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

Oct 10, 2016

$\cot \left(\frac{\pi}{2} - \theta\right) = \sqrt{35}$

Please see the explanation for the process.

#### Explanation:

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

$\cos \left(\frac{\pi}{2} - \theta\right) = \cos \left(\frac{\pi}{2}\right) \cos \left(\theta\right) + \sin \left(\frac{\pi}{2}\right) \sin \left(\theta\right)$

Use $\cos \left(\frac{\pi}{2}\right) = 0$ and $\sin \left(\frac{\pi}{2}\right) = 1$ to simplify:

$\cos \left(\frac{\pi}{2} - \theta\right) = \sin \left(\theta\right)$

Substitute sqrt(1 - cos²(theta)) for $\sin \left(\theta\right)$:

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

Substitute $\frac{1}{6}$ for $\cos \left(\theta\right)$:

cos(pi/2 - theta) = sqrt(1 - (1/6)²)

$\cos \left(\frac{\pi}{2} - \theta\right) = \frac{\sqrt{35}}{6}$

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

$\sin \left(\frac{\pi}{2} - \theta\right) = \sin \left(\frac{\pi}{2}\right) \cos \left(\theta\right) - \cos \left(\frac{\pi}{2}\right) \sin \left(\theta\right)$

Use $\cos \left(\frac{\pi}{2}\right) = 0$ and $\sin \left(\frac{\pi}{2}\right) = 1$ to simplify:

$\sin \left(\frac{\pi}{2} - \theta\right) = \cos \left(\theta\right)$

Substitute $\frac{1}{6}$ for $\cos \left(\theta\right)$:

$\sin \left(\frac{\pi}{2} - \theta\right) = \frac{1}{6}$

Use $\cot \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$:

$\cot \left(\frac{\pi}{2} - \theta\right) = \cos \frac{\frac{\pi}{2} - \theta}{\sin} \left(\frac{\pi}{2} - \theta\right)$

Substitute in the values from above:

$\cot \left(\frac{\pi}{2} - \theta\right) = \frac{\frac{\sqrt{35}}{6}}{\frac{1}{6}}$

$\cot \left(\frac{\pi}{2} - \theta\right) = \sqrt{35}$