Question

If $y=4{\sin }^2\theta -\cos 2\theta$, then $y$ lies in the interval .........

Answer 1

Ans. $[-1,5]$. We have
$y=4{\sin }^2\theta -\cos 2\theta$
$=4{\sin }^2\theta -(1-2{\sin }^2\theta )=6{\sin }^2\theta -1$
$\therefore \frac{y+1}{6}=si{n}^2\theta$
But $0\le {\sin }^2\theta \le 1$. Hence $0\le \frac{y+1}{6}\le 1$,
or $0\le y+1\le 6$ or $-1\le y\le 5\Rightarrow y\in [-1,5]$