Answer:
[tex]y=\dfrac{\pi}{3}[/tex]
Step-by-step explanation:
Given equation:
[tex]\sin (x+y) =\dfrac{1}{2} \sin(x) +\dfrac{\sqrt{3}}{2} \cos(x)[/tex]
[tex]\boxed{\begin{minipage}{6 cm}\underline{Trigonometric Identity}\\\\$\sin (A \pm B) \equiv \sin A \cos B \pm \cos A \sin B$\\\end{minipage}}[/tex]
Use the sine trigonometric identity to rewrite sin(x+y) in terms of sin and cos:
[tex]\implies \sin(x+y)= \sin (x) \cos (y) + \cos (x) \sin (y)[/tex]
Substitute this into the given equation:
[tex]\begin{aligned}\sin (x+y) & =\dfrac{1}{2} \sin(x) +\dfrac{\sqrt{3}}{2} \cos(x)\\\\\implies \sin (x) \cos (y) + \cos (x) \sin (y) & =\dfrac{1}{2} \sin(x) +\dfrac{\sqrt{3}}{2} \cos(x)\end{aligned}[/tex]
Compare the left side of the equation with the right side:
[tex]\implies \cos (y) = \dfrac{1}{2}[/tex]
[tex]\implies \sin (y) = \dfrac{\sqrt{3}}{2}[/tex]
Therefore, solving for y:
[tex]\begin{aligned}\implies \cos (y) & = \dfrac{1}{2}\\y & = \cos^{-1}\left(\dfrac{1}{2}\right)\\y & = \dfrac{\pi}{3}\end{aligned}[/tex]
[tex]\begin{aligned}\implies \sin (y) & = \dfrac{\sqrt{3}}{2}\\y & = \sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)\\y & = \dfrac{\pi}{3}\end{aligned}[/tex]
