Using trigonometric identities

In a nutshell

You can convert trigonometric parametric equations into Cartesian form using trigonometric identities.  Make sure that your calculator is set to work with angles in radians.

Example 1

A curve has parametric equations $$x(t)=\cos(t) +4$$ and $$y(t) = \sin(t) -5$$, $$0 \lt t \lt 2\pi$$.

i) Find the Cartesian equation of the curve given by the parametric equations.

ii) Hence sketch the curve.

i) Write the expressions for $$\sin(t)$$ and $$\cos(t)$$ in terms of $$x$$ and $$y$$.

$$\begin{aligned}x& =\cos(t) +4 \\\cos(t) &= x - 4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \textcircled{1} \\y&=\sin(t) -5 \\\sin(t )&= y + 5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \textcircled{2} \end{aligned}$$​​

Substitute $$\textcircled{1}$$ and $$\textcircled{2}$$ into $$\sin^2(t) + \cos^2(t) \equiv 1$$.

$$\underline{(y+5)^2+(x-4)^2=1}$$​​

ii) $$(x-a)^2+(y-b)^2=r^2$$ is the equation of a circle with centre $$(a,b)$$ and radius $$r$$.  Hence this curve is a circle with centre $$(4,-5)$$ and radius $$1$$.

Example 2

A curve has parametric equations $$x=2\cot^2(2t)$$ and $$y = 2\sin^2(2t)$$, $$0 \lt t \le \dfrac{\pi}{4}$$.

i) Find a Cartesian equation of the curve in the form $$y = f(x)$$.

ii) State the domain on which $$f(x)$$ is defined.

i) Rewrite $$\cot^2(2t)$$ in terms of $$\sin(2t)$$ and $$\cos(2t)$$​.

$$\begin{aligned}y&=2\sin^2(2t) \\x&=2\cot^2(2t) = \dfrac{2\cos^2 (2t)}{\sin^2 (2t)} \end{aligned}$$​​

Eliminate the $$\sin^2(2t)$$ from $$x$$ by multiplying $$x$$ and $$y$$.

$$xy = 2\sin^2(2t) \times \dfrac{2 \cos^2(2t)}{\sin^2(2t)} = 4 \cos^2 (2t)$$​​

Use the trigonometric identity $$\sin^2 (x)+ \cos^2 (x) \equiv 1$$.

$$2\sin^2 (2t) + 2\cos^2(2t) \equiv 2$$​​

Substitute in $$y$$ and $$xy$$ and manipulate algebraically.

$$\begin{aligned} y+\dfrac{xy}{2}&=2 \\2y+xy &= 4 \\y(x+2) &=4 \\y&= \underline{\dfrac{4}{x+2}} \end{aligned}$$​​

ii) The domain is the range of $$x(t)$$​: 

$$\underline{x \gt 0}$$​​