How can we sum up sin and cos series when the angles are in A.P (arithmetic progression) ?For example here is the sum of cos series:

∑k=0n−1cos(a+k⋅d)=sin(n×d2)sin(d2)×cos(2a+(n−1)⋅d2)
There is a slight difference in case of sin ,which is:

∑k=0n−1sin(a+k⋅d)=sin(n×d2)sin(d2)×sin(2a+(n−1)⋅d2)
How do we prove the above two identities?

A

**NSWER:-**
Let

S=sin(a)+sin(a+d)+⋯+sin(a+nd)
Now multiply both sides by sind2 . Then you have
S×sin(d2)=sin(a)sin(d2)+sin(a+d)⋅sin(d2)+⋯+sin(a+nd)⋅sin(d2)
Now, note that

sin(a)sin(d2)=12⋅[cos(a−d2)−cos(a+d2)]
and
sin(a+d)⋅sin(d2)=12⋅[cos(a+d−d2)−cos(a+d+d2)]
Then by doing the same thing you will have some terms cancelled out.
You can easily see which terms are going to get Cancelled. Proceed and
you should be able to get the formula.

I tried this by seeing this post. This has been worked for the case whend=1 . Just take a look here:

I tried this by seeing this post. This has been worked for the case when

- http://mathforum.org/library/drmath/view/72777.html

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