It is apparent by inspection that $\map f x$ is an odd function over $\openint {-\pi} \pi$. It follows from Fourier Series for Odd Function over Symmetric Range: :$\displaystyle \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$ where for all $n \in \Z_{> 0}$: :$b_n = \displaystyle \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x$ for all $n \in \Z_{>0}$. Thus by definition of $f$: :$\displaystyle b_n = -\frac 2 \pi \int_0^\pi \cos x \sin n x \rd x$ When $n \ne 1$, we have: {{begin-eqn}} {{eqn | l = b_n       | r = -\frac 2 \pi \int_0^\pi \cos x \sin n x \rd x       | c =  }} {{eqn | r = -\frac 2 \pi \intlimits {\frac {-\cos \paren {n - 1} x} {2 \paren {n - 1} } - \frac {\cos \paren {n + 1} x} {2 \paren {n + 1} } } 0 \pi       | c = Primitive of $\cos x \sin n x$ }} {{eqn | r = -\frac 2 \pi \paren {\paren {\frac {-\cos \paren {n - 1} \pi} {2 \paren {n - 1} } - \frac {\cos \paren {n + 1} \pi} {2 \paren {n + 1} } } - \paren {\frac {-\cos \paren {n - 1} 0} {2 \paren {n - 1} } - \frac {\cos \paren {n + 1} 0} {2 \paren {n + 1} } } }       | c =  }} {{eqn | r = \frac 1 \pi \paren {\paren {\frac {\cos \paren {n - 1} \pi} {n - 1} + \frac {\cos \paren {n + 1} \pi} {n + 1} } - \paren {\frac 1 {n - 1} + \frac 1 {n + 1} } }       | c = Cosine of Zero is One and simplifying }} {{eqn | r = \frac 1 \pi \paren {\frac {\cos \paren {n - 1} \pi} {n - 1} - \frac 1 {n - 1} + \frac {\cos \paren {n + 1} \pi} {n + 1} - \frac 1 {n + 1} }       | c = simplifying }} {{end-eqn}} When $n = 1$, we have: {{begin-eqn}} {{eqn | l = b_n       | r = -\frac 2 \pi \int_0^\pi \cos x \sin x \rd x       | c =  }} {{eqn | r = -\frac 2 \pi \intlimits {\frac {\sin^2 x} 2} 0 \pi       | c = Primitive of $\cos x \sin x$ }} {{eqn | r = -\frac 2 \pi \paren {\frac {\sin^2 \pi} 2 - \frac {\sin^2 0} 2}       | c =  }} {{eqn | r = 0       | c = Sine of Multiple of Pi }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map f x       | o = \sim       | r = \sum_{n \mathop = 1}^\infty b_n \sin n x       | c =  }} {{eqn | r = \frac 1 \pi \sum_{n \mathop = 2}^\infty \paren {\frac {\cos \paren {n - 1} \pi} {n - 1} - \frac 1 {n - 1} + \frac {\cos \paren {n + 1} \pi} {n + 1} - \frac 1 {n + 1} } \sin n x       | c = substituting for $b_n$ from above and rearranging }} {{end-eqn}} When $n$ is odd, we have $n = 2 r + 1$ for $r \ge 1$, and so: {{begin-eqn}} {{eqn | o =       | r = \frac 1 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\cos \paren {\paren {2 r + 1} - 1} \pi} {\paren {2 r + 1} - 1} - \frac 1 {\paren {2 r + 1} - 1} + \frac {\cos \paren {\paren {2 r + 1} + 1} \pi} {\paren {2 r + 1} + 1} - \frac 1 {\paren {2 r + 1} + 1} } \sin \paren {2 r + 1} x       | c =  }} {{eqn | r = \frac 1 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\cos 2 r \pi} {2 r} - \frac 1 {2 r} + \frac {\cos \paren {2 r + 2} \pi} {2 r + 2} - \frac 1 {2 r + 2} } \sin \paren {2 r + 1} x       | c =  }} {{eqn | r = \frac 1 \pi \sum_{r \mathop = 1}^\infty \paren {\frac 1 {2 r} - \frac 1 {2 r} + \frac 1 {2 r + 2} - \frac 1 {2 r + 2} } \sin \paren {2 r + 1} x       | c = Cosine of Multiple of Pi }} {{eqn | r = 0       | c = everything vanishes }} {{end-eqn}} When $n$ is even, we have $n = 2 r$ for $r \ge 1$, and so: {{begin-eqn}} {{eqn | o =       | r = \frac 1 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {\cos \paren {2 r - 1} \pi} {2 r - 1} - \frac 1 {2 r - 1} + \frac {\cos \paren {2 r + 1} \pi} {2 r + 1} - \frac 1 {2 r + 1} } \sin 2 r x       | c =  }} {{eqn | r = \frac 1 \pi \sum_{r \mathop = 1}^\infty \paren {\frac {-1} {2 r - 1} - \frac 1 {2 r - 1} + \frac {-1} {2 r + 1} - \frac 1 {2 r + 1} } \sin 2 r x       | c = Cosine of Multiple of Pi }} {{eqn | r = -\frac 2 \pi \sum_{r \mathop = 1}^\infty \paren {\frac 1 {2 r - 1} + \frac 1 {2 r + 1} } \sin 2 r x       | c =  }} {{eqn | r = -\frac 2 \pi \sum_{r \mathop = 1}^\infty \frac {\paren {2 r + 1} + \paren {2 r - 1} } {\paren {2 r - 1} \paren {2 r + 1} } \sin 2 r x       | c =  }} {{eqn | r = -\frac 2 \pi \sum_{r \mathop = 1}^\infty \frac {4 r} {4 r^2 - 1} \sin 2 r x       | c = Difference of Two Squares and simplifying }} {{eqn | r = -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r x} {4 r^2 - 1}       | c = simplifying }} {{end-eqn}} {{qed}}