Matzic|

This trigonometric formula may not useful in entire trigonometry. Sometimes, this type of mathematical expression could be useful in trigonometry problem solving. So, we are going to derive this advanced trigonometric formula. Actually, we have successfully derived two angles formula but here three angles A, B and C are come into picture.

However, in the same way we can also prove this theorem. In order to simpler this function we have to consider two angles as one angle, which means consider A + B as one angle and the resultant angle is θ. In other words, here θ is equal to A + B.

Sin(A + B + C) = Sin(θ + C)

We have already proved one trigonometric formula previously. Now, we are going to apply that mathematical application here in order to simplify this function.

Sin(A + B) = SinA.CosB + CosA.SinB

By using this mathematical formula, we can now expand it easily.

Sin(θ + C) = Sinθ.CosC + Cosθ.SinC

Now, we have to again further expand this mathematical function by replacing θ with actual value, which means A + B.

Sin(A + B + C) = Sin(A + B).CosC + Cos(A + B).SinC

We have already written Sin(A + B) expansion above. However, we have to write Cos(A + B) expansion here. Previously, we derived expansion of Cos(A +B) in Sine an Cosine terms.

Cos(A + B) = CosA.CosB – SinA.SinB

Let us substitute these two proven formulas in actual trigonometric function to get final result.

Sin(A + B + C) = Sin(A + B).CosC + Cos(A + B).SinC

» Sin(A + B + C) = [SinA.CosB + CosA.SinB].CosC +[ CosA.CosB – SinA.SinB].SinC

= SinA.CosB.CosC + CosA.SinB.CosC + CosA.CosB.SinC – SinA.SinB.SinC

In mathematics, Σ is summation of terms. So, we can express the above proven trigonometric equation in simpler way.

Sin(A + B + C) = Σ(SinA.CosB.CosC) – SinA.SinB.SinC

This is the final and required solution and expansion of the Sine function with three different angles.