I'm stumped

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• Sep 9th 2007, 12:49 PM
Freaky-Person
I'm stumped
A few question on my trig assignment... It's not going all that well.. I worked hard on the LaTex...

1. Simplify: $\displaystyle \frac{\tan(\frac{\pi}{4} - x) - \tan(\frac{\pi}{4} + x)}{\tan x}$

2. Solve for x, $\displaystyle 0 \leq x \leq 2\pi$

a) $\displaystyle \sin 2x + \cos x = 0$

b) $\displaystyle 3\tan x = \tan 2x$

THANKSIES!!
• Sep 9th 2007, 01:10 PM
Jhevon
Quote:

Originally Posted by Freaky-Person
A few question on my trig assignment... It's not going all that well.. I worked hard on the LaTex...

1. Simplify: $\displaystyle \frac{\tan(\frac{\pi}{4} - x) - \tan(\frac{\pi}{4} + x)}{\tan x}$

Hint: Use the addition formulas for tangent

recall: $\displaystyle \tan ( \alpha + \beta ) = \frac {\tan \alpha + \tan \beta }{1 - \tan \alpha \tan \beta}$ and $\displaystyle \tan ( \alpha - \beta ) = \frac {\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

Quote:

2. Solve for x, $\displaystyle 0 \leq x \leq 2\pi$

a) $\displaystyle \sin 2x + \cos x = 0$
use the double angle formula for sine

$\displaystyle \sin 2x + \cos x = 0$

$\displaystyle \Rightarrow 2 \sin x \cos x + \cos x = 0$

$\displaystyle \Rightarrow \cos x ( 2 \sin x + 1 ) = 0$

Now continue

Quote:

b) $\displaystyle 3\tan x = \tan 2x$
use the double angle formula for tangent

$\displaystyle 3 \tan x = \tan 2x$

$\displaystyle \Rightarrow 3 \tan x = \frac {2 \tan x }{1 - \tan^2 x}$

Now continue.

(alternatively, you could change everything to be in terms of sine and cosine here, but I think that would be more trouble than it's worth -- maybe not)

By the way, good job with the LaTex!
• Sep 9th 2007, 02:09 PM
Freaky-Person
Quote:

Originally Posted by Jhevon
Hint: Use the addition formulas for tangent

recall: $\displaystyle \tan ( \alpha + \beta ) = \frac {\tan \alpha + \tan \beta }{1 - \tan \alpha \tan \beta}$ and $\displaystyle \tan ( \alpha - \beta ) = \frac {\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

use the double angle formula for sine

$\displaystyle \sin 2x + \cos x = 0$

$\displaystyle \Rightarrow 2 \sin x \cos x + \cos x = 0$

$\displaystyle \Rightarrow \cos x ( 2 \sin x + 1 ) = 0$

Now continue

use the double angle formula for tangent

$\displaystyle 3 \tan x = \tan 2x$

$\displaystyle \Rightarrow 3 \tan x = \frac {2 \tan x }{1 - \tan^2 x}$

Now continue.

That's great and all, but I have a sheet filled to the brim with all that. I got to those steps already, but I don't know where to go from there :(
• Sep 9th 2007, 02:38 PM
Jhevon
Quote:

Originally Posted by Freaky-Person
A few question on my trig assignment... It's not going all that well.. I worked hard on the LaTex...

1. Simplify: $\displaystyle \frac{\tan(\frac{\pi}{4} - x) - \tan(\frac{\pi}{4} + x)}{\tan x}$

what have you tried here? you must have gotten farther than I showed you. if nothing else you could follow your nose, add the fractions and attempt to simplify. show me where you got stuck

Quote:

2. Solve for x, $\displaystyle 0 \leq x \leq 2\pi$

a) $\displaystyle \sin 2x + \cos x = 0$
picking up where i left off.

$\displaystyle \cos x ( 2 \sin x + 1) = 0$

$\displaystyle \Rightarrow \cos x = 0 \mbox { or } \sin x = - \frac {1}{2}$

$\displaystyle \Rightarrow x = \frac {\pi}{2}, \frac {3 \pi}{2}, \frac {7 \pi}{6}, \mbox { and } \frac {11 \pi}{6}$ for $\displaystyle 0 \leq x \leq 2 \pi$

Now check all those solutions to make sure they work

Quote:

b) $\displaystyle 3\tan x = \tan 2x$
again, picking up where i left off

$\displaystyle 3 \tan x = \frac {2 \tan x}{1 - \tan^2 x}$

$\displaystyle \Rightarrow 2 \tan x = 3 \tan x - 3 \tan^3 x$ ............i cross-multiplied

$\displaystyle \Rightarrow 3 \tan^3 x - \tan x = 0$

$\displaystyle \Rightarrow \tan x \left( 3 \tan^2 x - 1 \right) = 0$ .................i factored out the common tan(x)

$\displaystyle \Rightarrow \tan x = 0 \mbox { or } \tan x = \frac {1}{\sqrt {3}}$

$\displaystyle \Rightarrow x = 0, \pi, 2 \pi, \frac {\pi}{6} \mbox { and } \frac {7 \pi}{6}$ for $\displaystyle 0 \leq x \leq 2 \pi$

Check those solutions as well.