Evaluate each expression.cot(11π6)\cot\left(\frac{11\pi}{6}\right)cot(611π)

− 3 -\sqrt3 − 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">

Evaluate each expression. cot ( 11 π 6 ) \cot\left(\frac{11\pi}{6}\right) cot ( 6 11 π )

this problem, we're asked to evaluate the expression and the expression we're given here is the cotangent of 11 pi over 6. Now remember the cotangent is just the reciprocal of tangent and on the unit circle, I know that the tangent is just y over x, so that means that the cotangent is just that flipped, x over y. So I can locate this angle on my unit circle and divide those values accordingly. So on my unit circle, I see that 11 pi over 6 is right down here in quadrant 4, and I have my x and my y values. Now when dividing these, remember that when we have the same denominator, we're effectively just dividing those numerators. So here, in order to do that, I'm taking my numerator of my x value, dividing it by my numerator of my y value to give me the square root of 3 over negative one. Now dividing that out just gives me the negative square root of 3 and I'm done here. The cotangent of 11 pi over 6 is negative square root of of 3. Thanks for watching and let me know if you have any questions.

Evaluate each expression. cot ⁡ ( 11 π 6 ) \cot\left(\frac{11\pi}{6}\right) cot ( 6 11 π )

− 3 -\sqrt3 − 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">

− 3 3 -\frac{\sqrt3}{3} − 3 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">

9. Unit Circle

Reciprocal Trigonometric Functions on the Unit Circle

Precalculus

9. Unit Circle

Reciprocal Trigonometric Functions on the Unit Circle

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Multiple Choice

Evaluate each expression.
$\cot\left(\frac{11\pi}{6}\right)$

$\frac12$

$-\frac{\sqrt3}{3}$

$-\sqrt3$

$2$

Verified step by step guidance

Identify the angle in the expression $ \cot\left(\frac{11\pi}{6}\right) $. This angle is in radians and is equivalent to $ 330^\circ $.

Determine the reference angle for $ 330^\circ $. The reference angle is $ 360^\circ - 330^\circ = 30^\circ $.

Recall that the cotangent function is the reciprocal of the tangent function. Therefore, $ \cot(\theta) = \frac{1}{\tan(\theta)} $.

Use the identity $ \tan(30^\circ) = \frac{\sqrt{3}}{3} $. Therefore, $ \cot(30^\circ) = \frac{1}{\frac{\sqrt{3}}{3}} = \sqrt{3} $.

Since $ 330^\circ $ is in the fourth quadrant where cotangent is positive, $ \cot(330^\circ) = \sqrt{3} $. Thus, $ \cot\left(\frac{11\pi}{6}\right) = \sqrt{3} $.

3:23m

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