Question

Suppose S = sin(x) + sin(x + α) + sin(x + 2α) + ... + sin(x + nα), n ∈N. What is the value of S?

Answer 1

`S=(cos(x-(\alpha)/(2))-cos(x+n\alpha-(\alpha)/(2)))/(2sin(\alpha)/(2))`

Explanation:

We are given that
`S=sin(x) +sin(x+\alpha)+sin(x+2\alpha)+....+sin(x+n\alpha),n\in N`

We have to find the value of S

We know that

`\sum_(k=0)^(n-1)sin(x+k.d)=(sinn* (d)/(2))/(sin(d)/(2))* sin((2x+(n-1)d)/(2))`

We have d=
`\alpha`

Substitute the values then we get

`\sum_(k=0)^(n-1)sin(x+k.\alpha)=(sin(n\alpha)/(2))/(sin(\alpha)/(2))* sin((2x+(n-1)\alpha)/(2))`

`\sum_(k=0)^(n-1)sin(x+k.\alpha)=(sin(n\alpha)/(2)\cdot sin((2x+(n-1)\alpha)/(2)))/(sin(\alpha)/(2))`

`S=(sin(n\alpha)/(2)\cdot sin((2x+(n-1)\alpha)/(2)))/(sin(\alpha)/(2))`

`S=(2sin(n\alpha)/(2)\cdot sin((2x+(n-1)\alpha)/(2)))/(2sin(\alpha)/(2))`

`S=(cos(x+(n\alpha)/(2)-(\alpha)/(2)-(n\alpha)/(2))-cos(x+(n\alpha)/(2)-(\alpha)/(2)+(n\alpha)/(2)))/(2sin(\alpha)/(2))`

Because
`cos(x-y)-cos(x+y)=2 sinxsiny`

`S=(cos(x-(\alpha)/(2))-cos(x+n\alpha-(\alpha)/(2)))/(2sin(\alpha)/(2))`