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

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 $x$.

• i have got an other formula for $\tan(y)$ Feb 18, 2018 at 8:08
• You're almost there. See my answer.
– robjohn
Feb 18, 2018 at 9:14

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 $x$ • Why is the whole equation not differentiated? Feb 19, 2018 at 10:36
• Which equation you mean? I have differentiated whole $\tan y$ respect to $x$ Feb 19, 2018 at 10:39
• 6cos2x(5−3cos2x)=6sin2x⋅3sin2x So this means that there is no need to solve for the denominator since our goal is to make the numerator equal to 0. Is my understanding correct? Feb 19, 2018 at 10:42

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 $\dfrac34$.

• I got it until the part of tan y = (3 sin x cos x) / (1+ 3sin^2 x) Could you please explain how you got sin (2 x) ? Feb 18, 2018 at 8:30
• @Janjan $\sin(2x) = 2 \sin(x) \cos(x)$. You can see this by, for instance, considering $e^{2i x} = (e^{i x})^2$. Feb 18, 2018 at 9:30

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 $\tan(y)=\frac34$

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 $1+4\tan^2(x)=4\tan(x)+(2\tan(x)-1)^2$.

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 $\tan y=3/4$.