In fact, the calculation process of this kind of topic may be a bit complicated, but you can remember the conclusion. The following process is attached:

S=sina+sin2a+sin3a+...+ Sinna

S = sinna+sin (n-1) a+...+sin2a+Sina.

Add the above two formulas and use the formula Sina+sinβ = 2sin (a+β)/2 * cos (a-β)/2.

Then 2s = [sin (1+n) a/2cos (1-n)/2+sin (1+n) a/2cos (3-n) a/2+...+sin (1)

Through the formula sinacos β =1/2 [sin (a-β)+sin (a+β)]

(Note that the numerator of the trigonometric function above is dizzy with brackets, which are all like this, such as sin( 1+n)a/2, and the denominator 2 is below (1+n)a, which means sin[( 1+n)a/2)].

Then multiply both sides of the above formula by sina/2 to get 2s * Sina/2 = sin (n+1) a/2 [Sina/2+sin (1-n/2) a+sin (n/2-1) a+sin (2

If the sin function is odd function, then SIN (n/2-1) a =-SIN (1-n/2) a. Therefore, all brackets are removed except the first and last items.

So 2s * Sina/2 = 2sin (1+n) a/2sina/2, that is, sn = (sin (na/2) * sin ((n+1) a/2))/(sina/2).

Besides, let me expand it for you. If all the above functions are changed to cos, the result is

s =[cos((n+ 1)a/2)* sinna/2]/sin(a/2)