# Find the range of sinx(sinx+cosx)?

Apr 1, 2017

Range of $\sin x \left(\sin x + \cos x\right)$ is $\left[\frac{1}{2} - \frac{1}{\sqrt{2}} , \frac{1}{2} + \frac{1}{\sqrt{2}}\right]$

#### Explanation:

$\sin x \left(\sin x + \cos x\right)$

= $\sin x \sqrt{2} \left(\sin x \frac{1}{\sqrt{2}} + \cos x \frac{1}{\sqrt{2}}\right)$

= $\sqrt{2} \sin x \left(\sin x \cos {45}^{\circ} + \cos x \sin {45}^{\circ}\right)$

= $\sqrt{2} \sin x \sin \left(x + {45}^{\circ}\right)$

= $\frac{\sqrt{2}}{2} \left(2 \sin x \sin \left(x + {45}^{\circ}\right)\right)$

= $\frac{\sqrt{2}}{2} \left(2 \sin x \sin \left(x + {45}^{\circ}\right)\right)$

= 1/sqrt2[cos(x-(x+45^@)-cos(x+(x+45^@)]

= $\frac{1}{\sqrt{2}} \left[\cos \left(- {45}^{\circ}\right) - \cos \left(2 x + {45}^{\circ}\right)\right]$

= $\frac{1}{\sqrt{2}} \left[\frac{1}{\sqrt{2}} - \cos \left(2 x + {45}^{\circ}\right)\right]$

= $\frac{1}{2} - \frac{1}{\sqrt{2}} \cos \left(2 x + {45}^{\circ}\right)$

As range of $\cos \left(2 x + {45}^{\circ}\right)$ is $\left[- 1 , 1\right]$

range of $\sin x \left(\sin x + \cos x\right)$ or $\frac{1}{2} - \frac{1}{\sqrt{2}} \cos \left(2 x + {45}^{\circ}\right)$ is

$\left[\frac{1}{2} - \frac{1}{\sqrt{2}} , \frac{1}{2} + \frac{1}{\sqrt{2}}\right]$