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

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Answer 1 · 2016-10-10T18:51:21

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

Please see the explanation for the process.

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:

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

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

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