Chapter 8.4, Problem 8QR

Solution

To find: The value of $\cos \theta$ for the equation, $\cot 2\theta =\frac{3}{\sqrt{7}}$ .

The required value is , $\cos \theta =\frac{\sqrt{7}}{2\sqrt{2}}$ ..

Given Information:

The given identity is $\cot 2\theta =\frac{3}{\sqrt{7}}$ , $0<\theta <\frac{\pi }{2}$ .

Formula:

The Pythagorean theorem is ${\left(\text{Opposite}\right)}^2+{\left(\text{Adjacent}\right)}^2={\left(\text{Hypotenuse}\right)}^2$ .

The trigonometric identity $\cos 2\theta =2{\cos }^2\theta -1$ .

Calculation:

Since $\cot x=\frac{\text{Adjacent}}{\text{Opposite}}$ . Therefore, the adjacent and opposite are 3 and $\sqrt{7}$ , respectively. Substitute the values into the Pythagorean theorem and solve for hypotenuse.

  $\begin{array}{c}{\left(\text{Hypotenuse}\right)}^2={\left(\sqrt{7}\right)}^2+3^2\\ {\left(\text{Hypotenuse}\right)}^2=7+9\\ {\left(\text{Hypotenuse}\right)}^2=16\\ \text{Hypotenuse}=4\end{array}$

Use the definition of the cosine function and find the value of $\cos 2\theta$ .

  $\begin{array}{c}\cos 2\theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}\\ =\frac{3}{4}\end{array}$

Substitute the obtained value into the formula $\cos 2\theta =2{\cos }^2\theta -1$ and solve for $\cos \theta$ .

Apply ${\cot }^{-1}$ on both sides of the given equation.

  $\begin{array}{c}\frac{3}{4}=2{\cos }^2\theta -1\\ 2{\cos }^2\theta =1+\frac{3}{4}\\ {\cos }^2\theta =\frac{7}{8}\\ \cos \theta =\frac{\sqrt{7}}{2\sqrt{2}}\end{array}$

Therefore, the required value is , $\cos \theta =\frac{\sqrt{7}}{2\sqrt{2}}$ .