Convert the point to rectangular coordinates.(4,π6)(4,\frac{\pi}{6})(4,6π)

( 2 3 , 2 ) (2\sqrt{3},2) ( 2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 2 )

Convert the point to rectangular coordinates. ( 4 , π 6 ) (4,\frac{\pi}{6}) ( 4 , 6 π )

Here, we're asked to convert the point given in polar coordinates into rectangular coordinates. Now the point that we're given here is 4pi over 6. So let's go ahead and use our formulas here in order to get our x and y values. Now x is going to be equal to r times the cosine of theta and here our r value is 4, our theta value is pi over 6, so this is going to give me 4 times the cosine of pi over 6. Now the cosine of pi over 6 is root 3 over 2. So this is going to be 4 times the square root of 3 over 2. Now doing some simplification here, 4 over 2 just gives me 2, so this simplifies to 2 root 3 for that x value. Now for y, y is equal to r times the sine of theta. So plugging in r and theta again here, we get 4 times the sine of pi over 6. Now the sine of pi over 6 is 1 half so this gives me 4 times 1 half. Now 4 times 1 half is just 2, so that gives me my final y value, and this point in rectangular coordinates is therefore 2 root 3 comma 2. Now this is our final answer here, having converted our point from polar coordinates into rectangular coordinates. Thanks for watching, and I'll see you in the next one.

Convert the point to rectangular coordinates. ( 4 , π 6 ) (4,\frac{\pi}{6}) ( 4 , 6 π )

( 2 3 , 2 ) (2\sqrt{3},2) ( 2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 2 )

( 4 3 , 4 ) (4\sqrt{3},4) ( 4 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 4 )

( 2 , 2 3 ) (2,2\sqrt{3}) ( 2 , 2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

( 2 , 3 ) (2,\sqrt{3}) ( 2 , 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

15. Polar Equations

Convert Points Between Polar and Rectangular Coordinates

Precalculus

15. Polar Equations

Convert Points Between Polar and Rectangular Coordinates

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Multiple Choice

Convert the point to rectangular coordinates.
$(4,\frac{\pi}{6})$

$(2\sqrt{3},2)$

$(4\sqrt{3},4)$

$(2,2\sqrt{3})$

$(2,\sqrt{3})$

Verified step by step guidance

Identify the given polar coordinates: (r, θ) = (4, \frac{\pi}{6}).

Recall the formulas to convert polar coordinates to rectangular coordinates: x = r \cdot \cos(θ) and y = r \cdot \sin(θ).

Substitute the given values into the formulas: x = 4 \cdot \cos(\frac{\pi}{6}) and y = 4 \cdot \sin(\frac{\pi}{6}).

Calculate \cos(\frac{\pi}{6}) and \sin(\frac{\pi}{6}). These are standard trigonometric values: \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} and \sin(\frac{\pi}{6}) = \frac{1}{2}.

Substitute these trigonometric values back into the equations to find x and y: x = 4 \cdot \frac{\sqrt{3}}{2} and y = 4 \cdot \frac{1}{2}.

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