Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.

Question

Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.

Attempt at a solution:

$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$

$$\tan y = (3 \cos x - 3 \sin x \tan y) \sin x$$

$$\tan y + 3 \sin^2 x + \tan y = 3 \sin x \cos x$$

$$\tan y = \frac{3 \sin x \cos x}{1 + 3 \sin^2 x}$$

I have also tried substituting $0$, $30$, $45$, $60$, $90$ to the values of $x$.

Answer 1

It is a typo. It must be $$\tan y=\dfrac{3\sin x\cos x}{1+3\sin^2 x}=\dfrac{3\sin 2x}{5-3\cos 2x}$$for minimizing it we should have the 1st-order derivation of $\tan y$ equal to $0$ or $$\dfrac{d\tan y}{dx}=0$$which yields to $$6\cos 2x(5-3\cos 2x)=6\sin 2x\cdot 3\sin 2x$$which yields to $\cos 2x=\dfrac{3}{5}$ and $\sin 2x=\dfrac{4}{5}$. Substituting these values in the expression of $\tan y$ we attain the maximum:$$\max_{0\le x\le\dfrac{\pi}{2}} \tan y=\dfrac{3}{4}$$

Here is a sketch of $\tan y$ respect to $x$

Answer 2

Let's play.

$\sin y = 3\cos(x+y)\sin x = 3(\cos x \cos y-\sin x \sin y)\sin x $

Divide by $\cos x \cos y$

$\dfrac{\tan y}{\cos x} = 3(\cos x -\sin x \tan y)\tan x = 3\cos x\tan x -3\sin x\tan x \tan y $

$\tan y(\dfrac{1}{\cos x}+3\sin x\tan x) = 3\cos x\tan x =3\sin x $

$\tan y(1+3\sin^2 x) =3\sin x\cos x $

$\tan y =\dfrac{3\sin x\cos x}{1+3\sin^2 x} =\dfrac{\frac32\sin(2 x)}{1+3(1-\cos(2x))/2} =\dfrac{3\sin(2 x)}{2+3(1-\cos(2x))} =\dfrac{3\sin(2 x)}{5-3\cos(2x)} $

At this point, it being about midnight here, I threw this at Wolfy which said that its max value was $\dfrac34$ at $\pi n +\arctan(1/2)$.

So the max value is $\dfrac34$.

Answer 3

Starting from marty cohen's answer $$\tan (y)=\dfrac{3\sin(2 x)}{5-3\cos(2x)}$$ let $t=\tan(x)$ to get $$\tan (y)=\frac{3 t}{4 t^2+1}=f(t)$$ So $$f'(t)=\frac{3-12 t^2}{\left(4 t^2+1\right)^2}\qquad \text{and} \qquad f''(t)=\frac{24 t \left(4 t^2-3\right)}{\left(4 t^2+1\right)^3}$$ The first derivative cancels for $t_{\pm}=\pm \frac 12$. For $t_+$, $f''(t+)=-3$ corresponding to a maximum and for $t_-$, $f''(t_-)=3$ corresponding to a minimum.

So, $t=\frac 12$ and $\tan(y)=\frac34$

Answer 4

Hint:

Since $$ \begin{align} \sin(y) &=3\cos(x+y)\sin(x)\\ &=3\cos(x)\cos(y)\sin(x)-3\sin(x)\sin(y)\sin(x) \end{align} $$ we get $$ \begin{align} \tan(y) &=\frac{3\sin(x)\cos(x)}{1+3\sin^2(x)}\\ &=\frac{3\tan(x)}{\sec^2(x)+3\tan^2(x)}\\[3pt] &=\frac{3\tan(x)}{1+4\tan^2(x)} \end{align} $$ Then note that $1+4\tan^2(x)=4\tan(x)+(2\tan(x)-1)^2$.

Answer 5

I start, like everyone else with $3\sin2x/(5-3\cos2x)$.

Consider a circle, radius 3, around 5+0i in the complex plane. The highest value for tan y is for the line through the origin, tangent to the circle. That forms a 3-4-5 right-angled triangle, so $\tan y=3/4$.