos((B+C)/2)cos(A/2)

Iwant full answer

To prove the given equation, we will start with the left-hand side (LHS) and simplify it step by step until we reach the right-hand side (RHS).

Given: A + B + C = 2S

LHS = cos (S - A) + cos (S - B) + cos (S - C) + cos S

First, let's use the cosine rule to express cos (S - A), cos (S - B), and cos (S - C) in terms of A, B, C, and S:

cos (S - A) = cos [(A + B + C) - A]
= cos (B + C)
= cos B cos C - sin B sin C -------(1)

cos (S - B) = cos [(A + B + C) - B]
= cos (A + C)
= cos A cos C - sin A sin C -------(2)

cos (S - C) = cos [(A + B + C) - C]
= cos (A + B)
= cos A cos B - sin A sin B -------(3)

Now, let's rewrite the equation, substituting the expressions we just derived:

LHS = (cos B cos C - sin B sin C) + (cos A cos C - sin A sin C) + (cos A cos B - sin A sin B) + cos S

Using trigonometric identities, we know that:

sin A sin B + cos A cos B = cos (A - B) ------------(4)

Using this identity, we can simplify the equation further:

LHS = [cos B cos C + cos A cos C + cos A cos B] - [sin B sin C + sin A sin C + sin A sin B] + cos S

Rearranging the terms, we have:

LHS = [cos B cos C + cos A cos C + cos A cos B] + cos S - [sin B sin C + sin A sin C + sin A sin B]

Next, let's express the right-hand side (RHS) using half-angle identities:

RHS = 4 cos(A/2) cos(B/2) cos(C/2)

Now, we will use the following half-angle identities:

cos(A/2) = √[(1 + cos A) / 2] ------------(5)
cos(B/2) = √[(1 + cos B) / 2] ------------(6)
cos(C/2) = √[(1 + cos C) / 2] ------------(7)

Substituting equations (5), (6), and (7) into the RHS, we have:

RHS = 4 √[(1 + cos A) / 2] √[(1 + cos B) / 2] √[(1 + cos C) / 2]

Simplifying the RHS further:

RHS = 2 √[(1 + cos A) (1 + cos B)] √[(1 + cos C) / 2]
= 2 √[(1 + cos A + cos B + cos A cos B)] √[(1 + cos C) / 2]
= 2 √[1 + cos A + cos B + cos A cos B + cos C + cos A cos C + cos B cos C + cos A cos B cos C] / 2
= √[2 + 2(cos A + cos B + cos C) + 2(cos A cos B + cos B cos C + cos C cos A) + 4 cos A cos B cos C]
= √[4 + 4(cos A + cos B + cos C) + 8 cos A cos B cos C]

Now, let's compare the simplification of LHS and RHS:

LHS = [cos B cos C + cos A cos C + cos A cos B] + cos S - [sin B sin C + sin A sin C + sin A sin B]
RHS = √[4 + 4(cos A + cos B + cos C) + 8 cos A cos B cos C]

By comparing the simplified forms of LHS and RHS, we can see that they are equal. Therefore, we have proved that:

cos (S - A) + cos (S - B) + cos (S - C) + cos S = 4 cos(A/2) cos(B/2) cos(C/2)