As in other people's answers, you can do this by complex methods. However if (as suggested by the algebra-precalculus tag) you have not yet met complex numbers...
Let $S$ be the sum. Using a "products to sums" formula, $$\eqalign{(2\sin1)S &=\sum_{n=1}^{99}\bigl(\cos(n-1)-\cos(n+1)\bigr)\cr &=(\cos0+\cos1+\cdots+\cos98)-(\cos2+\cdots+\cos99+\cos100)\cr &=\cos0+\cos1-\cos99-\cos100\cr}$$$$\eqalign{(2\sin{\textstyle\frac{1}{2}})S &=\sum_{n=1}^{99}2\sin n\sin{\textstyle\frac{1}{2}}\cr &=\sum_{n=1}^{99}\bigl(\cos(n-{\textstyle\frac{1}{2}})-\cos(n+{\textstyle\frac{1}{2}})\bigr)\cr &=(\cos{\textstyle\frac{1}{2}}+\cos(1{\textstyle\frac{1}{2}})+\cdots +\cos(98{\textstyle\frac{1}{2}}))\cr &\qquad\qquad{}-(\cos(1{\textstyle\frac{1}{2}})+\cdots +\cos(98{\textstyle\frac{1}{2}})+\cos(99{\textstyle\frac{1}{2}}))\cr &=\cos{\textstyle\frac{1}{2}}-\cos(99{\textstyle\frac{1}{2}})\cr}$$ and so $$S=\frac{1+\cos1-\cos99-\cos100}{2\sin1}\ .$$$$S=\frac{\cos{\textstyle\frac{1}{2}}-\cos(99{\textstyle\frac{1}{2}})}{2\sin{\textstyle\frac{1}{2}}}\ .$$
Comment. For alternative answers you can follow the same basic idea starting with $(2\sin1)S$, $(2\cos1)S$ etc.