Half Angle Formula for Tangent

[sleepingMantis]

Hi all,

I am a self learner (graduated very long ago and rusty at math) working through the Riley, Hobson and Bence text, chapter 1.

1. Homework Statement

Use the fact that $sin(\pi/6) = 1/2$ to prove that $tan(π/12) = 2 − \sqrt{3}.$

Homework Equations

$tan(2x) = \frac { 2 tan(x)} {1 - tan^2(x)}$

The Attempt at a Solution

I set up the equation:

$$ tan(\frac {\pi} {12}) = \frac { 2 tan(\frac {\pi}{6})} {1 - tan^2(\frac{\pi}{6})} $$

Edit (more details):

I let $tan(\frac{\pi}{12}) = t^2$

Then rearranging I get

$$t^2 - 4t + 1 = 0$$

Simplifying and using the quadratic formula I end up with:

$$ t = 2 \pm \sqrt{3} $$

plugging in the value for $t$

$$ tan(\frac{\pi}{12}) = 2 \pm \sqrt{3} $$

Now my question is, how do I know what solution to pick?

In the book it says:

"To resolve the ambiguity, we note that, since $\frac{\pi}{12} < \frac{\pi}{4}$ and $tan(\frac{\pi}{4}) = 1$, we must have $tan(\frac{\pi}{12}) < 1$; hence, the negative sign is the appropriate choice."

But I do not understand what this means. Could someone help me out? Specifically how did the author deduce that the correct solution is the one with the negative sign?

Thanks!

 

[Ray Vickson]

Hi all,

I am a self learner (graduated very long ago and rusty at math) working through the Riley, Hobson and Bence text, chapter 1.

1. Homework Statement

Use the fact that $sin(π/6) = 1/2$ to prove that $tan(π/12) = 2 − \sqrt{3}.$

Homework Equations

$tan(2x) = \frac { 2 tan(x)} {1 - tan^2(x)}$

The Attempt at a Solution

I set up the equation:

$$ tan(\frac {\pi} {12}) = \frac { 2 tan(\frac {\pi}{6})} {1 - tan^2(\frac{\pi}{6})} $$

Simplifying and using the quadratic formula I end up with:

$$ tan(\frac{\pi}{12}) = 2 \pm \sqrt{3} $$

Now my question is, how do I know what solution to pick?

In the book it says:

"To resolve the ambiguity, we note that, since $\frac{\pi}{12} < \frac{\pi}{4}$ and $tan(\frac{\pi}{4}) = 1$, we must have $tan(\frac{\pi}{12}) < 1$; hence, the negative sign is the appropriate choice."

But I do not understand what this means. Could someone help me out? Specifically how did the author deduce that the correct solution is the one with the negative sign?

Thanks!

Evaluate the root with the "+" sign and see what you get. Is the result <1?

 

[Orodruin]

Simplifying and using the quadratic formula I end up with:

Don't just describe what you do in words and give the final result you get. Then there is no way we can see where your problem lies.

 

[sleepingMantis]

Evaluate the root with the "+" sign and see what you get. Is the result <1?

Thanks Ray, indeed the solution with the negative sign gives a result < 1.

I guess I am confused as to how the author knew that $tan (\frac{\pi}{4}) = 1$ was the right comparison to use? From my understanding tanx can be + or - infinity, so how did the author know we can bound it using $tan (\frac{\pi}{4}) = 1$?

 

[Orodruin]

You could have used any known angle with a tangent between $2-\sqrt 3$ and $2+\sqrt 3$ to the same effect. There is nothing special about $\pi/4$ other than that it has a tangent that is very easy to remember. Also, between 0 and $\pi/2$, the tangent goes between 0 and infinity and is continuous.

 

[sleepingMantis]

You could have used any known angle with a tangent between $2-\sqrt 3$ and $2+\sqrt 3$ to the same effect. There is nothing special about $\pi/4$ other than that it has a tangent that is very easy to remember. Also, between 0 and $\pi/2$, the tangent goes between 0 and infinity and is continuous.

Thank you,

Would I be correct in saying that if $x < y \implies tan(x) < tan(y)$?

 

[Orodruin]

Thank you,

Would I be correct in saying that if $x < y \implies tan(x) < tan(y)$?

If $-\pi/2 < x < y < \pi/2$, yes. Not in general.

Edit: You can see this using the unit circle. As you increase the angle, the sine increases and the cosine decreases monotonically (while both are positive in the first quadrant). Hence, their ratio, i.e., the tangent, increases.

 

[sleepingMantis]

If $-\pi/2 < x < y < \pi/2$, yes. Not in general.

Ahh, perfect now I understand - I think I will need to review basic trig functions in more depth (perhaps from another book). Thank you for the help!

 

[neilparker62]

If sin(θ) = 1/2 , then θ can be either π/6 or (π-π/6) = 5π/6. So 2 + √3 is the tan of 1/2 x (5π/6) = 5π/12 (75 degrees).

The equation set up by the op doesn't seem correct and doesn't use sin(π/6) = 1/2 . Also I don't know how it leads to t^2 - 4t +1 which is the correct quadratic

 

[sleepingMantis]

If sin(θ) = 1/2 , then θ can be either π/6 or (π-π/6) = 5π/6. So 2 + √3 is the tan of 1/2 x (5π/6) = 5π/12 (75 degrees).

The equation set up by the op doesn't seem correct and doesn't use sin(π/6) = 1/2 . Also I don't know how it leads to t^2 - 4t +1 which is the correct quadratic

Yes you are right, I seemed to have written the wrong double angle formula down. The equation I set up should be:

$$ sin(\frac{\pi}{6}) = \frac{1}{2} = \frac{2tan^2(\frac{\pi}{12})}{1 + tan(\frac{\pi}{12})} $$

Also in case another learner is looking at this thread, in the double angle formula I did right down, there is a major typo. It should have read:

$$ tan(\frac{\pi}{6}) = \frac{2tan(\frac{\pi}{12})}{1 - tan^2(\frac{\pi}{12})} $$
I cannot seem to edit my original post, apologies!.

 

[neilparker62]

Yes you are right, I seemed to have written the wrong double angle formula down. The equation I set up should be:

$$ sin(\frac{\pi}{6}) = \frac{1}{2} = \frac{2tan^2(\frac{\pi}{12})}{1 + tan(\frac{\pi}{12})} $$
$$ sin(\frac{\pi}{6}) = \frac{1}{2} = \frac{2tan(\frac{\pi}{12})}{1 + tan^2(\frac{\pi}{12})} $$
Also in case another learner is looking at this thread, in the double angle formula I did right down, there is a major typo. It should have read:

$$ tan(\frac{\pi}{6}) = \frac{2tan(\frac{\pi}{12})}{1 - tan^2(\frac{\pi}{12})} $$
I cannot seem to edit my original post, apologies!.

 

[neilparker62]

What's interesting about the triplet 2t, 1-t^2 and 1+t^2 is that with t rational, you can use it to generate rational Pythagorean triplets satisfying x^2 + y^2 = z^2. I've always had the hunch that it will not be possible to find a similar 'generator' for sets of cubes x^3 + y^3 = z^3 or more generally x^n + y^n = z^n. If someone could prove this 'hunch' , I suspect they would have proved Fermat's last theorem.

 

[Orodruin]

What's interesting about the triplet 2t, 1-t^2 and 1+t^2 is that with t rational, you can use it to generate rational Pythagorean triplets satisfying x^2 + y^2 = z^2. I've always had the hunch that it will not be possible to find a similar 'generator' for sets of cubes x^3 + y^3 = z^3 or more generally x^n + y^n = z^n. If someone could prove this 'hunch' , I suspect they would have proved Fermat's last theorem.

By Fermat’s theorem, no such generator can exist. However, proving that no such generator exists does not show Fermat’s theorem unless you can prove that any numbers satisfying the equation must be possible to generate like that.

 

[neilparker62]

By Fermat’s theorem, no such generator can exist. However, proving that no such generator exists does not show Fermat’s theorem unless you can prove that any numbers satisfying the equation must be possible to generate like that.

Well to begin with, there are no Pythagorean triplets which are not derivatives (similar triangles) of 2t, 1-t^2 and 1+t^2 anyway.

 

[Orodruin]

Well to begin with, there are no Pythagorean triplets which are not derivatives (similar triangles) of 2t, 1-t^2 and 1+t^2 anyway.

Which does not imply that the same must hold for the corresponding Fermat case. You would have to show this.

 

[neilparker62]

Which does not imply that the same must hold for the corresponding Fermat case. You would have to show this.

How to show that there are no exceptions to a non-existent 'generator rule' ? Hmm!

However since Fermat wrote his famous 'theorem' next to the text of Diophantus II VIII "How to divide a square into a sum of two squares" , we should perhaps take a careful look at the Diophantine method to see if there might be any clues which could have prompted Fermat's thinking/intuition.

As it turns out, the method described by Diophantus leads us to the afore-mentioned triplet generator and his 'arbitrary multiple of x' corresponds to the t-parameter within that triplet. The triplet generator contains terms in t and t^2 which are 'homogenous' for rational values of t - perhaps not for other powers:

$$ a^3 + b^3 = c^3 ⇒({a^{3/2}})^2+({b^{3/2}})^2=({c^{3/2}})^2 $$ $$t=\frac { c^{3/2}-a^{3/2}}{b^{3/2} } $$ $$ t^2 =\frac{a^3+c^3-2a^{1.5}c^{1.5}}{b^3} $$

But I digress - such speculation is admittedly somewhat distant from the OP's question.