- #1

sleepingMantis

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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.

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

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

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!

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 Statement1. 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!

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