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Using trigonometric identities

Using trigonometric identities

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Explainer Video

Tutor: Alice

Summary

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)+4x(t)=\cos(t) +4x(t)=cos(t)+4 and y(t)=sin⁡(t)−5y(t) = \sin(t) -5y(t)=sin(t)−5, 0<t<2π0 \lt t \lt 2\pi0<t<2π.

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)\sin(t)sin(t) and cos⁡(t)\cos(t)cos(t) in terms of xxx and yyy.

x=cos⁡(t)+4cos⁡(t)=x−4 1◯y=sin⁡(t)−5sin⁡(t)=y+5 2◯\begin{aligned}x& =\cos(t) +4 \\\cos(t) &= x - 4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \textcircled{1} \\y&=\sin(t) -5 \\\sin(t )&= y + 5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \textcircled{2} \end{aligned}xcos(t)ysin(t)​=cos(t)+4=x−41◯=sin(t)−5=y+52◯​​​


Substitute 1◯\textcircled{1}1◯ and 2◯\textcircled{2}2◯ into sin⁡2(t)+cos⁡2(t)≡1\sin^2(t) + \cos^2(t) \equiv 1sin2(t)+cos2(t)≡1.

(y+5)2+(x−4)2=1‾\underline{(y+5)^2+(x-4)^2=1}(y+5)2+(x−4)2=1​​​


ii) (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2(x−a)2+(y−b)2=r2 is the equation of a circle with centre (a,b)(a,b)(a,b) and radius rrr.  Hence this curve is a circle with centre (4,−5)(4,-5)(4,−5) and radius 111.

Maths; Parametric equations; KS5 Year 13; Using trigonometric identities


Example 2

A curve has parametric equations x=2cot⁡2(2t)x=2\cot^2(2t) x=2cot2(2t) and y=2sin⁡2(2t)y = 2\sin^2(2t)y=2sin2(2t), 0<t≤π40 \lt t \le \dfrac{\pi}{4}0<t≤4π​.

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

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


i) Rewrite cot⁡2(2t)\cot^2(2t)cot2(2t) in terms of sin⁡(2t)\sin(2t)sin(2t) and cos⁡(2t)\cos(2t)cos(2t)​.

y=2sin⁡2(2t)x=2cot⁡2(2t)=2cos⁡2(2t)sin⁡2(2t)\begin{aligned}y&=2\sin^2(2t) \\x&=2\cot^2(2t) = \dfrac{2\cos^2 (2t)}{\sin^2 (2t)} \end{aligned}yx​=2sin2(2t)=2cot2(2t)=sin2(2t)2cos2(2t)​​​​


Eliminate the sin⁡2(2t)\sin^2(2t) sin2(2t) from xxx by multiplying xxx and yyy.

xy=2sin⁡2(2t)×2cos⁡2(2t)sin⁡2(2t)=4cos⁡2(2t)xy = 2\sin^2(2t) \times \dfrac{2 \cos^2(2t)}{\sin^2(2t)} = 4 \cos^2 (2t)xy=2sin2(2t)×sin2(2t)2cos2(2t)​=4cos2(2t)​​


Use the trigonometric identity sin⁡2(x)+cos⁡2(x)≡1\sin^2 (x)+ \cos^2 (x) \equiv 1sin2(x)+cos2(x)≡1.

2sin⁡2(2t)+2cos⁡2(2t)≡22\sin^2 (2t) + 2\cos^2(2t) \equiv 22sin2(2t)+2cos2(2t)≡2​​


Substitute in yyy and xyxyxy and manipulate algebraically.

y+xy2=22y+xy=4y(x+2)=4y=4x+2‾\begin{aligned} y+\dfrac{xy}{2}&=2 \\2y+xy &= 4 \\y(x+2) &=4 \\y&= \underline{\dfrac{4}{x+2}} \end{aligned}y+2xy​2y+xyy(x+2)y​=2=4=4=x+24​​​​​


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

x>0‾\underline{x \gt 0}x>0​​​



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FAQs - Frequently Asked Questions

Which trigonometric identities should you use when working with parametric equations?

You can use any of the trigonometric identities you have learnt, such as sin2x=2sinxcosx.

Should you use degrees or radians when working with trigonometric parametric equations?

You should use radians when working with trigonometric parametric equations.

How do you convert trigonometric parametric equations into Cartesian form?

You can convert trigonometric parametric equations into Cartesian form using trigonometric identities.

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