Chapter 5.4, Problem 56E

To determine

To evaluate: The given trigonometric expression using right triangles.

Answer to Problem 56E

The value of the given trigonometric expression is $\frac{33}{25}$ .

Explanation of Solution

Given information:

The following expression:

  $\sin \left({\cos }^{-1}\frac{3}{5}-{\sin }^{-1}\frac{5}{13}\right)$

Formula used:

The following trigonometric relations and identities:

  $\begin{array}{l}\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}\\ \sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}\\ {\text{Hypotenuse}}^2={\text{Adjacent}}^{\text{2}}{\text{+opposite}}^{\text{2}}\end{array}$

The following difference formula:

  $\sin (u-v)=\sin u\cos v-\cos u\sin v$

Calculation:

This expression fits the formula:

  $\sin (u-v)$

Angles $u={\cos }^{-1}\frac{3}{5}$ and $v={\sin }^{-1}\frac{5}{13}$ are shown in the following figures respectively

  $\begin{array}{l}\text{For figure 1,}\\ u={\cos }^{-1}\frac{3}{5}\\ \cos u=\frac{3}{5}\text{ [Inverse function]}\\ \cos u=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{3}{5}\text{ [Pythagoras relation]}\\ \text{opposite = 4 [Pythagoras identity]}\\ \\ \text{For figure 2,}\\ v={\sin }^{-1}\frac{5}{13}\\ \sin v=\frac{5}{13}\text{ [Inverse function]}\\ \sin v=\frac{\text{Opposite}}{\text{Hypotenuse}}=\frac{5}{13}\text{ [Pythagoras relation]}\\ \text{adjacent = 12 [Pythagoras identity]}\\ \end{array}$

  

Therefore, the given equation can be written as:

  $\begin{array}{l}\sin (u-v)=\sin ({\cos }^{-1}\frac{3}{5})\cos ({\sin }^{-1}\frac{5}{13})-\cos ({\cos }^{-1}\frac{3}{5})\sin ({\sin }^{-1}\frac{5}{13})\text{ [Using difference formula]}\\ \text{ =}\frac{4}{5}\cdot \frac{12}{13}-\frac{3}{5}\cdot \frac{5}{13}\text{ [Derived from the repective diagrams]}\\ \text{ =}\frac{48}{65}-\frac{15}{65}\text{ [Multiply] }\\ \text{ =}\frac{33}{25}\text{ [Subtract] }\\ \end{array}$