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

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In summary, the equation tan2theta = -1 has a complementary angle of alpha and can be solved for $\theta$ using the tangent functions as gradients.
  • #1
DanielBW
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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?
 
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  • #2
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 $\tan \alpha = \frac y x$)?
 
  • #3
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...
 
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  • #4
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 \(\displaystyle tan(2 \theta)\) 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
 
  • #5
I would view the two tangent functions as gradients, and since their product is -1, they must be perpendicular. :D
 
  • #6
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
 
  • #7
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:
$$\theta = \frac\pi 4 + \frac \alpha 2 + k\frac \pi 2$$
 

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

1. What is the equation "tan2θ · tan α = -1" used for?

The equation "tan2θ · tan α = -1" is used to find the angle θ when the tangent of twice the angle (2θ) is multiplied by the tangent of angle α and the result is -1.

2. What is the relationship between the angles in "tan2θ · tan α = -1"?

The angles in "tan2θ · tan α = -1" are related by the fact that the tangent of twice an angle is equal to the product of the tangents of the two individual angles, which results in a negative value of -1.

3. How do you solve for θ in the equation "tan2θ · tan α = -1"?

To solve for θ in the equation "tan2θ · tan α = -1", you can follow these steps:

  1. Divide both sides of the equation by tan α.
  2. Take the inverse tangent of both sides to isolate 2θ.
  3. Divide the resulting angle by 2 to find the value of θ.

4. Can the equation "tan2θ · tan α = -1" have multiple solutions?

Yes, the equation "tan2θ · tan α = -1" can have multiple solutions. This is because the tangent function has a periodic nature, meaning it repeats its values at regular intervals. Therefore, there can be multiple values of θ that satisfy the equation, depending on the given value of α.

5. What are some practical applications of the equation "tan2θ · tan α = -1"?

The equation "tan2θ · tan α = -1" has various practical applications in fields such as surveying, engineering, and physics. For example, it can be used to calculate the angle of elevation or depression when measuring the height of an object or the slope of a ramp. It is also used in trigonometric identities and equations, which are important in many mathematical and scientific calculations.

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