If #cos theta =1/6# and theta is acute, how do you find #cot(pi/2 - theta)#?

1 Answer
Oct 10, 2016

Answer:

#cot(pi/2 - theta) = sqrt(35)#

Please see the explanation for the process.

Explanation:

Start with the identity for the cosine of the difference of two angles:

#cos(pi/2 - theta) = cos(pi/2)cos(theta) + sin(pi/2)sin(theta)#

Use #cos(pi/2) = 0# and #sin(pi/2) = 1# to simplify:

#cos(pi/2 - theta) = sin(theta)#

Substitute #sqrt(1 - cos²(theta))# for #sin(theta)#:

#cos(pi/2 - theta) = sqrt(1 - cos²(theta))#

Substitute #1/6# for #cos(theta)#:

#cos(pi/2 - theta) = sqrt(1 - (1/6)²)#

#cos(pi/2 - theta) = sqrt(35)/6#

Use the identity for the sine of the difference of two angles:

#sin(pi/2 - theta) = sin(pi/2)cos(theta) - cos(pi/2)sin(theta)#

Use #cos(pi/2) = 0# and #sin(pi/2) = 1# to simplify:

#sin(pi/2 - theta) = cos(theta)#

Substitute #1/6# for #cos(theta)#:

#sin(pi/2 - theta) = 1/6#

Use #cot(x) = cos(x)/sin(x)#:

#cot(pi/2 - theta) = cos(pi/2 - theta)/sin(pi/2 - theta)#

Substitute in the values from above:

#cot(pi/2 - theta) = (sqrt(35)/6)/(1/6)#

#cot(pi/2 - theta) = sqrt(35)#