[tex]\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})
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\textit{also recall that }\qquad sin^2(\theta)+cos^2(\theta)=1\\\\
-------------------------------\\\\
\begin{cases}
x=4cos\left( t+\frac{\pi }{6} \right)&\impliedby \textit{let's solve for cos(t)}
\\\\
y=2sin(t)&\impliedby \textit{let's solve for sin(t)}
\end{cases}\\\\
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y=2sin(t)\implies \cfrac{y}{2}=sin(t)\\\\
-------------------------------\\\\[/tex]
[tex]\bf x=4cos\left( t+\frac{\pi }{6} \right)\implies \cfrac{x}{4}=cos\left( t+\frac{\pi }{6} \right)
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\cfrac{x}{4}=cos(t)cos\left(\frac{\pi }{6} \right)-sin(t)sin\left(\frac{\pi }{6} \right)\implies \cfrac{x}{4}=cos(t)\cdot \cfrac{\sqrt{3}}{2}~-~sin(t)\cdot \cfrac{1}{2}
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\cfrac{x}{4}=\cfrac{cos(t)\sqrt{3}~-~sin(t)}{2}\implies \cfrac{x}{2}=cos(t)\sqrt{3}~-~sin(t)
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\cfrac{x}{2}+sin(t)=cos(t)\sqrt{3}\implies \cfrac{\frac{x}{2}+sin(t)}{\sqrt{3}}=cos(t)[/tex]
[tex]\bf \cfrac{\frac{x}{2}+\stackrel{\downarrow }{\frac{y}{2}}}{\sqrt{3}}=cos(t)\\\\
-------------------------------\\\\
\stackrel{cos^2(\theta )}{\left( \cfrac{\frac{x}{2}+\frac{y}{2}}{\sqrt{3}} \right)^2}+\stackrel{sin^2(\theta )}{\left( \cfrac{y}{2} \right)^2} ~~=~~ 1
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\left( \cfrac{\frac{x+y}{2}}{\sqrt{3}} \right)^2+\left( \cfrac{y}{2} \right)^2=1\implies \left( \cfrac{x+y}{2\sqrt{3}} \right)^2+\left( \cfrac{y}{2} \right)^2=1[/tex]
[tex]\bf \cfrac{(x+y)^2}{(2\sqrt{3})^2}+\cfrac{y^2}{2^2}=1\implies \cfrac{(x+y)^2}{2^2(3)}+\cfrac{y^2}{4}=1\implies \cfrac{(x+y)^2}{12}+\cfrac{y^2}{4}=1
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\textit{now we'll multiply both sides by 12, to do away with the denominators}
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(x+y)^2+\stackrel{a}{3}y^2=\stackrel{b}{12}[/tex]
