Question

Part b if you have time please

Answer 1

`\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}}) \\\\\\ \textit{also recall that }\qquad sin^2(\theta)+cos^2(\theta)=1\\\\ -------------------------------\\\\ \begin{cases} x=4cos\left( t+(\pi )/(6) \right)&\impliedby \textit{let's solve for cos(t)} \\\\ y=2sin(t)&\impliedby \textit{let's solve for sin(t)} \end{cases}\\\\ -------------------------------\\\\ y=2sin(t)\implies \cfrac{y}{2}=sin(t)\\\\ -------------------------------\\\\`


`\bf x=4cos\left( t+(\pi )/(6) \right)\implies \cfrac{x}{4}=cos\left( t+(\pi )/(6) \right) \\\\\\ \cfrac{x}{4}=cos(t)cos\left((\pi )/(6) \right)-sin(t)sin\left((\pi )/(6) \right)\implies \cfrac{x}{4}=cos(t)\cdot \cfrac{√(3)}{2}~-~sin(t)\cdot \cfrac{1}{2} \\\\\\ \cfrac{x}{4}=\cfrac{cos(t)√(3)~-~sin(t)}{2}\implies \cfrac{x}{2}=cos(t)√(3)~-~sin(t) \\\\\\ \cfrac{x}{2}+sin(t)=cos(t)√(3)\implies \cfrac{(x)/(2)+sin(t)}{√(3)}=cos(t)`


`\bf \cfrac{(x)/(2)+\stackrel{\downarrow }{(y)/(2)}}{√(3)}=cos(t)\\\\ -------------------------------\\\\ \stackrel{cos^2(\theta )}{\left( \cfrac{(x)/(2)+(y)/(2)}{√(3)} \right)^2}+\stackrel{sin^2(\theta )}{\left( \cfrac{y}{2} \right)^2} ~~=~~ 1 \\\\\\ \left( \cfrac{(x+y)/(2)}{√(3)} \right)^2+\left( \cfrac{y}{2} \right)^2=1\implies \left( \cfrac{x+y}{2√(3)} \right)^2+\left( \cfrac{y}{2} \right)^2=1`


`\bf \cfrac{(x+y)^2}{(2√(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 \\\\\\ \textit{now we'll multiply both sides by 12, to do away with the denominators} \\\\\\ (x+y)^2+\stackrel{a}{3}y^2=\stackrel{b}{12}`