Matzic|

We have learned previously the summation of two Sine functions with two different angles. This concept is called as multiple and sub multiple angles’ trigonometry. In this article, we are going to express the subtraction of two Sine functions with two different angles. In order to prove this, we have to assume C and D are two different angles of a right angle triangle. And the subtraction is as stated below mathematically.

SinC – SinD

We can convert the above trigonometric equaion with mathematical fundamentals. However, there is an alternative method to simplify this funtion easily. We have to consider one assumption in order to covert multiple angles into its sub multiple angles.

C = A + B and D = A – B

Now, substitute C and D values in terms of A and B in order to simply our sine function easily. In other words, our actual trigonometric function will be like this.

Sin(A + B) – Sin(A – B)

We have already proved mathematically that

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

And

Sin(A – B) = SinA.CosB – CosA.SinB

Now, we have to submit these two trigonometric formulas in our converted function for further simplification, which means we can write above function as stated below with the help of above two trigonometric formulas.

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

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

» Sin(A + B) – Sin(A – B) = 2.CosA.SinB

We have successfully simplified the trigonometric function in terms of A and B. However, it is not our required solution, which means we have to convert this function in terms of C and D angles.

As per our assumption, C = A + B and D = A – B. We can convert these assumptions in reverse direction, which means A = (C + D)/2 and B = (C – D)/2. Now substitute these values in order to get our required solution.

Sin(A + B) – Sin(A – B) = 2.CosA.SinB

» SinC – SinD = 2.Cos[(C + D)/2].Sin[(C – D)/2]

This is the actual solution for this function. We use this formula to deal trigonometry problems.

Conclusion:

SinC – SinD = 2.Cos[(C + D)/2].Sin[(C – D)/2]