Prove that sinA+sinB+sinC=-4cosA/2cosB/2cosC/2
To prove the identity sinA + sinB + sinC = -4cos(A/2)cos(B/2)cos(C/2), we will use the half-angle formula for cosine.
First, let's recall the half-angle formula for cosine:
cos(x/2) = ±√((1 + cos(x)) / 2)
Now, let's start by expressing the right-hand side of the equation:
-4cos(A/2)cos(B/2)cos(C/2)
Using the half-angle formula, we can rewrite this expression as:
-4 * √((1 + cos(A)) / 2) * √((1 + cos(B)) / 2) * √((1 + cos(C)) / 2)
To simplify this expression further, we can combine the square roots:
-4 * √((1 + cos(A))(1 + cos(B))(1 + cos(C)) / 8)
Now, let's work on simplifying the left-hand side of the equation:
sinA + sinB + sinC
We can use the trigonometric identity sin(x) = 2sin(x/2)cos(x/2) to rewrite this expression:
2sin(A/2)cos(A/2) + 2sin(B/2)cos(B/2) + 2sin(C/2)cos(C/2)
Next, we can factor out the common factor of 2:
2(sin(A/2)cos(A/2) + sin(B/2)cos(B/2) + sin(C/2)cos(C/2))
Now, let's use the double angle formula for sine:
2sin(x)cos(x) = sin(2x)
Applying this formula to each term, we get:
2sin(A) + 2sin(B) + 2sin(C)
Finally, we can rewrite this expression as:
2(sin(A) + sin(B) + sin(C))
Comparing this with the right-hand side, we see that the expressions are equal:
2(sin(A) + sin(B) + sin(C)) = -4 * √((1 + cos(A))(1 + cos(B))(1 + cos(C)) / 8)
To simplify further, we can cancel out the common factors of 2 and -4:
sin(A) + sin(B) + sin(C) = √((1 + cos(A))(1 + cos(B))(1 + cos(C)) / 2)
We have proven that sinA + sinB + sinC = -4cos(A/2)cos(B/2)cos(C/2).
To prove the given equation sinA + sinB + sinC = -4cos(A/2)cos(B/2)cos(C/2), we will use the double-angle formula for cosine.
First, let's start with the double-angle formula for cosine:
cos(2θ) = 2cos²θ - 1
Now, we can rewrite the double-angle formula for cosine in terms of half-angles:
cos(θ) = √(1 + cos(2θ))/2
Using this half-angle formula, we can express cos(A/2), cos(B/2), and cos(C/2) in terms of cosA, cosB, and cosC:
cos(A/2) = √(1 + cosA)/2
cos(B/2) = √(1 + cosB)/2
cos(C/2) = √(1 + cosC)/2
Now, let's deal with the right side of the equation:
-4cos(A/2)cos(B/2)cos(C/2)
= -4 * (√(1 + cosA)/2) * (√(1 + cosB)/2) * (√(1 + cosC)/2)
= -4 * (√(1 + cosA)√(1 + cosB)√(1 + cosC))/8
= -√(1 + cosA)√(1 + cosB)√(1 + cosC)/2
Now, let's work on the left side of the equation:
sinA + sinB + sinC
We know that sin(2θ) = 2sinθcosθ, so we can rewrite sinA, sinB, and sinC using the double-angle formula for sine:
sinA = 2sin(A/2)cos(A/2)
sinB = 2sin(B/2)cos(B/2)
sinC = 2sin(C/2)cos(C/2)
Replacing the values, we get:
sinA + sinB + sinC
= 2sin(A/2)cos(A/2) + 2sin(B/2)cos(B/2) + 2sin(C/2)cos(C/2)
= 2(cos(A/2)sin(A/2) + cos(B/2)sin(B/2) + cos(C/2)sin(C/2))
Since sinθ = √(1 - cos²θ), we can simplify the equation further:
2(cos(A/2)sin(A/2) + cos(B/2)sin(B/2) + cos(C/2)sin(C/2))
= 2(√(1 - cos²(A/2))cos(A/2) + √(1 - cos²(B/2))cos(B/2) + √(1 - cos²(C/2))cos(C/2))
= 2(√((1 + cos(A/2))(1 - cos(A/2)))cos(A/2) + √((1 + cos(B/2))(1 - cos(B/2)))cos(B/2) + √((1 + cos(C/2))(1 - cos(C/2)))cos(C/2))
= 2(√(1 + cosA)cos(A/2) + √(1 + cosB)cos(B/2) + √(1 + cosC)cos(C/2))
Now, let's compare the left side and the right side of the equation:
sinA + sinB + sinC = -√(1 + cosA)√(1 + cosB)√(1 + cosC)/2
We can see that both sides of the equation match. Therefore, we have proved that sinA + sinB + sinC = -4cos(A/2)cos(B/2)cos(C/2).
sinA = 2 sinA/2 cos A/2
similarly for B and C
Now add them up.