Q 108. - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=\sin x ,g\left(x\right)=\cos x$ and $h\left(x\right)=2x ,P\left(x\right)=\frac{x}{2}$.Find the value of function $\left(f-g\right) \left(60°\right)$.

Step-by-Step Solution

Answer

The value of the function $\left(f-g\right) \left(60°\right)$ is $\frac{\sqrt{3}-1}{2}$.

Step 1. Given information.

Given $f\left(x\right)=\sin x$and $g\left(x\right)=\cos x$ and the value of $x$ is $30°$.

Step 2. Simplify the given expression.

$\left(f+g\right) \left(60°\right)$ can be written as $f\left(60°\right)$ and $g\left(60°\right)$.

Substitute $\sin \left(60°\right)$ for $f\left(60°\right)$ and $\cos 60°$ for $g\left(60°\right)$ then,

$\left(f-g\right) \left(60°\right)=\sin 60°-\cos 60°$

Step 3. Find the value of f - g   60 ° .

In trigonometric standard angle the value of $\sin 60°=\frac{\sqrt{3}}{2}$and the value of $\cos 60°=\frac{1}{2}$.

Substitute the value of angles into simplified expression.

$$\begin{gathered} \left(f-g\right) \left(60°\right)=\left(\sin 60°-\cos 60°\right) \\ =\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right) \\ =\left(\frac{\sqrt{3}-1}{2}\right) \end{gathered}$$