What is the value of cot(113pi/12)?

Question

What is the value of cot(113pi/12)?

Answer 1

cot(113π/12) = cot(9π + 5π/12) = cot5π/12 = tanπ/12 = sinπ/6/(1 + cosπ/6) = 1/2/(1 + √3/2) = 2 - √3

Answer 2

cot 113 pi /12=cot ((113/12)pi)=cot(9pi+5pi/12). This gives a reference angle of 5pi/12 in the third quadrant.

Therefore cot 113pi/12 =cot 5pi/12 with a positive sign as cot is positive in the third quadrant.

cot5pi/12=(2-sqrt(3))=0.268.

Answer 3

a few different ways to work this problem. here are a few of them:

cot(113pi/12)

convert to degrees if you feel more comfortable or your calculator is set for degrees

113pi/12 = 113pi/12 x 180/pi = 1,695 degrees

1695/360 = 4.708333....

.708333(360)= 255 degrees

cot(255) = about 0.268

cot(113pi/12) = about 0.268

113pi/12 = 9 5/12 pi

angle on a unit circle < 2pi, >0 = 1 5/12 pi = 17pi/12 = 255 degrees

reference angle = 255-180=75 degrees= 5pi/12

cot75 = about 0.268

tan75 = 3.73

cot75 = 1/tan75 = 1/3.73 = .268

use sum formulas for trig functions

sin75 = sin(30+45) = sin30cos45+ sin45cos30

= (.5)(.5sqr2)+.5sqr2(.5sqr3)

=.25sqr2 +.25sqr6= (sqr2+sqr6)/4 = about .966

cos75 = cos(30+45) = cos30cos45 -sin30sin45= .5sqr3(.5sqr2) - .5(.5sqr2) = .25sqr6-.25sqr2 = .25(sqr6-sqr2)

= (sqr6-sqr2)/4 = about .2588

cot75 = cos75/sin75 = (sqr6-sqr2)/(sqr6+sqr2) = about .2588/.966 = about 0.268

cot(a+b) = cotacotb-1)/(cota+cotb)

cot(30+45) = (cot30cot45 -1)/(cot30+cot45) = (sqr3(1) -1)/(sqr3 +1) = (sqr3-1)/(sqr3 +1)

= (sqr3 -1)^2/(3-1) = (3-2sqr3 +1)/2 = 2-sqr3

= about 2-1.732 = 0.268

cot75 = (sqr6-sqr2)/(sqr6+sqr2)

rationalize the denominator

= (sqr6-sqr2)^2/(6-2) = (6- 2sqr12 +2)/4 = (8-4sqr3)/4 = 2-sqr3 = about 2-1.732 = about 0.268

cot(113pi/12) = exactly 2-sqr3 = about 0.268

if you work the problem a couple different ways and get the same answer, it's virtually certain to be the correct answer

exact solution= 2-sqr3

approximate answer rounded to 3 decimal places = 0.268 = about 1/4

113pi/12 is in quadrant 3 where cotangents are always positive