# Half Angle Formula for Tangent

• sleepingMantis
In summary: The OP's equation is correct. They used the double angle formula for tangent to simplify the expression. The quadratic equation follows when they substitute t^2 for tan(π/12) and solve for t.
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!

Last edited:
berkeman
sleepingMantis said:
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?

sleepingMantis
sleepingMantis said:
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.

Ray Vickson said:
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##?

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
Orodruin said:
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) ##?

sleepingMantis said:
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
Orodruin said:
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!

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
neilparker62 said:
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!.

Last edited:
sleepingMantis said:
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!.

sleepingMantis
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.

sleepingMantis
neilparker62 said:
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.

sleepingMantis
Orodruin said:
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.

sleepingMantis
neilparker62 said:
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.

SammyS and sleepingMantis
Orodruin said:
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.

Last edited:
sleepingMantis

## 1. What is the Half Angle Formula for Tangent?

The Half Angle Formula for Tangent is a trigonometric identity that relates the tangent of half an angle to the tangent of the full angle. It is commonly used to simplify trigonometric expressions and solve equations involving half angles.

## 2. How is the Half Angle Formula for Tangent derived?

The Half Angle Formula for Tangent can be derived using the double angle formula for tangent and some basic algebraic manipulations. The derivation process is often taught in high school or college trigonometry courses.

## 3. What is the purpose of the Half Angle Formula for Tangent?

The Half Angle Formula for Tangent is used to simplify trigonometric expressions and solve equations involving half angles. It can also be used to find the exact values of trigonometric functions for special angles, such as 15 degrees or 75 degrees, which are half of the common angles (30, 45, 60, etc.).

## 4. Can the Half Angle Formula for Tangent be applied to other trigonometric functions?

Yes, the Half Angle Formula for Tangent can be applied to other trigonometric functions such as sine and cosine. However, the formula will be different for each function. For example, the half angle formula for sine is √[(1-cosθ)/2], while the half angle formula for cosine is √[(1+cosθ)/2].

## 5. What is the importance of the Half Angle Formula for Tangent in scientific research?

The Half Angle Formula for Tangent is important in scientific research as it allows for the simplification of complex trigonometric expressions, making calculations and equations easier to work with. It is also used in fields such as engineering, physics, and astronomy to solve problems involving angles and trigonometric functions.

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