# trigonometry, inequation

• Oct 27th 2006, 12:02 AM
scorpion007
trigonometry, inequation
This is the question from memory, i may have potentially made a mistake, but i do believe it is correct. This should all be done without a calculator.

For the equation $\displaystyle y = \arctan(x-1) - a\tan(\frac{\pi}{8})$, find the minimum value for 'a' where y > 0 for all of x. Leave in exact answer form.

Note, on a previous question i was asked to show (without a calculator) that $\displaystyle \tan(\frac{\pi}{8}) = \sqrt{2} - 1$, so that may be relevant.

In fact, i would appreciate if you could explain the second question too.

Thanks
• Oct 27th 2006, 04:16 AM
ThePerfectHacker
Quote:

Originally Posted by scorpion007
This is the question from memory, i may have potentially made a mistake, but i do believe it is correct. This should all be done without a calculator.

For the equation $\displaystyle y = \arctan(x-1) - a\tan(\frac{\pi}{8})$, find the minimum value for 'a' where y > 0 for all of x. Leave in exact answer form.

Note, on a previous question i was asked to show (without a calculator) that $\displaystyle \tan(\frac{\pi}{8}) = \sqrt{2} - 1$, so that may be relevant.

In fact, i would appreciate if you could explain the second question too.

Thanks

Let $\displaystyle b=a\tan (\pi/8)$
You want the range of the strictly monotomic function (an increasing function) of $\displaystyle \tan^{-1}(x-1)+b>0$

The regular arc-tangent function $\displaystyle -\pi/2<y<\pi/2$ has this range. If you add $\displaystyle \pi/2$ you have, $\displaystyle 0<y+\pi/2<\pi$
Thus,
$\displaystyle \tan^{-1}(x-1)+\pi/2$
That means,
$\displaystyle b=\pi/2=a(\sqrt{2}-1)$
Thus,
$\displaystyle a=\frac{\pi}{2(\sqrt{2}-1)}$
• Oct 27th 2006, 11:19 PM
scorpion007
could you help me prove that
$\displaystyle \tan(\frac{\pi}{8}) = \sqrt{2} - 1$ ?
• Oct 28th 2006, 12:16 AM
CaptainBlack
Quote:

Originally Posted by scorpion007
could you help me prove that
$\displaystyle \tan(\frac{\pi}{8}) = \sqrt{2} - 1$ ?

You need to know that $\displaystyle \tan(\pi/4)=1,\ \sin(\pi/4)=\cos(\pi/4)=1/\sqrt{2}$, then use the 1/2-angle formula for 1st quadrant angles:

$\displaystyle \tan(A/2) = \frac{\sqrt{1-\cos(A)}}{\sqrt{1+\cos(A)}}=\frac{\sin(A)}{1+\cos( A)}$

Put $\displaystyle A=\pi/4$ and substitute in the know values of the trig functions for this angle, and then rationalise.

RonL
• Oct 28th 2006, 12:34 AM
scorpion007
hmm, that method works wonderfully.
Unfortunately, i never learned that formula (its not in my textbook). Is it anything like this one?

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

Is it possible to use that one to solve it?
• Oct 29th 2006, 12:01 AM
earboth
Quote:

Originally Posted by scorpion007
...Is it anything like this one?

$\displaystyle \tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$
Is it possible to use that one to solve it?

No,

that's the double-angle formula. CaptBlack used the half-angle formula:
$\displaystyle \tan(A/2) = \frac{\sqrt{1-\cos(A)}}{\sqrt{1+\cos(A)}}=\frac{\sin(A)}{1+\cos( A)}$=$\displaystyle \frac{1-\cos(A)}{\sin(A)}$

Your formula can be completed to:

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

(If you solve your equation for tan(x) you will probably come out with CaptBlack's formula)

EB