show that x = sin t and y = sin (t + pi/6)
can be written in the form y = ax + b√1-x^2 stating the values of a and b
im really stuck
$\displaystyle y = \sin \left( t + \frac {\pi}6 \right)$
$\displaystyle = \sin t \cos \frac {\pi}6 + \sin \frac {\pi}6 \cos t$ ........by the addition formula for sine
$\displaystyle = \frac {\sqrt{3}}2 \sin t + \frac 12 \cos t$
$\displaystyle = \frac {\sqrt{3}}2 \sin t + \frac 12 \sqrt{ 1 - \sin^2 t }$
can you finish? in particular, pay attention to the $\displaystyle \sin t$'s you see in the above expression. what do you know about $\displaystyle \sin t$ here?