{
"query": {
"display": "polar to cartesian $$\\left(2\\sqrt{3},\\:\\frac{2π}{3}\\right)$$",
"symbolab_question": "POLAR#cartesian (2\\sqrt{3},\\frac{2π}{3})"
},
"solution": {
"level": "PERFORMED",
"subject": "Pre Calculus",
"topic": "Polar Coordinates",
"subTopic": "Cartesian",
"default": "(-\\sqrt{3},3)"
},
"steps": {
"type": "interim",
"title": "Convert $$\\left(2\\sqrt{3},\\:\\frac{2π}{3}\\right)\\:$$to Cartesian coordinates:$${\\quad}\\left(-\\sqrt{3},\\:3\\right)$$",
"steps": [
{
"type": "definition",
"title": "Definition",
"text": "To convert Polar coordinates $$\\left(r,\\:\\theta\\right)\\:$$to Cartesian coordinates $$\\left(x,\\:y\\right)\\:$$apply:<br/>$$x=r\\cdot\\cos\\left(\\theta\\right)\\quad\\:y=r\\cdot\\sin\\left(\\theta\\right)$$",
"secondary": [
"$$r=2\\sqrt{3}$$",
"$$θ=\\frac{2π}{3}$$"
]
},
{
"type": "step",
"primary": "$$x=r\\cdot\\cos\\left(\\theta\\right)$$",
"result": "x=2\\sqrt{3}\\cos\\left(\\frac{2π}{3}\\right)"
},
{
"type": "interim",
"title": "$$2\\sqrt{3}\\cos\\left(\\frac{2π}{3}\\right)=-\\sqrt{3}$$",
"input": "2\\sqrt{3}\\cos\\left(\\frac{2π}{3}\\right)",
"steps": [
{
"type": "interim",
"title": "Simplify $$\\cos\\left(\\frac{2π}{3}\\right):{\\quad}-\\frac{1}{2}$$",
"input": "\\cos\\left(\\frac{2π}{3}\\right)",
"result": "=2\\sqrt{3}\\left(-\\frac{1}{2}\\right)",
"steps": [
{
"type": "step",
"primary": "Use the following trivial identity:$${\\quad}\\cos\\left(\\frac{2π}{3}\\right)=-\\frac{1}{2}$$",
"secondary": [
"$$\\cos\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\cos(x)&x&\\cos(x)\\\\\\hline 0&1&π&-1\\\\\\hline \\frac{π}{6}&\\frac{\\sqrt{3}}{2}&\\frac{7π}{6}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{1}{2}&\\frac{4π}{3}&-\\frac{1}{2}\\\\\\hline \\frac{π}{2}&0&\\frac{3π}{2}&0\\\\\\hline \\frac{2π}{3}&-\\frac{1}{2}&\\frac{5π}{3}&\\frac{1}{2}\\\\\\hline \\frac{3π}{4}&-\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&-\\frac{\\sqrt{3}}{2}&\\frac{11π}{6}&\\frac{\\sqrt{3}}{2}\\\\\\hline \\end{array}$$"
],
"result": "=-\\frac{1}{2}"
}
],
"meta": {
"interimType": "Algebraic Manipulation Simplify Title 1Eq"
}
},
{
"type": "interim",
"title": "Simplify",
"input": "2\\sqrt{3}\\left(-\\frac{1}{2}\\right)",
"steps": [
{
"type": "step",
"primary": "Remove parentheses: $$\\left(-a\\right)=-a$$",
"result": "=-2\\sqrt{3}\\frac{1}{2}"
},
{
"type": "step",
"primary": "Multiply fractions: $$a\\cdot\\frac{b}{c}=\\frac{a\\:\\cdot\\:b}{c}$$",
"result": "=-\\frac{1\\cdot\\:2\\sqrt{3}}{2}"
},
{
"type": "step",
"primary": "Cancel the common factor: $$2$$",
"result": "=-1\\cdot\\:\\sqrt{3}"
},
{
"type": "step",
"primary": "Multiply: $$1\\cdot\\:\\sqrt{3}=\\sqrt{3}$$",
"result": "=-\\sqrt{3}"
}
],
"meta": {
"interimType": "Generic Simplify 0Eq"
}
},
{
"type": "step",
"result": "=-\\sqrt{3}"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver"
}
},
{
"type": "step",
"result": "x=-\\sqrt{3}"
},
{
"type": "step",
"primary": "$$y=r\\cdot\\sin\\left(\\theta\\right)$$",
"result": "y=2\\sqrt{3}\\sin\\left(\\frac{2π}{3}\\right)"
},
{
"type": "interim",
"title": "$$2\\sqrt{3}\\sin\\left(\\frac{2π}{3}\\right)=3$$",
"input": "2\\sqrt{3}\\sin\\left(\\frac{2π}{3}\\right)",
"steps": [
{
"type": "interim",
"title": "Simplify $$\\sin\\left(\\frac{2π}{3}\\right):{\\quad}\\frac{\\sqrt{3}}{2}$$",
"input": "\\sin\\left(\\frac{2π}{3}\\right)",
"result": "=2\\sqrt{3}\\frac{\\sqrt{3}}{2}",
"steps": [
{
"type": "step",
"primary": "Use the following trivial identity:$${\\quad}\\sin\\left(\\frac{2π}{3}\\right)=\\frac{\\sqrt{3}}{2}$$",
"secondary": [
"$$\\sin\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\sin(x)&x&\\sin(x)\\\\\\hline 0&0&π&0\\\\\\hline \\frac{π}{6}&\\frac{1}{2}&\\frac{7π}{6}&-\\frac{1}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{\\sqrt{3}}{2}&\\frac{4π}{3}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{2}&1&\\frac{3π}{2}&-1\\\\\\hline \\frac{2π}{3}&\\frac{\\sqrt{3}}{2}&\\frac{5π}{3}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{3π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&\\frac{1}{2}&\\frac{11π}{6}&-\\frac{1}{2}\\\\\\hline \\end{array}$$"
],
"result": "=\\frac{\\sqrt{3}}{2}"
}
],
"meta": {
"interimType": "Algebraic Manipulation Simplify Title 1Eq"
}
},
{
"type": "interim",
"title": "Simplify",
"input": "2\\sqrt{3}\\frac{\\sqrt{3}}{2}",
"steps": [
{
"type": "step",
"primary": "Multiply fractions: $$a\\cdot\\frac{b}{c}=\\frac{a\\:\\cdot\\:b}{c}$$",
"result": "=\\frac{\\sqrt{3}\\cdot\\:2\\sqrt{3}}{2}"
},
{
"type": "step",
"primary": "Cancel the common factor: $$2$$",
"result": "=\\sqrt{3}\\sqrt{3}"
},
{
"type": "step",
"primary": "Apply radical rule: $$\\sqrt{a}\\sqrt{a}=a$$",
"secondary": [
"$$\\sqrt{3}\\sqrt{3}=3$$"
],
"result": "=3",
"meta": {
"practiceLink": "/practice/radicals-practice",
"practiceTopic": "Radical Rules"
}
}
],
"meta": {
"interimType": "Generic Simplify 0Eq"
}
},
{
"type": "step",
"result": "=3"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver"
}
},
{
"type": "step",
"result": "y=3"
},
{
"type": "step",
"primary": "The Cartesian coordinates of $$\\left(2\\sqrt{3},\\:\\frac{2π}{3}\\right)$$",
"result": "\\left(-\\sqrt{3},\\:3\\right)"
}
]
}
}
Solution
polar to cartesian
Solution
Solution steps
The Cartesian coordinates of