Author Topic: Evaluate cos(pi/8) (Read 12299 times) Tweet Share

tram · « **Reply #15 on:** April 17, 2010, 11:16:44 pm »

dammit, TT you have e-books of everything, do you have essentials???

jasopan · « **Reply #16 on:** April 18, 2010, 05:28:54 pm »

Hey can someone re-explain the
$tan ( \frac {\pi}{8})$

And one more;
Show that: $sin3x=3sinx-4sin^3x$

tA

Yitzi_K · « **Reply #17 on:** April 18, 2010, 06:01:02 pm »

Quote from: jasopan on April 18, 2010, 05:28:54 pm

And one more;
Show that: $sin3x=3sinx-4sin^3x$

tA

$sin(3x) = sin(2x+x)$

$sin(2x+x) = sin(2x)cos(x) + cos(2x)sin(x)$

$=2(sin(x)cos(x))(cos(x)) + (1 - 2sin^2(x))(sin(x))$

$=2sin(x)cos^2(x) + sin(x) - 2sin^2(x)sin(x)$

$=2(sin(x))(1-sin^2(x)) + sin(x) - 2sin^2(x)sin(x)$

$=2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x)$

$=3sin(x)-4sin^3(x)$ as required.

brightsky · « **Reply #18 on:** April 18, 2010, 06:12:23 pm »

$\tan\left(\frac{\pi}{8}\right) = \tan \left(\frac{\frac{\pi}{4}}{2}\right)$

Use the identity $\tan \left(\frac{x}{2}\right) = \frac{1 - \cos(x)}{\sin(x)}$ .

The rest is trivial.

jasopan · « **Reply #19 on:** April 18, 2010, 06:33:11 pm »

^ Whoa, my book doesn't even have that identity ...

Gotta google some now $:-\$

@Yitzi_K yes! got it! thanks alotalotalotalot

Another one since everyone's so helpful

Simplify $\sqrt{2-\sqrt{2}$ divide $\sqrt{2+ \sqrt{2}$

brightsky · « **Reply #20 on:** April 18, 2010, 06:41:29 pm »

You can derive the formula from this image:

http://upload.wikimedia.org/wikipedia/commons/2/21/Weierstrass_substitution.png

Look at $\triangle ACE$ , then use $\tan(x) = \frac{\text{Opposite}}{\text{Hypotenuse}}$ as normal.

TrueTears · « **Reply #21 on:** April 18, 2010, 06:45:35 pm »

Quote from: tram on April 17, 2010, 11:16:44 pm

dammit, TT you have e-books of everything, do you have essentials???

yeah i do...

brightsky · « **Reply #22 on:** April 18, 2010, 07:02:57 pm »

$\frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2 + \sqrt{2}}}$

$\implies \sqrt{\frac{2 - \sqrt{2}}{2 + \sqrt{2}}}$

$\implies \sqrt{\frac{(\sqrt{2})^2 - \sqrt{2}}{(\sqrt{2})^2 + \sqrt{2}}}$

Let $\sqrt{2} = x$ .

$\implies \sqrt{\frac{x^2 - x}{x^2 + x}}$

$\implies \sqrt{\frac{x(x - 1)}{x(x + 1)}$

$\implies \sqrt{\frac{x - 1}{x + 1}$

So:

$\implies \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}$

$\implies \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1}}$

$\implies \sqrt{\frac{(\sqrt{2} - 1)^2}{(\sqrt{2})^2 - 1^2}}$

$\implies \sqrt{\frac{(\sqrt{2} - 1)^2}{2 - 1}}$

$\implies \sqrt{(\sqrt{2} - 1)^2}$

$\implies \sqrt{2} - 1$

jasopan · « **Reply #23 on:** April 18, 2010, 07:06:14 pm »

^ Holy crap. Is it that hard?
I tried to use tanx=sinx/cosx to get tanx as the book didn't have the right formula and thats what i got stuck on :S

Thanks again brightsky, this is gold

superflya · « **Reply #24 on:** April 18, 2010, 07:17:54 pm »

or u could use the double angle formula

$\tan (\frac{\pi}{4})=\frac{2\tan \frac{\pi}{8}}{1-\tan ^{2}(\frac{\pi}{8})}$

let $t=\tan (\frac{\pi}{8})$

$\Rightarrow \tan (\frac{\pi}{4})=\frac{2t}{1-t^{2}}$ and note $\tan (\frac{\pi}{4})$

rearrange to get a quadratic $\Rightarrow t^{2}+2t-1=0$

solve and u get $t=-1\pm \sqrt{2}$

as $\tan (\frac{\pi}{8})> 0$

$\therefore \tan (\frac{\pi}{8})=\sqrt{2}-1$

jasopan · « **Reply #25 on:** April 18, 2010, 09:01:45 pm »

What does it mean if they go 'if x is an acute angle' ? does that mean its only in the first quadrant ? so all sin cos tan are positive?

And just to confirm if domain is pi > x > pi/2, then domain of x/2 is (pi/4, pi/2)

brightsky · « **Reply #26 on:** April 18, 2010, 09:03:42 pm »

Yep. Acute means $0^{\circ} < \theta < 90^{\circ}$ for angle $\theta$ , hence in the first quadrant.

Yes for your second question as well (I'm pretty sure).

superflya · « **Reply #27 on:** April 18, 2010, 09:04:49 pm »

jasopan · « **Reply #28 on:** April 18, 2010, 09:16:08 pm »

for $4\pi > x > 2\pi$

Is the domain written;

$2\pi > x > \pi$ or $2\pi > \frac {x}{2} > \pi$

m@tty · « **Reply #29 on:** April 18, 2010, 09:22:59 pm »

The second one.

$4\pi>x>2\pi$ is two statements combined.

$x<4\pi$ and $x>2\pi$

So $\frac{x}{2}<2\pi$ and $\frac{x}{2}>\pi$ .

Or $2\pi>\frac{x}{2}>\pi$ .

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