Find the range of $sin x (sin x + cos x)$ $sinx(sinx+cosx)$ ?

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Find the range of $sin x (sin x + cos x)$ $sinx(sinx+cosx)$ ?

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Answer 1 · 2017-04-01T17:53:21

Range of $sin x (sin x + cos x)$ $sinx(sinx+cosx)$ is $[\frac{1}{2} - \frac{1}{\sqrt{2}}, \frac{1}{2} + \frac{1}{\sqrt{2}}]$ $[1/2-1/sqrt2,1/2+1/sqrt2]$

Explanation:

$sin x (sin x + cos x)$ $sinx(sinx+cosx)$

= $sin x \sqrt{2} (sin x \frac{1}{\sqrt{2}} + cos x \frac{1}{\sqrt{2}})$ $sinxsqrt2(sinx1/sqrt2+cosx1/sqrt2)$

= $\sqrt{2} sin x (sin x {cos 45}^{\circ} + cos x {sin 45}^{\circ})$ $sqrt2sinx(sinxcos45^@+cosxsin45^@)$

= $\sqrt{2} sin x sin (x + 45^{\circ})$ $sqrt2sinxsin(x+45^@)$

= $\frac{\sqrt{2}}{2} (2 sin x sin (x + 45^{\circ}))$ $sqrt2/2(2sinxsin(x+45^@))$

= $\frac{1}{\sqrt{2}} [cos (x - (x + 45^{\circ}) - cos (x + (x + 45^{\circ})]$ $1/sqrt2[cos(x-(x+45^@)-cos(x+(x+45^@)]$

= $\frac{1}{\sqrt{2}} [cos (- 45^{\circ}) - cos (2 x + 45^{\circ})]$ $1/sqrt2[cos(-45^@)-cos(2x+45^@)]$

= $\frac{1}{\sqrt{2}} [\frac{1}{\sqrt{2}} - cos (2 x + 45^{\circ})]$ $1/sqrt2[1/sqrt2-cos(2x+45^@)]$

= $\frac{1}{2} - \frac{1}{\sqrt{2}} cos (2 x + 45^{\circ})$ $1/2-1/sqrt2cos(2x+45^@)$

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

range of $sin x (sin x + cos x)$ $sinx(sinx+cosx)$ or $\frac{1}{2} - \frac{1}{\sqrt{2}} cos (2 x + 45^{\circ})$ $1/2-1/sqrt2cos(2x+45^@)$ is

$[\frac{1}{2} - \frac{1}{\sqrt{2}}, \frac{1}{2} + \frac{1}{\sqrt{2}}]$ $[1/2-1/sqrt2,1/2+1/sqrt2]$

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Find the range of $sin x (sin x + cos x)$ $sinx(sinx+cosx)$ ?

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

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Find the range of $sin x (sin x + cos x)$ $sinx(sinx+cosx)$ ?