on which interval is y = sin(x) strictly increasing?\n(-π, -\\frac{π}{2})\n(0, π)\n(\\frac{π}{2}…

on which interval is y = sin(x) strictly increasing?\n(-π, -\\frac{π}{2})\n(0, π)\n(\\frac{π}{2}, \\frac{3π}{2})\n(0, \\frac{π}{2})

on which interval is y = sin(x) strictly increasing?\n(-π, -\\frac{π}{2})\n(0, π)\n(\\frac{π}{2}, \\frac{3π}{2})\n(0, \\frac{π}{2})

Answer

Explanation:

Step1: Recall sine - function property

The derivative of $y = \sin(x)$ is $y'=\cos(x)$. A function $y = f(x)$ is strictly increasing when $y'>0$. So we need to find where $\cos(x)>0$.

Step2: Analyze cosine - function intervals

The cosine function $\cos(x)>0$ in the intervals $\left(-\frac{\pi}{2}+ 2k\pi,\frac{\pi}{2}+2k\pi\right),k\in\mathbb{Z}$. When $k = 0$, the interval is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Among the given options, the interval $\left(0,\frac{\pi}{2}\right)$ is a sub - interval of $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ where $\sin(x)$ is strictly increasing.

Answer:

D. $\left(0,\frac{\pi}{2}\right)$