Question

Suppose that sin theta + cos theta  = 1/2 and 0 < theta 12 years ago

Answer 1

\[2\sin \theta+2\cos \theta-1=0\]
\[2\sin \theta-1=-2\sqrt{1-\sin ^{2}\theta }\]
squaring both sides
\[4\sin ^{2}\theta+1-4\sin \theta=4\left( 1-\sin ^{2}\theta\right) \]
\[8\sin ^{2}\theta-4\sin \theta-3=0\]
\[\sin \theta=\frac{ 4\pm \sqrt{16-4*8*-3} }{2*4 }=\frac{ 4\pm \sqrt{112} }{8 }=\frac{ 4\pm4\sqrt{7} }{8 }\]
\[=\frac{ 1+\sqrt{7} }{2 }only \because~\in~0<\theta<\pi,\sin \theta~ is~ positive.\]