Maximum value of $ y = 3 \cos\alpha \sin\alpha + \sqrt 3 \sin^2\alpha$

Question

Can somebody point me in the right direction in finding the maximum value of $y$ for a given value of $\alpha$ for the following trigonometric expression.

$$ y = 3 \cos\alpha \sin\alpha + \sqrt 3 \sin^2\alpha$$

Thanks

Many thanks to all those that contributed to the solution to this problem. Below is how I solved this question.

The "trick" is to use the double angle formulas.

$$ y = \bigl(\frac{3}{2}\bigr) 2 \cos\alpha \sin\alpha + \sqrt3 \bigl(\frac{1}{2}[1 - \cos 2\alpha]\bigr)$$

$$ y = \bigl(\frac{3}{2}\bigr) \sin(2\alpha) + \frac{\sqrt3}{2} - \frac{\sqrt3}{2} \cos (2\alpha)$$

$$ y = \frac{\sqrt3}{2} +\bigl(\frac{3}{2}\bigr) \sin(2\alpha) - \frac{\sqrt3}{2} \cos (2\alpha)$$

$$ y = \frac{\sqrt3}{2} +\frac{1}{2} \bigl(3\sin(2\alpha) - \sqrt3 \cos (2\alpha)\bigl)$$

Now let $$ x = 3\sin(2\alpha) - \sqrt3 \cos (2\alpha)$$

Differentiate $x$ with respect to $\alpha$ and set

$$\frac{dx}{d\alpha} = 0 $$

$$\frac{dx}{d\alpha} = 6\cos(2\alpha) + 2\sqrt3\sin(2\alpha) = 0$$

$$6\cos(2\alpha) = -2\sqrt3\sin(2\alpha) $$

$$\tan(2\alpha) = -\frac{6}{2\sqrt3}$$

Simplifying

$$\tan(2\alpha) = -\sqrt3$$

This angle is in the 2nd quadrant.

$$2\alpha = 180^{\circ}- \tan^{-1}\sqrt3$$

$$2\alpha = 180^{\circ}- 60^{\circ}$$

$$2\alpha = 120^{\circ}$$

$$\alpha = 60^{\circ}$$

So maximum value of original equation $y = 3 \cos\alpha \sin\alpha + \sqrt 3 \sin^2\alpha$ occurs when $\alpha = 60^{\circ}$.

Answer 1

$y = \dfrac{3\sin(2\alpha)}{2}+\sqrt{3}\cdot \dfrac{1-\cos(2\alpha)}{2}= \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\left(3\sin(2\alpha)-\sqrt{3}\cos(2\alpha)\right)\le \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\cdot \sqrt{3^2+(-\sqrt{3})^2}\cdot \sqrt{\sin^2(2\alpha)+\cos^2(2\alpha)}=\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{12}}{2}=\dfrac{3\sqrt{3}}{2}\implies y_{\text{max}} = \dfrac{3\sqrt{3}}{2}$ .

Answer 2

Hint

Divide both sides by $\cos^2\alpha$ to form a quadratic equation in $\tan\alpha$ which is real

So the discriminant must be $\ge0$

Answer 3

$$2y-\sqrt{3}=3\sin(2\alpha)-\sqrt{3}\cos(2\alpha)$$

$$=2\sqrt{3}\sin(2\alpha-\frac{\pi}{6})$$ the maximum of $ y $ is attained when

$$2\alpha-\frac{\pi}{6}=\frac{\pi}{2}+2k\pi$$

Answer 4

$$\dfrac{2y}{\sqrt3}=2\sin A(\sin A+\sqrt3\cos A)$$

$$=2\sin A\sin(A+\pi/3)$$

$$=\cos\pi/3-\cos(2A+\pi/3)$$

using https://mathworld.wolfram.com/WernerFormulas.html

Now for real $x,$ $$-1\le\cos x\le1$$

Answer 5

$$ \begin{aligned} y &=3 \cos \alpha \sin \alpha+\sqrt{3} \sin ^2 \alpha \\ &=\frac{3}{2} \sin 2 \alpha+\sqrt{3}\left(\frac{1-\cos 2 \alpha}{2}\right) \\ &=\frac{3}{2} \sin 2 \alpha-\frac{\sqrt{3}}{2} \cos 2 \alpha+\frac{\sqrt{3}}{2}\\&= \sqrt{3} \sin \left(2 \alpha-\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}, \end{aligned} $$ We conclude that $$y_{max}= \sqrt3+\frac{\sqrt3}{2} = \frac{3\sqrt3}{2} $$ at $\alpha=n \pi+\frac{5 \pi}{3}\textrm{ where } n\in Z$.