If cos theta =1/6cosθ=16 and theta is acute, how do you find cot(pi/2 - theta)cot(π2θ)?

1 Answer
Oct 10, 2016

cot(pi/2 - theta) = sqrt(35)cot(π2θ)=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)cos(π2θ)=cos(π2)cos(θ)+sin(π2)sin(θ)

Use cos(pi/2) = 0cos(π2)=0 and sin(pi/2) = 1sin(π2)=1 to simplify:

cos(pi/2 - theta) = sin(theta)cos(π2θ)=sin(θ)

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)