- #1
sleepingMantis
- 25
- 5
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}.##
##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 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: