{
"query": {
"display": "polar to cartesian $$\\left(-3,\\:-\\frac{π}{6}\\right)$$",
"symbolab_question": "POLAR#cartesian (-3,-\\frac{π}{6})"
},
"solution": {
"level": "PERFORMED",
"subject": "Pre Calculus",
"topic": "Polar Coordinates",
"subTopic": "Cartesian",
"default": "(-\\frac{3\\sqrt{3}}{2},\\frac{3}{2})"
},
"steps": {
"type": "interim",
"title": "Convert $$\\left(-3,\\:-\\frac{π}{6}\\right)\\:$$to Cartesian coordinates:$${\\quad}\\left(-\\frac{3\\sqrt{3}}{2},\\:\\frac{3}{2}\\right)$$",
"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": [
"$$r=-3$$",
"$$θ=-\\frac{π}{6}$$"
]
},
{
"type": "step",
"primary": "$$x=r\\cdot\\cos\\left(\\theta\\right)$$",
"result": "x=-3\\cos\\left(-\\frac{π}{6}\\right)"
},
{
"type": "interim",
"title": "$$-3\\cos\\left(-\\frac{π}{6}\\right)=-\\frac{3\\sqrt{3}}{2}$$",
"input": "-3\\cos\\left(-\\frac{π}{6}\\right)",
"steps": [
{
"type": "interim",
"title": "$$\\cos\\left(-\\frac{π}{6}\\right)=\\frac{\\sqrt{3}}{2}$$",
"input": "\\cos\\left(-\\frac{π}{6}\\right)",
"result": "=-3\\cdot\\:\\frac{\\sqrt{3}}{2}",
"steps": [
{
"type": "step",
"primary": "Use the following property: $$\\cos\\left(-x\\right)=\\cos\\left(x\\right)$$",
"secondary": [
"$$\\cos\\left(-\\frac{π}{6}\\right)=\\cos\\left(\\frac{π}{6}\\right)$$"
],
"result": "=\\cos\\left(\\frac{π}{6}\\right)"
},
{
"type": "interim",
"title": "Use the following trivial identity:$${\\quad}\\cos\\left(\\frac{π}{6}\\right)=\\frac{\\sqrt{3}}{2}$$",
"input": "\\cos\\left(\\frac{π}{6}\\right)",
"steps": [
{
"type": "step",
"primary": "$$\\cos\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\cos(x)&x&\\cos(x)\\\\\\hline 0&1&π&-1\\\\\\hline \\frac{π}{6}&\\frac{\\sqrt{3}}{2}&\\frac{7π}{6}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{1}{2}&\\frac{4π}{3}&-\\frac{1}{2}\\\\\\hline \\frac{π}{2}&0&\\frac{3π}{2}&0\\\\\\hline \\frac{2π}{3}&-\\frac{1}{2}&\\frac{5π}{3}&\\frac{1}{2}\\\\\\hline \\frac{3π}{4}&-\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&-\\frac{\\sqrt{3}}{2}&\\frac{11π}{6}&\\frac{\\sqrt{3}}{2}\\\\\\hline \\end{array}$$"
},
{
"type": "step",
"result": "=\\frac{\\sqrt{3}}{2}"
}
],
"meta": {
"interimType": "Trig Trivial Angle Value Title 0Eq"
}
},
{
"type": "step",
"result": "=\\frac{\\sqrt{3}}{2}"
}
],
"meta": {
"solvingClass": "Trig Evaluate",
"interimType": "Trig Evaluate"
}
},
{
"type": "step",
"primary": "Multiply fractions: $$a\\cdot\\frac{b}{c}=\\frac{a\\:\\cdot\\:b}{c}$$",
"result": "=-\\frac{\\sqrt{3}\\cdot\\:3}{2}"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7+T/7f5Y4cSVLGv6oviwE/LObx3lB/e8mjzvwUPWUAP8DnzlbPZjyKgy1eUCFsLd5AL4Fn9Y7zU1+NUpPXbjHobLzqGyu80FJzbSR5eEPyc1eTZMykILbMC5S4vTIC/oK9g+VE3n0AMXZXnCGWbHeZvj4C+/2rrllAyl0vtXekzIk96XE2/4ak2RL8kN6VIeWialcV/dI5TH4fXyp+ncwuA=="
}
},
{
"type": "step",
"result": "x=-\\frac{3\\sqrt{3}}{2}"
},
{
"type": "step",
"primary": "$$y=r\\cdot\\sin\\left(\\theta\\right)$$",
"result": "y=-3\\sin\\left(-\\frac{π}{6}\\right)"
},
{
"type": "interim",
"title": "$$-3\\sin\\left(-\\frac{π}{6}\\right)=\\frac{3}{2}$$",
"input": "-3\\sin\\left(-\\frac{π}{6}\\right)",
"steps": [
{
"type": "interim",
"title": "$$\\sin\\left(-\\frac{π}{6}\\right)=-\\frac{1}{2}$$",
"input": "\\sin\\left(-\\frac{π}{6}\\right)",
"result": "=-3\\left(-\\frac{1}{2}\\right)",
"steps": [
{
"type": "step",
"primary": "Use the following property: $$\\sin\\left(-x\\right)=-\\sin\\left(x\\right)$$",
"secondary": [
"$$\\sin\\left(-\\frac{π}{6}\\right)=-\\sin\\left(\\frac{π}{6}\\right)$$"
],
"result": "=-\\sin\\left(\\frac{π}{6}\\right)"
},
{
"type": "interim",
"title": "Use the following trivial identity:$${\\quad}\\sin\\left(\\frac{π}{6}\\right)=\\frac{1}{2}$$",
"input": "\\sin\\left(\\frac{π}{6}\\right)",
"steps": [
{
"type": "step",
"primary": "$$\\sin\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\sin(x)&x&\\sin(x)\\\\\\hline 0&0&π&0\\\\\\hline \\frac{π}{6}&\\frac{1}{2}&\\frac{7π}{6}&-\\frac{1}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{\\sqrt{3}}{2}&\\frac{4π}{3}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{2}&1&\\frac{3π}{2}&-1\\\\\\hline \\frac{2π}{3}&\\frac{\\sqrt{3}}{2}&\\frac{5π}{3}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{3π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&\\frac{1}{2}&\\frac{11π}{6}&-\\frac{1}{2}\\\\\\hline \\end{array}$$"
},
{
"type": "step",
"result": "=\\frac{1}{2}"
}
],
"meta": {
"interimType": "Trig Trivial Angle Value Title 0Eq"
}
},
{
"type": "step",
"result": "=-\\frac{1}{2}"
}
],
"meta": {
"solvingClass": "Trig Evaluate",
"interimType": "Trig Evaluate"
}
},
{
"type": "interim",
"title": "Simplify",
"input": "-3\\left(-\\frac{1}{2}\\right)",
"steps": [
{
"type": "step",
"primary": "Apply rule $$-\\left(-a\\right)=a$$",
"result": "=3\\cdot\\:\\frac{1}{2}"
},
{
"type": "step",
"primary": "Multiply fractions: $$a\\cdot\\frac{b}{c}=\\frac{a\\:\\cdot\\:b}{c}$$",
"result": "=\\frac{1\\cdot\\:3}{2}"
},
{
"type": "step",
"primary": "Multiply the numbers: $$1\\cdot\\:3=3$$",
"result": "=\\frac{3}{2}"
}
],
"meta": {
"interimType": "Generic Simplify 0Eq"
}
},
{
"type": "step",
"result": "=\\frac{3}{2}"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver"
}
},
{
"type": "step",
"result": "y=\\frac{3}{2}"
},
{
"type": "step",
"primary": "The Cartesian coordinates of $$\\left(-3,\\:-\\frac{π}{6}\\right)$$",
"result": "\\left(-\\frac{3\\sqrt{3}}{2},\\:\\frac{3}{2}\\right)"
}
]
}
}
Solution
polar to cartesian
Solution
Solution steps
The Cartesian coordinates of