The curve shown in Figure 2 has parametric equations $x = ext{sin}(t), \, y = ext{sin}\left(t + \frac{\pi}{6}\right), \quad \frac{\pi}{2} < t < \frac{\pi}{2}.$ (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}\sqrt{(1 - x^2)}, \quad -1 < x < 1. - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 6

To find the tangent line, we need to compute the derivatives of $x$ and $y$ with respect to $t$:

1. Differentiate both parametric equations:

   • $x = \text{sin}(t) \Rightarrow \frac{dx}{dt} = \text{cos}(t)$
   • $y = \text{sin}\left(t + \frac{\pi}{6}\right) \Rightarrow \frac{dy}{dt} = \text{cos}\left(t + \frac{\pi}{6}\right)$

2. Find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\text{cos}\left(t + \frac{\pi}{6}\right)}{\text{cos}(t)}$

3. Evaluate at $t = \frac{\pi}{6}$:

   • $\frac{dx}{dt} = \text{cos}\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
   • $\frac{dy}{dt} = \text{cos}\left(\frac{\pi}{6} + \frac{\pi}{6}\right) = \text{cos}\left(\frac{\pi}{3}\right) = \frac{1}{2}$
   • Thus, $\frac{dy}{dx} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$.

4. Find the point on the curve where $t = \frac{\pi}{6}$:

   • $x = \text{sin}\left(\frac{\pi}{6}\right) = \frac{1}{2}$
   • $y = \text{sin}\left(\frac{\pi}{6} + \frac{\pi}{6}\right) = \text{sin}\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.
   • Therefore, the point is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.

5. Now we can form the equation of the tangent line using the point-slope form: $y - y_1 = m(x - x_1)$ where $m = \frac{1}{\sqrt{3}}, \; (x_1, y_1) = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.

   • This leads to: $y - \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(x - \frac{1}{2}\right).$