The Trigonometric Identity
In a triangle ABC, angles A, B, and C satisfy A + B + C = π. A student is asked to prove that sin A + sin B + sin C = 4 cos(A/2) cos(B/2) cos(C/2). The student writes: 'Using the sum-to-product formula, sin A + sin B = 2 sin((A+B)/2) cos((A-B)/2). Since A+B = π - C, sin((A+B)/2) = sin(π/2 - C/2) = cos(C/2). Then sin A + sin B + sin C = 2 cos(C/2) cos((A-B)/2) + sin C. Now sin C = 2 sin(C/2) cos(C/2). So the expression becomes 2 cos(C/2)[cos((A-B)/2) + sin(C/2)]. Then use cos((A-B)/2) + sin(C/2) = 2 cos(A/2) cos(B/2). Hence the identity holds.' The teacher asks you to verify each step and to find the value of the expression when A = 60°, B = 60°, C = 60°.
1. Critically examine the student's proof. Is every step correct? If there is an error, identify and correct it.
Read the worked answer
The student's proof is mostly correct but has a subtle error in the last step. Let's verify:
- (\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}) is correct.
- Since (A+B = \pi - C), (\sin\frac{A+B}{2} = \sin\left(\frac{\pi}{2} - \frac{C}{2}\right) = \cos\frac{C}{2}). Correct.
- So (\sin A + \sin B = 2\cos\frac{C}{2}\cos\frac{A-B}{2}). Correct.
- (\sin C = 2\sin\frac{C}{2}\cos\frac{C}{2}). Correct.
- Sum: (2\cos\frac{C}{2}\cos\frac{A-B}{2} + 2\sin\frac{C}{2}\cos\frac{C}{2} = 2\cos\frac{C}{2}\left(\cos\frac{A-B}{2} + \sin\frac{C}{2}\right)). Correct.
- Now the student claims (\cos\frac{A-B}{2} + \sin\frac{C}{2} = 2\cos\frac{A}{2}\cos\frac{B}{2}). This is incorrect. The correct identity is (\cos\frac{A-B}{2} - \cos\frac{A+B}{2} = 2\sin\frac{A}{2}\sin\frac{B}{2}), but here we have (\sin\frac{C}{2}). Since (\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}), we have (\sin\frac{C}{2} = \cos\frac{A+B}{2}). So (\cos\frac{A-B}{2} + \sin\frac{C}{2} = \cos\frac{A-B}{2} + \cos\frac{A+B}{2} = 2\cos\frac{A}{2}\cos\frac{B}{2}) using the sum-to-product formula (\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}). Indeed, with (x = \frac{A-B}{2}) and (y = \frac{A+B}{2}), we get (x+y = A), (x-y = -B), so (\cos\frac{A-B}{2} + \cos\frac{A+B}{2} = 2\cos\frac{A}{2}\cos\frac{B}{2}). So the student's step is actually correct, but they didn't justify it. The proof is valid if we fill in that (\sin\frac{C}{2} = \cos\frac{A+B}{2}). So the student's proof is correct but missing a justification. Thus, no error, but incomplete explanation.
Alternatively, we can prove the identity directly by using (\sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}) which is a known identity. The student's derivation is fine with the added note.
2. Compute the value of sin A + sin B + sin C when A = B = C = 60°. Verify that it matches the right-hand side of the identity.
Read the worked answer
When (A = B = C = 60^\circ = \frac{\pi}{3}), we have (\sin A = \sin B = \sin C = \sin 60^\circ = \frac{\sqrt{3}}{2}). So the left-hand side is (3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}). The right-hand side: (4 \cos 30^\circ \cos 30^\circ \cos 30^\circ = 4 \left(\frac{\sqrt{3}}{2}\right)^3 = 4 \cdot \frac{3\sqrt{3}}{8} = \frac{3\sqrt{3}}{2}). They match. So the identity holds for the equilateral triangle.