Simplifying a curve parametrization

Question

A parametric curve is described by $$ x=\sin t,\,\,\,y=\sin (t+\frac{\pi}{6}). $$ Is there a way to show that $$ y=\frac{\sqrt{3}}{2}x+\frac{\sqrt{1-x^2}}{2} $$ without using the property $\sin(a+b)=\sin a\cos b+\sin b \cos a$?

Answer 1

Since $\sin(t+\frac{\pi}{6})=\frac{\sqrt{3}}{2}\sin t+\frac{1}{2}\cos t$, we have $y=\frac{\sqrt{3}}{2}x+\frac{1}{2}\sqrt{1-x^2}$.

Without using the formula, I feel like $$\arcsin y-\frac{\pi}{6}=t=\arcsin x.$$