The curve shown in Figure 2 has parametric equations $x = ext{sin}(t), \, y = ext{sin}\left(t + \frac{\pi}{6}\right), \quad \frac{\pi}{2} < t < \frac{\pi}{6}.$ (a) Find an equation of the tangent to the curve at the point where $t = \frac{\pi}{6}.$ (b) Show that a cartesian equation of the curve is $y = \frac{\sqrt{3}}{2} + \frac{1}{2}(1 - x^{2}), \, -1 < x < 1.$ - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 6

To find the equation of the tangent, we first need to differentiate the parametric equations.

1. Differentiate:

   $\frac{dx}{dt} = \cos(t)$

   $\frac{dy}{dt} = \cos\left(t + \frac{\pi}{6}\right)$

   Using the compound angle formula:

   $\cos\left(t + \frac{\pi}{6}\right) = \cos(t)\cos\left(\frac{\pi}{6}\right) - \sin(t)\sin\left(\frac{\pi}{6}\right) = \cos(t)\frac{\sqrt{3}}{2} - \sin(t)\frac{1}{2}$

2. Find the derivatives at $t = \frac{\pi}{6}$:

   At $t = \frac{\pi}{6}$,

   $x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

   $y = \sin\left(\frac{\pi}{6} + \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

   So, the point is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.

3. Calculate the slope of the tangent:

   $$$$

4. Now, using point-slope form, the equation of the tangent line is:

   $y - \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(x - \frac{1}{2}\right)$

   Rearranging gives:

   $y = \frac{1}{\sqrt{3}} x + \frac{\sqrt{3}}{2} - \frac{1}{2\sqrt{3}}$.