Chapter 8.4, Problem 10QR

Solution

To find: The value of $\sin \theta$ for the equation, $\cot 2\theta =\frac{45}{28}$ .

The required value is , $\sin \theta =\frac{2}{\sqrt{53}}$ .

Given Information:

The given identity is $\cot 2\theta =\frac{45}{28}$ , $0<\theta <\frac{\pi }{2}$ .

Formula:

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

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

Calculation:

Since $\cot x=\frac{\text{Adjacent}}{\text{Opposite}}$ . Therefore, the adjacent and opposite are 45 and 28, respectively. Substitute the values into the Pythagorean theorem and solve for hypotenuse.

  $\begin{array}{c}{\left(\text{Hypotenuse}\right)}^2={45}^2+{28}^2\\ {\left(\text{Hypotenuse}\right)}^2=2025+784\\ {\left(\text{Hypotenuse}\right)}^2=2809\\ \text{Hypotenuse}=53\end{array}$

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

  $\begin{array}{c}\cos 2\theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}\\ =\frac{45}{53}\end{array}$

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

  $\begin{array}{c}\frac{45}{53}=1-2{\sin }^2\theta \\ 2{\sin }^2\theta =1-\frac{45}{53}\\ {\sin }^2\theta =\frac{8}{106}\\ \sin \theta =\frac{2}{\sqrt{53}}\end{array}$

Therefore, the required value is , $\sin \theta =\frac{2}{\sqrt{53}}$ .