Find the greatest possible value of tan B.

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In summary, we are given that $A$ and $B$ are acute angles and that $\tan B = 2015\sin A \cos A - 2015\sin^2 A \tan B$. We are asked to find the greatest possible value of $\tan B$. Using algebraic manipulation and differentiation, we can find that the maximum value of $\tan B$ is approximately 22.4388.
  • #1
anemone
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Let $A,\,B$ be acute angles such that $\tan B=2015\sin A \cos A-2015\sin^2 A \tan B$.

Find the greatest possible value of $\tan B$.
 
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  • #2
anemone said:
Let $A,\,B$ be acute angles such that $\tan B=2015\sin A \cos A-2015\sin^2 A \tan B$.

Find the greatest possible value of $\tan B$.
[sp]Let $n = 2015.$ Then $\tan B(1 + n\sin^2 A) = n\sin A\cos A$. Therefore $\tan B = \dfrac{n\sin A\cos A}{1 + n\sin^2 A}$ and so $$\cot B = \frac{1 + n\sin^2 A}{n\sin A\cos A} = \frac2{n\sin (2A)} + \tan A.$$ To maximise $\tan B$ we need to minimise $\cot B$ (the reason for doing this is that $\cot B$ is easier to differentiate than $\tan B$), so let's differentiate it: $$\frac d{dA}\cot B = -\frac{4\cos(2A)}{n\sin^2(2A)} + \frac1{\cos^2 A} = \frac{-\cos(2A) + n\sin^2A}{n\sin^2A\cos^2A}.$$ That is zero when $0 = -\cos(2A) + n\sin^2A = -1 + (n+2)\sin^2A$, so that $\sin A = \dfrac1{\sqrt{n+2}}$ and $\cos A = \dfrac{\sqrt{n+1}}{\sqrt{n+2}}.$

The corresponding (maximal) value of $\tan B$ is $$\frac{\frac{n\sqrt{n+1}}{n+2}}{1 + \frac{n}{n+2}} = \frac{n\sqrt{n+1}}{2(n+1)} = \frac n{2\sqrt{n+1}}.$$ With $n=2015$ that gives the maximum value of $\tan B$ as $\dfrac{2015}{2\sqrt{2016}} = \dfrac{2015}{24\sqrt{14}} \approx 22.4388.$[/sp]
 
  • #3
Opalg said:
[sp]Let $n = 2015.$ Then $\tan B(1 + n\sin^2 A) = n\sin A\cos A$. Therefore $\tan B = \dfrac{n\sin A\cos A}{1 + n\sin^2 A}$ and so $$\cot B = \frac{1 + n\sin^2 A}{n\sin A\cos A} = \frac2{n\sin (2A)} + \tan A.$$ To maximise $\tan B$ we need to minimise $\cot B$ (the reason for doing this is that $\cot B$ is easier to differentiate than $\tan B$), so let's differentiate it: $$\frac d{dA}\cot B = -\frac{4\cos(2A)}{n\sin^2(2A)} + \frac1{\cos^2 A} = \frac{-\cos(2A) + n\sin^2A}{n\sin^2A\cos^2A}.$$ That is zero when $0 = -\cos(2A) + n\sin^2A = -1 + (n+2)\sin^2A$, so that $\sin A = \dfrac1{\sqrt{n+2}}$ and $\cos A = \dfrac{\sqrt{n+1}}{\sqrt{n+2}}.$

The corresponding (maximal) value of $\tan B$ is $$\frac{\frac{n\sqrt{n+1}}{n+2}}{1 + \frac{n}{n+2}} = \frac{n\sqrt{n+1}}{2(n+1)} = \frac n{2\sqrt{n+1}}.$$ With $n=2015$ that gives the maximum value of $\tan B$ as $\dfrac{2015}{2\sqrt{2016}} = \dfrac{2015}{24\sqrt{14}} \approx 22.4388.$[/sp]

Aww Opalg! Thanks for participating in this other Olympiad Math problem from Hong Kong and your answer is of course correct! I will wait for a couple of days before posting the proposed solution.:)
 
  • #4
Solution of other:
$\tan B=2015\sin A \cos A-2015\sin^2 A \tan B$

$\dfrac{\sin B}{\cos B}=2015\sin A \cos A-2015\sin^2 A \left(\dfrac{\sin B}{\cos B}\right)$

$\begin{align*}\sin B&=2015\sin A \cos A\cos B-2015\sin^2 A\sin B\\&=2015\sin A( \cos A\cos B-\sin A\sin B)\\&=2015\sin A( \cos (A+B)\\&=2015\left(\dfrac{\sin (2A+B)-\sin B}{2}\right)\end{align*}$

$\therefore \left(1+\dfrac{2015}{2}\right)\sin B=\dfrac{2015}{2}\sin (2A+B)$

And we get

$\sin B=\dfrac{2015}{2017}\sin (2A+B)$

Since $\sin B\le \dfrac{2015}{2017}$, that implies $\tan B\le \dfrac{2015}{\sqrt{2017^2-2015^2}}=\dfrac{2015}{24\sqrt{14}}$.

Equality is attained when $\sin (2A+B)=1$, that is when $2A+B$ is a right angle.
 

FAQ: Find the greatest possible value of tan B.

What is the formula for finding the greatest possible value of tan B?

The formula for finding the greatest possible value of tan B is tan B = sin B/cos B.

How do you determine the greatest possible value of tan B using a calculator?

To determine the greatest possible value of tan B using a calculator, you would enter the value of B in degrees or radians and then press the "tan" button on your calculator. This will give you the value of tan B, which is the greatest possible value.

Can the greatest possible value of tan B be negative?

Yes, the greatest possible value of tan B can be negative. This will occur when the angle B is in the second or fourth quadrant, where both the sine and cosine functions are negative.

Is the greatest possible value of tan B always defined?

No, the greatest possible value of tan B is not always defined. It is only defined when the value of cos B is not equal to 0. If cos B is equal to 0, then the denominator of the formula will be 0, making the value of tan B undefined.

How can you use the unit circle to find the greatest possible value of tan B?

You can use the unit circle to find the greatest possible value of tan B by identifying the coordinates of the point where the terminal side of angle B intersects the unit circle. Then, you can use the formula tan B = y/x, where y is the y-coordinate and x is the x-coordinate, to calculate the value of tan B.

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