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

  • Thread starter Thread starter DanielBW
  • Start date Start date
  • Tags Tags
    Angle
AI Thread Summary
The equation tan(2θ) · tan(α) = -1 leads to the solution θ = π/4 + α/2, but there are complexities involved. The derivation reveals two potential solutions for tan(θ): tan(α) + sec(α) and tan(α) - sec(α), with the former yielding the known answer. However, substituting the solution back into the original equation results in an indeterminate form, raising questions about its validity. The discussion emphasizes that the two tangent functions are perpendicular, suggesting additional solutions exist, including θ = π/4 + α/2 + kπ/2 for integer k. Overall, the problem illustrates the intricacies of trigonometric identities and their implications.
DanielBW
Messages
7
Reaction score
0
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?
 
Mathematics news on Phys.org
DanielBW said:
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 $\tan \alpha = \frac y x$)?
 
I like Serena said:
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...
 
Last edited:
DanielBW said:
$tan2\theta \cdot tan \alpha =-1$

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

However there are two problems with this.

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

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
[math]tan \left ( 2 \left ( \frac{\pi}{4} + \frac{\alpha}{2} \right ) \right )[/math]

[math]= tan \left ( \frac{\pi}{2} + \alpha \right )[/math]

[math]= \frac{tan \left ( \frac{\pi}{2} \right ) + tan( \alpha )}{1 - tan \left ( \frac{\pi}{2} \right )~tan( \alpha )}[/math]

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

-Dan
 
I would view the two tangent functions as gradients, and since their product is -1, they must be perpendicular. :D
 
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: [math]tan(\theta) = tan(\alpha) \pm sec(\alpha)[/math]. The + sign generates the given answer. Does anyone know how to solve [math]tan(\theta) = tan(\alpha) - sec(\alpha)[/math]?

-Dan
 
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:
$$\theta = \frac\pi 4 + \frac \alpha 2 + k\frac \pi 2$$
 

Similar threads

Back
Top