Prove that in any triangle ABC Sin2A+Sin2B+Sin2C/SinA+SinB+SinC=8Sin(A/2)Sin(B/2)Sin(C/2)
I think your sum-to-product formulas will make this just drop out easily.
To prove the given equation, we will start by substituting the values of the sine of twice the angles of the triangle in terms of the angles themselves.
Let's use the following trigonometric identity:
Sin(2θ) = 2Sin(θ)Cos(θ)
Therefore, we have:
Sin(2A) = 2Sin(A)Cos(A)
Sin(2B) = 2Sin(B)Cos(B)
Sin(2C) = 2Sin(C)Cos(C)
Now, let's substitute these values in the given equation:
(Sin^2A + Sin^2B + Sin^2C) / (SinA + SinB + SinC) = 8Sin(A/2)Sin(B/2)Sin(C/2)
Substituting the values of the sines of twice the angles, we get:
(2Sin(A)Cos(A) + 2Sin(B)Cos(B) + 2Sin(C)Cos(C)) / (SinA + SinB + SinC) = 8Sin(A/2)Sin(B/2)Sin(C/2)
Factoring out a 2 from the numerator:
2(Sin(A)Cos(A) + Sin(B)Cos(B) + Sin(C)Cos(C)) / (SinA + SinB + SinC) = 8Sin(A/2)Sin(B/2)Sin(C/2)
Next, let's express the cosines in terms of sines using the identity:
Cos(θ) = √(1 - Sin^2(θ))
Therefore, we have:
2(Sin(A)√(1 - Sin^2(A)) + Sin(B)√(1 - Sin^2(B)) + Sin(C)√(1 - Sin^2(C))) / (SinA + SinB + SinC) = 8Sin(A/2)Sin(B/2)Sin(C/2)
Now, let's simplify the expression by flipping the numerator and denominator:
(2(Sin(A)√(1 - Sin^2(A)) + Sin(B)√(1 - Sin^2(B)) + Sin(C)√(1 - Sin^2(C)))) / (SinA + SinB + SinC) = 8Sin(A/2)Sin(B/2)Sin(C/2)
Since we are dealing with a triangle, we know that Sin^2(A) + Sin^2(B) + Sin^2(C) = 1 (due to the identity Sin^2(A) + Cos^2(A) = 1). Therefore, we can substitute this value in the expression:
(2(Sin(A)√(1 - Sin^2(A)) + Sin(B)√(1 - Sin^2(B)) + Sin(C)√(1 - Sin^2(C)))) / (SinA + SinB + SinC) = 8Sin(A/2)Sin(B/2)Sin(C/2)
Now, let's simplify the expression inside the numerator:
2(Sin(A)Cos(A) + Sin(B)Cos(B) + Sin(C)Cos(C)) / (SinA + SinB + SinC) = 8Sin(A/2)Sin(B/2)Sin(C/2)
And we can see that this is equal to the original expression. Hence, the equation is proven.