Chapter 5.5, Problem 109E

To determine

To prove:The trigonometric expression $\cos 3\beta ={\cos }^3\beta -3{\sin }^2\beta \cos \beta$ and check result with the help of graphing utility.

Explanation of Solution

Given information:

The given trigonometric expression is $\cos 3\beta ={\cos }^3\beta -3{\sin }^2\beta \cos \beta$ .

Formula Used:

Use the formulas

  $\begin{array}{c}\cos 3x=4{\cos }^{3}x-3\cos x\\ 1-{\cos }^{2}x={\sin }^{2}x\end{array}$

Proof:

Consider theleft hand side of the equation and simplify it.

  $\begin{array}{c}\cos 3\beta =4{\cos }^3\beta -3\cos \beta \\ ={\cos }^3\beta +3{\cos }^3\beta -3\cos \beta \\ ={\cos }^3\beta +3{\cos }^3\beta \left({\cos }^2\beta -1\right)\end{array}$

Solve the above equation.

  $\begin{array}{c}\cos 3\beta ={\cos }^3\beta -3\cos \beta \left(1-{\cos }^2\beta \right)\\ ={\cos }^3\beta -3\cos \beta {\sin }^2\beta \\ ={\cos }^3\beta -3{\sin }^2\beta \cos \beta \end{array}$

From the above, it is clear thatone side of the equation can be transformed into other side of the equation.

Check result by using graphing utility.

Start the graphing calculator.

Make sure that the calculator is in the radians mode.

  

Press $\boxed{\text{Y}=}$ ley and enter the data $\text{Y}1=\boxed{\cos }\boxed{3}\boxed{\cdot }\boxed{\text{X}}\boxed{)}$ then press $\boxed{\text{ENTER}}$ key and enter $\text{Y2}=\boxed{(}\boxed{\cos }\boxed{\text{X}}\boxed{)}\boxed{)}\boxed{\wedge }\boxed{3}\boxed{-}\boxed{(}\boxed{3}\boxed{\cdot }\boxed{(}\boxed{\sin }\boxed{\text{X}}\boxed{)}\boxed{)}\boxed{\wedge }\boxed{2}\boxed{\cdot }\boxed{\cos }\boxed{\text{X}}\boxed{)}\boxed{)}$ .

  

To insert $\pi$ in window setting, click $\boxed{2\text{nd}}\boxed{\wedge }$ .

Adjust the window setting: press $\boxed{\text{WINDOW}}$ key.

  $\boxed{{\text{X}}_{\min }=0}\boxed{{\text{X}}_{\max }=3\pi }\boxed{{\text{X}}_{\text{scl}}=\pi /2}\boxed{{\text{Y}}_{\min }=-3}\boxed{{\text{Y}}_{\max }=3}\boxed{{\text{Y}}_{\text{scl}}=1}\boxed{{\text{X}}_{\text{res}}=1}$

  

Press $\boxed{\text{GRAPH}}$ .

  

The above graph appears to coincide, the given equation appears to be an identity.

Hence, the result is true.