Can someone explain why I get wrong answer in simplifying this expression? $$\sin\frac{11\pi}{12}\sin\frac{29\pi}{12}-\cos\frac{13\pi}{12}\cos\frac{41\pi}{12}$$ If we rewrite the expression with new angles, $$\frac{29\pi}{12}\to\frac{5\pi}{12} \quad\text{and}\quad \frac{13\pi}{12}\to\frac{11\pi}{12}$$ we don't change the value of the expression. But, if we now use sine of sum of angles, we get $\frac{1}{2}$ instead of $0$.
Why does this happen? Do angles need to be in the same quadrant for the formula to work?
No, you don't get $\frac12$. You get $0$ too, because\begin{align}\sin\left(\frac{11\pi}{12}\right)\sin\left(\frac{5\pi}{12}\right)-\cos\left(\frac{11\pi}{12}\right)\cos\left(\frac{41\pi}{12}\right)&=\sin\left(\frac{11\pi}{12}\right)\sin\left(\frac{5\pi}{12}\right)-\cos\left(\frac{11\pi}{12}\right)\cos\left(3\pi+\frac{5\pi}{12}\right)\\&=\sin\left(\frac{11\pi}{12}\right)\sin\left(\frac{5\pi}{12}\right)+\cos\left(\frac{11\pi}{12}\right)\cos\left(\frac{5\pi}{12}\right)\\&=\cos\left(\frac{11\pi}{12}-\frac{5\pi}{12}\right)\\&=\cos\left(\frac\pi2\right)\\&=0.\end{align}