Question

lf $0\le x,y\le {180}^{\circ }$ and $\sin (x-y)=\cos (x+y)=\frac{1}{2}$, then thevalues of $x$ and $y$ are given by

• A. $x={45}^{\circ },y={15}^{\circ }$
• B. $x={45}^{\circ },y={135}^{\circ }$
• C. $x={165}^{\circ },y={15}^{\circ }$
• D. $x={165}^{\circ },y={135}^{\circ }$

Answer 1

Answer: (A), (D)

The correct options are
A $x={45}^{\circ },y={15}^{\circ }$
D $x={165}^{\circ },y={135}^{\circ }$

$\sin (x-y)=\cos (x+y)=\frac{1}{2}$

$\sin (x+y)=\cos (x-y)=\pm \frac{\sqrt{3}}{2}$

$\sin \left(2x\right)=\sin \left(\right(x+y)+(x-y\left)\right)$

$=\sin (x+y)\cos (x-y)+\cos (x+y)\sin (x-y)$

$\sin 2x=\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \frac{1}{2}=\frac{3}{4}+\frac{1}{4}=1$

$\Rightarrow 2x={90}^o\Rightarrow x={45}^o$

$\sin 2x=\frac{\sqrt{3}}{2}\cdot -\frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \frac{1}{2}=-\frac{3}{4}+\frac{1}{4}=-\frac{1}{2}$

$\Rightarrow 2x={330}^o\Rightarrow x={165}^o$

$\sin \left(2y\right)=\sin \left(\right(x+y)-(x-y\left)\right)$

$=\sin (x+y)\cos (x-y)-\cos (x+y)\sin (x-y)$

$\sin 2y=\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}-\frac{1}{2}\cdot \frac{1}{2}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$

$\Rightarrow 2y=30\Rightarrow y={15}^o$

$\sin 2y=\frac{\sqrt{3}}{2}\cdot -\frac{\sqrt{3}}{2}-\frac{1}{2}\cdot \frac{1}{2}=-\frac{3}{4}-\frac{1}{4}=-1$

$\Rightarrow 2y={270}^o\Rightarrow y={135}^0$