By definition of half-range Fourier sine series: :$\displaystyle \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \dfrac {n x} 2$ where: {{begin-eqn}} {{eqn | l = b_n       | r = \frac 2 {2 \pi} \int_0^{2 \pi} \map f x \sin \frac {n \pi x} {2 \pi} \rd x       | c =  }} {{eqn | r = \frac 1 \pi \int_0^{2 \pi} \map f x \sin \frac {n x} 2 \rd x       | c =  }} {{end-eqn}} for all $n \in \Z_{>0}$. Thus: {{begin-eqn}} {{eqn | l = b_n       | r = \frac 1 \pi \int_0^{2 \pi} \map f x \sin \dfrac {n x} 2 \rd x       | c =  }} {{eqn | r = \frac 1 \pi \int_0^\pi \sin \dfrac x 2 \sin \dfrac {n x} 2 \rd x + \frac 1 \pi \int_\pi^{2 \pi} -\sin \dfrac x 2 \sin \dfrac {n x} 2 \rd x       | c = Definition of $f$ }} {{end-eqn}} When $\dfrac {n x} 2 \ne \dfrac x 2$, that is, when $n \ne 1$, we have: {{begin-eqn}} {{eqn | l = \int \sin \dfrac x 2 \sin \dfrac {n x} 2 \rd x       | r = \frac {\sin \paren {\dfrac x 2 - \dfrac {n x} 2} } {2 \paren {\dfrac 1 2 - \dfrac n 2} } - \frac {\sin \paren {\dfrac x 2 + \dfrac {n x} 2} } {2 \paren {\dfrac 1 2 + \dfrac n 2} } + C       | c = Primitive of Sine of $\dfrac x 2$ by Sine of $\dfrac {n x} 2$ }} {{eqn | r = \frac {\sin \paren {\dfrac {\paren {n - 1} x} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} x} 2} } {n + 1} + C       | c =  }} {{end-eqn}} and so for $n \ne 1$: {{begin-eqn}} {{eqn | l = b_n       | r = \frac 1 \pi \int_0^{2 \pi} \map f x \sin \dfrac {n x} 2 \rd x       | c =  }} {{eqn | r = \frac 1 \pi \intlimits {\frac {\sin \paren {\dfrac {\paren {n - 1} x} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} x} 2} } {n + 1} } 0 \pi - \frac 1 \pi \intlimits {\frac {\sin \paren {\dfrac {\paren {n - 1} x} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} x} 2} } {n + 1} } \pi {2 \pi}       | c =  }} {{eqn | r = \frac 1 \pi \paren {\paren {\frac {\sin \paren {\dfrac {\paren {n - 1} \pi} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} \pi} 2} } {n + 1} } - \paren {\frac {\sin \paren {\dfrac {\paren {n - 1} 0} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} 0} 2} } {n + 1} } }       | c =  }} {{eqn | o =       | ro= -       | r = \frac 1 \pi \paren {\paren {\frac {\sin \paren {\dfrac {\paren {n - 1} 2 \pi} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} 2 \pi} 2} } {n + 1} } - \paren {\frac {\sin \paren {\dfrac {\paren {n - 1} \pi} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} \pi} 2} } {n + 1} } }       | c =  }} {{eqn | r = \frac 2 \pi \paren {\frac {\sin \paren {\dfrac {\paren {n - 1} \pi} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} \pi} 2} } {n + 1} }       | c = Sine of Multiple of Pi and simplifying }} {{end-eqn}} When $\dfrac {n x} 2 = \dfrac x 2$, that is, when $n = 1$, we have: {{begin-eqn}} {{eqn | l = \int \sin \dfrac x 2 \sin \dfrac x 2 \rd x       | r = \int \sin^2 \dfrac x 2 \rd x + C       | c =  }} {{eqn | r = \frac x 2 - \frac {\sin 2 \dfrac x 2} {4 \dfrac 1 2} + C       | c = Primitive of $\sin^2 \dfrac x 2$ }} {{eqn | r = \frac x 2 - \frac {\sin x} 2 + C       | c = simplifying }} {{end-eqn}} and so for $n = 1$: {{begin-eqn}} {{eqn | l = b_n       | r = \frac 1 \pi \int_0^{2 \pi} \map f x \sin \dfrac x 2 \rd x       | c =  }} {{eqn | r = \frac 1 \pi \intlimits {\frac x 2 - \frac {\sin x} 2} 0 \pi - \frac 1 \pi \intlimits {\frac x 2 - \frac {\sin x} 2} \pi {2 \pi}       | c =  }} {{eqn | r = \frac 1 \pi \paren {\paren {\frac \pi 2 - \frac {\sin \pi} 2} - \paren {\frac 0 2 - \frac {\sin 0} 2} }       | c =  }} {{eqn | o =       | ro= -       | r = \frac 1 \pi \paren {\paren {\frac {2 \pi} 2 - \frac {\sin 2 \pi} 2} - \paren {\frac \pi 2 - \frac {\sin \pi} 2} }       | c =  }} {{eqn | r = \frac 1 \pi \paren {\frac \pi 2 - \frac {2 \pi} 2 + \frac \pi 2}       | c = Sine of Multiple of Pi }} {{eqn | r = 0       | c = everything vanishes }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map f x       | o = \sim       | r = \sum_{n \mathop = 1}^\infty b_n \sin \dfrac {n x} 2       | c =  }} {{eqn | r = \sum_{n \mathop = 2}^\infty \frac 2 \pi \paren {\frac {\sin \paren {\dfrac {\paren {n - 1} \pi} 2} } {n - 1} - \frac {\sin \paren {\dfrac {\paren {n + 1} \pi} 2} } {n + 1} } \sin \dfrac {n x} 2       | c = substituting for $b_n$ from above }} {{end-eqn}} When $n$ is odd, we have $n = 2 r + 1$ for $r \ge 1$, and so: {{begin-eqn}} {{eqn | o =       | r = \sum_{r \mathop = 1}^\infty \frac 2 \pi \paren {\frac {\sin \paren {\dfrac {\paren {\paren {2 r + 1} - 1} \pi} 2} } {\paren {2 r + 1} - 1} - \frac {\sin \paren {\dfrac {\paren {\paren {2 r + 1} + 1} \pi} 2} } {\paren {2 r + 1} + 1} } \sin \dfrac {\paren {2 r + 1} x} 2       | c =  }} {{eqn | r = \sum_{r \mathop = 1}^\infty \frac 2 \pi \paren {\frac {\sin \paren {\dfrac {2 r \pi} 2} } {2 r} - \frac {\sin \paren {\dfrac {\paren {2 r + 2} \pi} 2} } {2 r + 2} } \sin \dfrac {\paren {2 r + 1} x} 2       | c =  }} {{eqn | r = \sum_{r \mathop = 1}^\infty \frac 2 \pi \paren {\frac {\sin r \pi} {2 r} - \frac {\sin \paren {r + 1} \pi} {2 r + 2} } \sin \dfrac {\paren {2 r + 1} x} 2       | c =  }} {{eqn | r = 0       | c = Sine of Multiple of Pi }} {{end-eqn}} When $n$ is even, we have $n = 2 r$ for $r \ge 1$, and so: {{begin-eqn}} {{eqn | o =       | r = \sum_{r \mathop = 1}^\infty \frac 2 \pi \paren {\frac {\sin \paren {\dfrac {\paren {2 r - 1} \pi} 2} } {2 r - 1} - \frac {\sin \paren {\dfrac {\paren {2 r + 1} \pi} 2} } {2 r + 1} } \sin \dfrac {2 r x} 2       | c =  }} {{eqn | r = \frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\sin \paren {r - \frac 1 2} \pi} {2 r - 1} - \frac {\sin \paren {r + \frac 1 2} \pi} {2 r + 1} } \sin r x       | c =  }} {{eqn | r = \frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\paren {-1}^{r - 1} } {2 r - 1} - \frac {\paren {-1}r} {2 r + 1} } \sin r x       | c = Sine of Half-Integer Multiple of Pi }} {{eqn | r = \frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\paren {-1}^{r - 1} } {2 r - 1} + \frac {\paren {-1}^{r - 1} } {2 r + 1} } \sin r x       | c =  }} {{eqn | r = \frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \paren {\frac {2 r + 1 + 2 r - 1} {\paren {2 r - 1} \paren {2 r + 1} } } \sin r x       | c =  }} {{eqn | r = \frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {4 r} {4 r^2 - 1} \sin r x       | c = Difference of Two Squares }} {{eqn | r = \frac 8 \pi \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r \sin r x} {4 r^2 - 1}        | c = simplifying }} {{end-eqn}} {{qed}}