Final answer:
The identity sinA + sinB + sinC = (4cos(A/2))(cos(B/2))(cos(C/2)) can be proven by utilizing trigonometric identities and formulas, including sum and difference identities for sine, and the formula for converting sin(180° - x). The proof is completed by expressing sinA+sinC and sinB in terms of half-angle components and simplifying.
Step-by-step explanation:
Proof of Trigonometric Identity
To prove the identity sinA + sinB + sinC = (4cos(A⁄2))(cos(B⁄2))(cos(C⁄2)), we can utilize trigonometric formulas and identities. Assuming A+B+C=180° (which holds for angles in a triangle), we use the following formulas for simplification:
- sin(a ± b) = sin(a)cos(b) ± cos(a)sin(b)
- cos(2a) = 1 - 2sin²(a) = 2cos²(a) - 1
- sin a + sin b = 2sin((a + b)⁄2)cos((a - b)⁄2)
Now, let's use point 15 and substitute a and b by A and (180°-C), respectively:
sinA + sin(180°-C) = sinA + sinC (since sin(180°-x)=sin(x))
= 2sin((A+(180°-C))⁄2)cos((A-(180°-C))⁄2)
Since A+B+C = 180°:
= 2sin((A+B)⁄2)cos((A-C)⁄2).
Similarly, sinB can be written using point 15 with a substituted by B and b by (180°-A), yielding:
sinB = 2sin((B+(180°-A))⁄2)cos((B-A)⁄2)
= 2sin((A+B)⁄2)cos((A-B)⁄2).
Now, using the sum of angles A and B:
sinA + sinB + sinC = 4sin((A+B)⁄2)cos((A-C)⁄2)cos((A-B)⁄2)
Applying the substitution (A+B)=180°-C:
= 4(sin(90°-C⁄2)cos((A-C)⁄2)cos((A-B)⁄2))
Using the trigonometric identity sin(90°-x) = cos(x), we get:
= 4(cos(C⁄2)cos(A⁄2)cos(B⁄2))
Hence, the identity holds true.