Chapter 7.3, Problem 47E

To determine

To find: the exact value of the given expression.

Answer to Problem 47E

  $\sin \left(2{\cos }^{-1}\frac{7}{25}\right)=\frac{336}{625}$ .

Explanation of Solution

Given information:

Given expression

  $\sin \left(2{\cos }^{-1}\frac{7}{25}\right)$

Calculation:

We need to express the given expression $\sin \left(2{\cos }^{-1}\frac{7}{25}\right)$

Let $\theta ={\cos }^{-1}\frac{7}{25}$ , then , $\cos \theta =\frac{7}{25}$

We sketch a right angle triangle $ABC$ , right angled at $B$ . Since $\cos \theta =\frac{7}{25}$ , we label one side and hypotenuse as

  

To find the trigonometric function, sine, we first need to find the value of $AB$ . Therefore, we use the Pythagorean Theorem in the triangle $ABC$ , to get

  $A{C}^2=A{B}^2+B{C}^2$

Substituting $BC=7$ and $AC=25$ , we get

  $\begin{array}{l}{25}^2=A{B}^2+7^2\\ 625=A{B}^2+49\end{array}$

Isolating $A{B}^2$ on the right hand side, we get

  $\begin{array}{c}625-49=A{B}^2\\ 576=A{B}^2\end{array}$

Switching the sides and taking square root on both sides, we get

  $\begin{array}{c}AB=\sqrt{576}\\ =24\end{array}$

Therefore, $\sin \theta =\frac{24}{25}$ .

Now, we use the double angle formula for sine as

  $\sin 2\theta =2\sin \theta \cos \theta$

Substituting the values of $\cos \theta$ and $\sin \theta$ , we get

  $\begin{array}{c}\sin 2\theta =2\times \frac{24}{25}\times \frac{7}{25}\\ =\frac{336}{625}\end{array}$

Substituting $\theta ={\cos }^{-1}\frac{7}{25}$ , we get

  $\sin \left(2{\cos }^{-1}\frac{7}{25}\right)=\frac{336}{625}$

Hence, the exact value of $\sin \left(2{\cos }^{-1}\frac{7}{25}\right)=\frac{336}{625}$ .