# Question 10e48

Dec 28, 2017

Write the given as two equations:

$\cos \left(\theta\right) = x \text{ [1]}$

$\theta = {\tan}^{-} 1 \left(\frac{4}{3}\right) \text{ [2]}$

and we want to find the value of x.

Use equation [2] to find the value of $\tan \left(\theta\right)$ by applying the tangent function to both sides:

$\tan \left(\theta\right) = \tan \left({\tan}^{-} 1 \left(\frac{4}{3}\right)\right) \text{ [2.1]}$

The tangent of its inverse reduces to $\frac{4}{3}$ on the right:

$\tan \left(\theta\right) = \frac{4}{3} \text{ [2.2]}$

We can use the identity

$1 + {\tan}^{2} \left(\theta\right) = {\sec}^{2} \left(\theta\right)$

to give us a relationship between $\tan \left(\theta\right)$ and $\cos \left(\theta\right)$:

$1 + {\left(\frac{4}{3}\right)}^{2} = {\sec}^{2} \left(\theta\right)$

$\frac{9}{9} + \frac{16}{9} = {\sec}^{2} \left(\theta\right)$

$\frac{25}{9} = {\sec}^{2} \left(\theta\right)$

We know that ${\sec}^{2} \left(\theta\right) = \frac{1}{\cos} ^ 2 \left(\theta\right)$:

$\frac{25}{9} = \frac{1}{\cos} ^ 2 \left(\theta\right)$

${\cos}^{2} \left(\theta\right) = \frac{9}{25}$

$\cos \left(\theta\right) = \pm \frac{3}{5}$

We do not know whether we are in the 1st or 3rd quadrant, therefore, we must leave the $\pm$ as is.

Dec 28, 2017

$\pm \frac{3}{5}$

#### Explanation:

We know that,

$53 \approx {\tan}^{- 1} \left(\frac{4}{3}\right)$ Using a calculator...

$\implies \tan 53 \approx \frac{4}{3}$

So, cos(tan^(-1)(4/3) ) approx cos(tan^(-1)(tan(53))#

$\cos \left(n \pi + 53\right) = \pm \frac{3}{5}$, $n \in Z$