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Let $x\in \left[0,\frac{\pi}{2}\right]$ and $k>0$, solve $\tan(2x)=-\frac{1}{4k}$.

My book solves the equation by saying this "Let $2\alpha \in \left(\frac{\pi}{2},\pi\right]$ the arc such that $2\alpha=\arctan \left(-\frac{1}{4k}\right)$, since $2x\in[0,\pi]$, that value of the tangent is acceptable. Relatively to that value, it is $\sin(2\alpha)=\frac{1}{\sqrt{16k^2+1}}$ and $\cos(2\alpha)=-\frac{4k}{\sqrt{16k^2+1}}$."

I don't get two things in particular: why has the author chose $2\alpha \in \left(\frac{\pi}{2},\pi\right]$? I think it is something about where the tangent is invertible, but I'm not sure...and the second one is: I know that $\sin \arctan \theta =\frac{\theta}{\sqrt{1+\theta^2}}$ and $\cos \arctan \theta =\frac{1}{\sqrt{1+\theta^2}}$ for all $\theta\in\mathbb{R}$ but I get the opposite signs

$$\sin(2\alpha)=\sin \arctan \left(-\frac{1}{4k}\right)=\frac{-\frac{1}{4k}}{\sqrt{1+\left(-\frac{1}{4k}\right)^2}}=-\frac{1}{\sqrt{1+16k^2}}$$

$$\cos(2\alpha)=\cos\arctan \left(-\frac{1}{4k}\right)=\frac{1}{\sqrt{1+\left(-\frac{1}{4k}\right)^2}}=\frac{4k}{\sqrt{1+16k^2}}$$

I understand intuitively that since $2\alpha \in \left(\frac{\pi}{2},\pi\right]$ it must be that sin is positive and cosine is negative, and since $k>0$ it is $\frac{1}{\sqrt{1+16k^2}} >0$ and $\frac{4k}{\sqrt{1+k^2}}>0$ then I can adjust the signs just by eye, but I would like to know if there is a more algebraic method to get the signs correctly. Thank you.

 Apr 5, 2021
 #1
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I don't think there is one.

 

It's much easier just to visualize it, as you did. With a square root in the denominator, there are always sign issues. 

 

For your first question, they did it for the sake of convenience (so that we could have signs on $\sin(2\alpha)$ and $\cos(2\alpha)$.)

 Apr 5, 2021
edited by thedudemanguyperson  Apr 5, 2021
 #2
avatar+29 
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Thank you, now the part on the signs is clear; however, I still don't get why $2\alpha \in \left(\frac{\pi}{2},\pi\right]$. For example, why not $2\alpha \in \left(0,\frac{\pi}{2}\right]$? That would change the sign of $\cos (2\alpha)$, so I don't believe that this other choice is equivalent, so I suppose that there must be a criterion to take the right choice. Can you help me again? Thanks.

Hitago  Apr 5, 2021
 #3
avatar+423 
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We can't have $2\alpha\in (0,\pi/2]$ because then $\tan(2\alpha)>0$ and $k>0$ so $-\frac{1}{4k}<0$!!!! 

 #4
avatar+29 
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Oh, I see. Maybe now is clear: $2\alpha$ denotes just a particular solution in $[0,\pi]$ for the equation $\tan(2x)=-\frac{1}{4k}$, in general I must find a solution in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ adding the period $h\pi$ ($h\in\mathbb{Z}$) because tan is defined there, but here $2x\in[0,\pi]$ and so I must restrict the interval where I'm searching the solutions to $[0,\pi] \setminus \{\frac{\pi}{2}\}$. So the only possibilities is that $2\alpha \in \left[0,\frac{\pi}{2}\right)$ or $2\alpha \in \left(\frac{\pi}{2},\pi\right]$; for what you've said in the answers, the only possibility is $2\alpha \in \left(\frac{\pi}{2},\pi\right]$ because is the only one that is coherent with the sign of $-\frac{1}{4k}$. Is this correct? Thank you.

Hitago  Apr 5, 2021
edited by Hitago  Apr 5, 2021
 #5
avatar+423 
+3

That's 100% correct


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