January 04 2010
Deriving CosC – CosD = – 2.Sin[(C + D)/2].Sin[(C - D)/2]
This is also one more important mathematical formula in multiple and sub-multiple angles’ trigonometry. We have previously developed summation of sine, summation of cosine and differentiation of sine function with two different angles. Similarly, we are going to derive differentiation of Cosine function.
First of all we have to assume, C and D are different angles of a right angle triangle. Now we are going to subtract these two angles with respect Cosine, which means we are developing simplification of below trigonometric function.
CosC – CosD
It is very complicated or not easy to simplify this trigonometric function directly by using mathematical applications. However, we can simplify this function by taking an assumption. Consider C and D are equal to A + B and A – B respectively for our convenience.
Now, convert the entire trigonometric function in A and B angles.
CosC – CosD = Cos(A + B) – Cos(A – B)
We have already developed expansion of cosine functions as stated below.
Cos(A + B) = CosA.CosB – SinA.SinB
And
Cos(A – B) = CosA.CosB + SinA.SinB
Add above two trigonometric formulas in order to further simply our actual function.
Cos(A + B) – Cos(A – B) = (CosA.CosB – SinA.SinB) – (CosA.CosB + SinA.SinB)
= CosA.CosB – SinA.SinB – CosA.CosB – SinA.SinB
= – 2.SinA.SinB
We have successfully simplified this mathematical function but this is not our actual solution. In other words, we have to convert this entire function in terms of C and D.
If C = A + B and D = A – B then A = (C + D)/2 and B = (C – D)/2.
With the help of these assumptions, we can now completely change above solution.
CosC – CosD = – 2.SinA.SinB = – 2.Sin[(C + D)/2].Sin[(C – D)/2]
This is actual solution of actual function. This mathematical fundamental helped us in solving trigonometry’s problems easily and developing advanced trigonometry formulas.
Conclusion:
- CosC – CosD = – 2.Sin[(C + D)/2].Sin[(C - D)/2]