Chapter 5.4, Problem 59E

To determine

To solve: The given trigonometric expression using right triangles.

Answer to Problem 59E

The value of the given trigonometric expression is $\frac{24}{25}$

Explanation of Solution

Given information:

The following expression:

  $\sin \left({\cos }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{3}{5}\right)$

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 sum formula:

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

Calculation:

This expression fits the formula:

  $\sin (u+v)$

Angles $u={\cos }^{-1}\frac{4}{5}$ and $v={\sin }^{-1}\frac{3}{5}$ are shown in the following figures respectively

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

  

Therefore, the given equation can be written as:

  $\begin{array}{l}\sin (u+v)=\sin ({\cos }^{-1}\frac{4}{5})\cos ({\sin }^{-1}\frac{3}{5})+\cos ({\cos }^{-1}\frac{4}{5})\sin ({\sin }^{-1}\frac{3}{5})\text{ [Using sum formula]}\\ \text{ =}\frac{3}{5}\cdot \frac{4}{5}\text{+}\frac{4}{5}\cdot \frac{3}{5}\text{ [Derived from the repective diagrams]}\\ \text{ =}\frac{12}{25}\text{+}\frac{12}{25}\text{ [Multiply] }\\ \text{ =}\frac{24}{25}\text{ [Add] }\\ \end{array}$