sinA.sin(B – C) + sinB.sin(C – A) + sinC.sin(A – B) = 0

1 answer

This equation is an identity in trigonometry known as the sum-to-product identity. It states that for any three angles A, B, and C, the sum of the products of the sines of the differences of pairs of these angles equals zero.

sinA.sin(B – C) + sinB.sin(C – A) + sinC.sin(A – B) = 0

This identity can be derived using the product-to-sum identity for sines:

sin(A + B) = sinA.cosB + cosA.sinB

By rearranging terms and substituting appropriate values, we can prove this identity:

Start with:
sinA.sin(B – C) + sinB.sin(C – A) + sinC.sin(A – B)

Using the product-to-sum identity:
= [sinA.cos(B – C) – cosA.sin(B – C)] + [sinB.cos(C – A) – cosB.sin(C – A)] + sinC.sin(A – B)

Expanding each product:
= sinA.cosB.cosC + sinA.sinB.sinC – cosA.sinB.cosC + cosA.cosB.sinC + sinB.cosA.cosC – sinB.sinA.sinC + sinC.sinA.cosB – cosC.sinA.sinB – sinC.cosA.cosB

Rearranging terms:
= (sinA.sinB.sinC – cosA.cosB.cosC) + (sinB.sinC.sinA – cosB.cosC.cosA) + (sinC.sinA.sinB – cosC.cosA.cosB)

Using the identity sin(x) = sin(pi - x) and cos(x) = -cos(pi - x):
= (sinA.sinB.sinC + cosA.cosB.cosC) + (sinB.sinC.sinA + cosB.cosC.cosA) + (sinC.sinA.sinB + cosC.cosA.cosB)

= sinA.cosB.cosC + sinA.sinB.sinC + cosA.sinB.cosC + cosA.cosB.sinC + sinB.cosA.cosC + sinB.sinA.sinC + sinC.cosA.cosB + cosC.sinA.sinB + sinC.sinB.sinA

= sinA.cosB.cosC + sinA.sinB.sinC – cosA.sinB.cosC + cosA.cosB.sinC + sinB.cosA.cosC – sinB.sinA.sinC + sinC.sinA.cosB – cosC.sinA.sinB – sinC.cosA.cosB

= sinA.sin(B – C) + sinB.sin(C – A) + sinC.sin(A – B)

Hence, sinA.sin(B – C) + sinB.sin(C – A) + sinC.sin(A – B) = 0

This identity is useful in various trigonometric calculations and proofs.