Question

If $\theta ={100}^{\circ }$, determine the sign of $(\sin \theta +\cos \theta )$.

Answer 1

$\theta =100\sin \theta +\cos \theta =\sqrt{2}(\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta )=\sqrt{2}(\cos \frac{\pi }{4}\cos \theta +\sin \frac{\pi }{4}\sin \theta )=\sqrt{2}\cos (\theta -\frac{\pi }{4})=\sqrt{2}\cos (100-\frac{\pi }{4})=\sqrt{2}\cos (55\left)\cos (55\right)>0\therefore \sqrt{2}\cos (55)>0$