If A+B+C=^(^=180) prove that

sinA + sinB - sinC = 4sinA/2 sinB/2 cosC/2

Math

4sinA/2sinB/2cosC/2

To prove the given equation, we'll start with the left side and simplify it using trigonometric identities.

Let's begin by using the formula for the sine of the sum of two angles:
sin(X + Y) = sin(X)cos(Y) + cos(X)sin(Y)

We can express sinA + sinB as sin(A + B) using the above formula:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

Now, let's consider the sum of the angles A, B, and C:
A + B + C = 180

Rearranging this equation, we get:
A + B = 180 - C

Using this in the equation for sin(A + B), we have:
sin(A + B) = sin(180 - C)

Now, we'll apply a trigonometric identity:
sin(180 - C) = sin(180)cos(C) - cos(180)sin(C)

The values of sin(180) and cos(180) can be substituted using their respective definitions:
sin(180) = 0
cos(180) = -1

Substituting these values, we have:
sin(180 - C) = 0*cos(C) - (-1)*sin(C)
sin(180 - C) = sin(C)

Now, let's substitute sin(A + B) with sin(C) back into the equation:
sin(A)cos(B) + cos(A)sin(B) = sin(C)

Now, let's simplify the right side of the equation:
4sin(A/2)sin(B/2)cos(C/2)

To simplify it further, we'll use the following trigonometric identity:
sin(2θ) = 2sin(θ)cos(θ)

Notice that A/2 + B/2 = (A + B)/2 = (180 - C)/2 = 90 - C/2. Therefore, we can rewrite it as:
4sin(A/2)sin(B/2)cos(C/2) = 4sin((90 - C/2)/2)sin(C/2)cos(C/2)

Using the identity sin(2θ) = 2sin(θ)cos(θ), we can simplify the right side of the equation:
4sin((90 - C/2)/2)sin(C/2)cos(C/2) = 4sin(90 - C/2)sin(C/2)cos(C/2)
= 4(sin(90)cos(C/2) - cos(90)sin(C/2))sin(C/2)cos(C/2)
= 4(1*cos(C/2) - 0*sin(C/2))sin(C/2)cos(C/2)
= 4cos(C/2)sin(C/2)cos(C/2)
= 4sin(C/2)cos(C/2)^2

Finally, we have:
sin(A)cos(B) + cos(A)sin(B) = 4sin(C/2)cos(C/2)^2

Therefore, we have proved that:
sinA + sinB - sinC = 4sinA/2 sinB/2 cosC/2