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 the function$\left(p·g\right)\left(315°\right)$.

Short answer

The value of the function$\left(p·g\right)\left(315°\right)$is$-\frac{1}{2}$.

Step-by-step solution

Step 1. Given information.

Given $f\left(x\right)=\sin x,g\left(x\right)=\cos x$and $h\left(x\right)=2x,P\left(x\right)=\frac{x}{2}$and the value of$x$is$315°$.

Step 2. Simplify the given expression.

$\left(p·g\right)\left(315°\right)$can be written as $p\left(315°\right)$and $g\left(315°\right)$.

Substitute $\sin 315°$for $p\left(315°\right)$and $\cos 315°$for $g\left(315°\right)$.

$\left(p·g\right)\left(315°\right)=\left(\sin 315°·\cos 315°\right)$

Step 3. Find the value of p·g 315°.

In trigonometric standard angle the value of $\sin 315°$is $-\frac{1}{\sqrt{2}}$and the value of $\cos 315°$is $\frac{1}{\sqrt{2}}$.

Substitute the value of angles in the simplified expression.

$$\begin{gathered} \left(p·g\right)\left(315°\right)=\left(\sin 315°·\cos 315°\right) \\ =\left(-\frac{1}{\sqrt{2}}·\frac{1}{\sqrt{2}}\right) \\ =-\frac{1}{2} \end{gathered}$$