# Thread: Trigonometric Equation involving Tangent and Arctangent

1. ## Trigonometric Equation involving Tangent and Arctangent

Dear all,

For the following equation which I have to solve for $\displaystyle t$, I am getting an expression that differs from the given answer and I do not understand why.

I have made two attempts, both posted below. Thank you very much for your help.

---

Show: $\displaystyle \tan(t\sqrt{5}) = \frac{-2 \times \sqrt{5}}{\alpha + 2} \Rightarrow t = \frac{1}{\sqrt{5}}\left(\pi - arctan\frac{2\sqrt{5}}{\alpha + 2}\right)$, where $\displaystyle \alpha \geq 0$

---

Solution 1: $\displaystyle \tan(t\sqrt{5}) = \frac{-2 \times \sqrt{5}}{\alpha + 2}$

$\displaystyle \Rightarrow t\sqrt{5} = \arctan\frac{-2\sqrt{5}}{\alpha + 2}$

$\displaystyle \Rightarrow t = -\frac{1}{\sqrt{5}}\arctan\frac{2\sqrt{5}}{\alpha + 2}$,

where I have used the facts that $\displaystyle \arctan(-x) = -x$ and took the arctangent of the original equation.

2. ## Re: Trigonometric Equation involving Tangent and Arctangent

Here's one way, though possibly not the best.

$\displaystyle tan(t\sqrt{5}) = \frac{-2\sqrt5}{a+2}$

$\displaystyle -tan(t\sqrt{5}) = \frac{2\sqrt5}{a+2}$

$\displaystyle tan(\pi-t\sqrt{5}) = \frac{2\sqrt5}{a+2}$

$\displaystyle \pi-t\sqrt{5} = arctan(\frac{2\sqrt5}{a+2})$

$\displaystyle -t\sqrt{5} = -\pi+arctan(\frac{2\sqrt5}{a+2})$

$\displaystyle t = \frac{1}{\sqrt{5}}(\pi - arctan(\frac{2\sqrt5}{a+2}))$

Also I think you meant $\displaystyle arctan(-x)=-arctan(x)$ for your fact.

But you may have forgot about $\displaystyle -tan(x) = tan(\pi-x)$

3. ## Re: Trigonometric Equation involving Tangent and Arctangent

Originally Posted by scherz0
Dear all,

For the following equation which I have to solve for $\displaystyle t$, I am getting an expression that differs from the given answer and I do not understand why.

I have made two attempts, both posted below. Thank you very much for your help.

---

Show: $\displaystyle \tan(t\sqrt{5}) = \frac{-2 \times \sqrt{5}}{\alpha + 2} \Rightarrow t = \frac{1}{\sqrt{5}}\left(\pi - arctan\frac{2\sqrt{5}}{\alpha + 2}\right)$, where $\displaystyle \alpha \geq 0$

---

Solution 1: $\displaystyle \tan(t\sqrt{5}) = \frac{-2 \times \sqrt{5}}{\alpha + 2}$

$\displaystyle \Rightarrow t\sqrt{5} = \arctan\frac{-2\sqrt{5}}{\alpha + 2}$

$\displaystyle \Rightarrow t = -\frac{1}{\sqrt{5}}\arctan\frac{2\sqrt{5}}{\alpha + 2}$,

where I have used the facts that $\displaystyle \arctan(-x) = -x$ and took the arctangent of the original equation.
What they have done is recognised that $\displaystyle \displaystyle \frac{-2\sqrt{5}}{\alpha + 2} < 0$, so $\displaystyle \displaystyle t\sqrt{5}$ must be in the second or fourth quadrants. Their solution only has the shift to the second quadrant (by subtracting from $\displaystyle \displaystyle \pi$).

4. ## Re: Trigonometric Equation involving Tangent and Arctangent

Originally Posted by Prove It
What they have done is recognised that $\displaystyle \displaystyle \frac{-2\sqrt{5}}{\alpha + 2} < 0$, so $\displaystyle \displaystyle t\sqrt{5}$ must be in the second or fourth quadrants. Their solution only has the shift to the second quadrant (by subtracting from $\displaystyle \displaystyle \pi$).
Thank you very much for your replies.

$\displaystyle tan(t\sqrt{5}) = \frac{-2\sqrt5}{a+2}$
$\displaystyle \Rightarrow -tan(t\sqrt{5}) = \frac{2\sqrt5}{a+2}$
$\displaystyle \Rightarrow tan(\pi-t\sqrt{5}) = \frac{2\sqrt5}{a+2}$
@Stro: I understand perfectly the algebra which you used. However, how did you think of using $\displaystyle \tan(\pi - x) = -\tan(x)$? Was it based on what member, Prove It, wrote afterwards? I'm trying to grasp the inner logic behind proving this equation.
By the way, thank you; I did mean $\displaystyle arctan(x) = -arctan(x)$.