Find Angle Given $\alpha$ in $tan2\theta \cdot tan \alpha =-1$

[DanielBW]

Hello! I have this equation:

$tan2\theta \cdot tan \alpha =-1$

where $\alpha$ is known. I need to find $\theta$. The answer is:

$\theta = \dfrac{\pi}{4}+\dfrac{\alpha}{2}$

How is that possible?

 

[I like Serena]

Hello! I have this equation:

$tan2\theta \cdot tan \alpha =-1$

where $\alpha$ is known. I need to find $\theta$. The answer is:

$\theta = \dfrac{\pi}{4}+\dfrac{\alpha}{2}$

How is that possible?

Hi DanielBW! Welcome to MHB! :)

Suppose you draw a right triangle with angle $\alpha$ in it.
And suppose the sides are $x$ and $y$, such that $\tan \alpha = \frac y x$.

Then what can you say about $\tan 2\theta$ (after substituting $\tan \alpha = \frac y x$)?

 

[DanielBW]

Hi DanielBW! Welcome to MHB! :)

Suppose you draw a right triangle with angle $\alpha$ in it.
And suppose the sides are $x$ and $y$, such that $\tan \alpha = \frac y x$.

Then what can you say about $\tan 2\theta$ (after substituting $\tan \alpha = \frac y x$)?

Hi I like Serena ! I'm not sure at all, but i think in that case $tan2\theta=-\dfrac{x}{y}$ which means that $2\theta$ is a complementary angle of $\alpha$ ... if that's correct, then $2\theta=C+\alpha$. Obviously C has to be $\pi/2$ but as i said, I'm not sure at all...

 

[topsquark]

$tan2\theta \cdot tan \alpha =-1$

$\theta = \dfrac{\pi}{4}+\dfrac{\alpha}{2}$

If you slog through the whole messy problem using the \(\displaystyle tan(2 \theta)\) trig expression you indeed (after a while) find that
\(\displaystyle \theta = \frac{\pi}{4} + \frac{\alpha}{2}\)

However there are two problems with this.

First, there is another possible solution. At some point in the derivation we get
\(\displaystyle tan( \theta ) = tan( \alpha ) \pm sec( \alpha )\)

The + sign generates the given answer. I do not know how to solve the case for the - sign.

Second, if you put the given solution into the original equation you get
\(\displaystyle tan \left ( 2 \left ( \frac{\pi}{4} + \frac{\alpha}{2} \right ) \right )\)

\(\displaystyle = tan \left ( \frac{\pi}{2} + \alpha \right )\)

\(\displaystyle = \frac{tan \left ( \frac{\pi}{2} \right ) + tan( \alpha )}{1 - tan \left ( \frac{\pi}{2} \right )~tan( \alpha )}\)

I don't care what WA says, this expression is indeterminate for arbitrary \(\displaystyle \alpha\). So the given answer does not work in the original equation. The only hope for a solution here is solving \(\displaystyle tan( \theta ) = tan(a) - sec(a)\). Does someone know how to solve this?

-Dan

 

[MarkFL]

I would view the two tangent functions as gradients, and since their product is -1, they must be perpendicular. :D

 

[topsquark]

Okay, I found out why the tan(x + y) formula doesn't work for me. When deriving the tan(x + y) formula we divide numerator and denominator by cos(x)cos(y) which is 0 in this case.

I still have a question. My derivation shows that there are two possible solutions: \(\displaystyle tan(\theta) = tan(\alpha) \pm sec(\alpha)\). The + sign generates the given answer. Does anyone know how to solve \(\displaystyle tan(\theta) = tan(\alpha) - sec(\alpha)\)?

-Dan

 

[I like Serena]

Not sure how you got there, but if $\tan \alpha = \frac yx$, then $\tan 2\theta=-\frac x y$.
As Mark said, this corresponds with any angle perpendicular to $\alpha$.
I'd make a drawing, but I'm currently not in a position to do so. Maybe later.

Anyway, it means we have:
$$2\theta = \frac\pi 2 + \alpha + k\pi$$
where $k$ can be any integer.

So the actual complete solution is:
$$\theta = \frac\pi 4 + \frac \alpha 2 + k\frac \pi 2$$