Question

over which interval is the graph of ( y = cos(x) ) strictly increasing?
( 0 < x < \frac{pi}{2} )
( 0 < x < pi )
( \frac{pi}{2} < x < \frac{3pi}{2} )
( pi < x < 2pi )

Explanation:

Step1: Recall the derivative of \(y = \cos(x)\)

The derivative of \(y=\cos(x)\) is \(y'=-\sin(x)\). A function \(y = f(x)\) is strictly increasing when \(y'>0\). So, we need to find when \(-\sin(x)>0\), which is equivalent to \(\sin(x)<0\).

Step2: Analyze the sine - function's sign

The sine function \(y = \sin(x)\) has the following properties:

• \(\sin(x)>0\) when \(2k\pi<x<(2k + 1)\pi,k\in\mathbb{Z}\)
• \(\sin(x)<0\) when \((2k+1)\pi<x<(2k + 2)\pi,k\in\mathbb{Z}\)

Let \(k = 0\), then \(\sin(x)<0\) when \(\pi

Answer:

\(\pi