Trigonometric Equation involving Tangent and Arctangent

**scherz0** · June 13th 2011, 12:17 PM

Dear all,

For the following equation which I have to solve for $t$ , 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.

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

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

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

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

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

**Stro** · June 13th 2011, 09:23 PM

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

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

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

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

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

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

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

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

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

**Prove It** · June 13th 2011, 09:33 PM

Originally Posted by scherz0

Dear all,

For the following equation which I have to solve for $t$ , 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: $\tan(t\sqrt{5}) = \frac{-2 \times \sqrt{5}}{\alpha + 2} \Rightarrow t = \frac{1}{\sqrt{5}}\left(\pi - arctan\frac{2\sqrt{5}}{\alpha + 2}\right)$ , where $\alpha \geq 0$

---

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

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

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

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

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

**scherz0** · June 17th 2011, 10:17 AM

Originally Posted by Prove It

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

Thank you very much for your replies.

@Prove It: Does your reply explain what member, Stro, did:

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

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

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

@Stro: I understand perfectly the algebra which you used. However, how did you think of using $\tan(\pi - x) = -\tan(x)$ ? 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 $arctan(x) = -arctan(x)$ .

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