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If sin(A)=cos(A) was is the angle measurement?

$\small{ \begin{array}{rcl|rcl|rcl} \sin(A)&=&\cos(A) \\ \cos(A)-\sin(A) &=& 0 & \sin(45^{\circ}-A) &=& \sin(45^{\circ})\cdot\cos(A)- \cos(45^{\circ})\cdot\sin(A) & \sin(45^{\circ})=\cos(45^{\circ}) = \frac{\sqrt{2}}{2} \\ & & & \sin(45^{\circ}-A) &=& \frac{\sqrt{2}}{2}\cdot\cos(A)- \frac{\sqrt{2}}{2}\cdot\sin(A) \\ & & & \sin(45^{\circ}-A) &=& \frac{\sqrt{2}}{2}\Big(\cdot\cos(A)- \sin(A) \Big) \\ \frac{2}{\sqrt{2}}\cdot \sin(45^{\circ}-A)&=& 0 & \frac{2}{\sqrt{2}}\cdot \sin(45^{\circ}-A) &=& \cos(A)- \sin(A) \\ \sin(45^{\circ}-A)&=& 0 \\ 45^{\circ}-A &=& \arcsin(0) \\ 45^{\circ}-A &=& 0\pm n\cdot 360^{\circ} \qquad n \in \mathbb{N} \\ \mathbf{A} & \mathbf{=} & \mathbf{45^{\circ} \pm n\cdot 360^{\circ}} \\\\ \sin(45^{\circ}-A)=\sin(180^{\circ}-(45^{\circ}-A) ) &=& 0 \\ 180^{\circ}-(45^{\circ}-A) &=& \arcsin(0) \\ 180^{\circ}-45^{\circ}+A &=& \arcsin(0) \\ 135^{\circ} + A &=& 0\pm n\cdot 360^{\circ} \qquad n \in \mathbb{N} \\\\ A &=& -135^{\circ} \pm n\cdot 360^{\circ} \\ A &=& -135^{\circ} + 360^{\circ} \pm n\cdot 360^{\circ} \\ \mathbf{A} & \mathbf{=} & \mathbf{225^{\circ} \pm n\cdot 360^{\circ}} \\ \end{array} }$