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

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

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

Answer

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<x<2\pi)

Answer:

(\pi<x<2\pi)