If sin θ = cos (θ − 45°), where θ and θ − 45° are acute angles

Given that: $\sin \theta=\cos \left(\theta-45^{\circ}\right)$ where $\theta$ and $\left(\theta-45^{\circ}\right)$ are acute angles

We have to find $\theta$

$\sin \theta=\cos \left(\theta-45^{\circ}\right)$

$\Rightarrow \cos \left(90^{\circ}-\theta\right)=\cos \left(\theta-45^{\circ}\right)$

$\Rightarrow 90^{\circ}-\theta=\theta-45^{\circ}$

$\Rightarrow-2 \theta=-125^{\circ}$

$\Rightarrow \theta=\frac{135^{\circ}}{2}$

Therefore $\theta=67 \frac{1}{2}$