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What is the range of the function

$$\frac{\sqrt{3}\sin x}{2+\cos x}$$

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  • $\begingroup$ I hope the answer is clear to you! $\endgroup$ – user80400 May 31 '13 at 16:04
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$$\sqrt3\sin x-\cos x=2\sin\left(x-\frac\pi6\right)\le 2$$ $$\implies \sqrt3\sin x\le 2+\cos x$$

$$\implies \frac{\sqrt3\sin x}{2+\cos x}\le1\text{ as }2+\cos x>0$$

Again, $$\sqrt3\sin x+\cos x=2\sin\left(x+\frac\pi6\right)\ge -2$$

$$\implies \sqrt3\sin x\ge -(2+\cos x)$$

$$\implies \frac{\sqrt3\sin x}{2+\cos x}\ge-1 \text{ as }2+\cos x>0$$

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  • 1
    $\begingroup$ A more compact approach could be $$\sqrt3\sin x\pm\cos x=2\sin\left(x\pm\frac\pi6\right)$$ $$\text{As }-1\le \sin\left(x\pm\frac\pi6\right)\le 1,$$ $$-2\le \sqrt3\sin x\pm\cos x\le2\implies -(2\pm \cos x)\le \sqrt3\sin x\le 2\mp\cos x $$ $$\text{As }2\pm \cos x>0, -1\le \frac{\sqrt3\sin x}{2\pm\cos x}\le 1$$ $\endgroup$ – lab bhattacharjee May 22 '13 at 10:19
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Transform it into $\tan\theta/2$ using $$\cos2\theta=(1-\tan^2\theta)/(1+\tan^2\theta)$$ and $$\sin2\theta=(2\tan\theta)/(1+\tan^2\theta)$$

to get $$y=\dfrac{2\sqrt 3 \tan\theta/2}{3+\tan^2\theta/2}$$

Now $\tan\theta/2\in \Bbb R$ so, the quad. wq. in $\tan\theta/2$ must have discriminant positive. You get the result $$-1\le y\le1$$

enter image description here

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  • $\begingroup$ @exploringnet: how did you made that graph? i mean is there some online source to plot these? $\endgroup$ – MathGeek May 5 '13 at 13:44
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Note that $y=f(x)=\dfrac{\sqrt 3 \sin x}{2+\cos x}$.

Note that $y^2=\dfrac{3 \sin ^2 x}{(2+\cos ^2 x)^2}=\dfrac{3 (1- \cos ^2 x)}{(2+\cos ^2 x)^2}$

Put $\cos x =u$, we have

\begin{align} y^2(2+u)^2 =3(1-u^2) \\ \implies (3+y^2)u^2+4y^2u+4y^2-3=0 \end{align} For this to have a solution, we need discriminant of the quadratic to be non-negative. Hence, we get $$(4y)^2-4(3+y^2)(4y^2-3) \ge 0 \\ \implies y^2 \le 1\\ \implies-1 \le y \le 1$$

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Consider that $(u,v)=(\cos x,\sin x)$ is a point on the unit circle. Let $\frac{\sqrt3\cos x}{2+\sin x}=\frac{\sqrt3u}{2+v}=t$. Then we have the straight line $L: v=\frac{\sqrt3}{t}u-2$ passing through $(0,-2)$, and $\frac{\sqrt3}{t}$ corresponds to the slope of $L$. Hence $t$ reaches extrema when $L$ is tangent to the unit circle.

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Let $$y = \frac{\sqrt{3}\sin x}{2+\cos x}\Rightarrow 2y+y\cos x = \sqrt{3}\sin x$$

So $$\sqrt{3}\sin x-y\cos x = 2y$$

Now Using $\bf{Cauchy\; Schwarz \; Inequality}$

$$\left[(\sqrt{3})^2+(-1)^2\right]\cdot \left[\sin^2 x+\cos^2 x\right]\geq \left[\sqrt{3}\sin x-y\cos x\right]^2$$

So $$4\geq 4y^2\Rightarrow y^2\leq 1\Rightarrow y \in \left[-1,1\right]$$

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Hint: You need to find the maximum and minimum values. To do this, find the $x$ values where the derivative of the function is zero, and then substitute these into the function itself. Note that, in general, you need to check the edges of the domain, and compare all turning points to find the absolute maxima and minima.

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