Trig

• Jun 16th 2008, 01:27 PM
FORK
Trig
Heres the question, a full solution step by step is great as i already know the process to answer a question like this but have trouble doing so.\

$\displaystyle tanx-sinx/tanxsinx = tanxsinx/tanx+sinx$
• Jun 16th 2008, 01:42 PM
Moo
Hi again !

Quote:

Originally Posted by FORK
Heres the question, a full solution step by step is great as i already know the process to answer a question like this but have trouble doing so.\

$\displaystyle tanx-sinx/tanxsinx = tanxsinx/tanx+sinx$

I must guess it's $\displaystyle \frac{\tan x-\sin x}{\tan x \sin x}=\frac{\tan x \sin x}{\tan x+\sin x}$

Please next time, put parenthesis or brackets...
Now, do you have to prove this equation or solve for x ?
• Jun 16th 2008, 02:02 PM
FORK
Quote:

Originally Posted by Moo
Hi again !

I must guess it's $\displaystyle \frac{\tan x-\sin x}{\tan x \sin x}=\frac{\tan x \sin x}{\tan x+\sin x}$

Please next time, put parenthesis or brackets...
Now, do you have to prove this equation or solve for x ?

Sorry, i didnt know how to do that. And that equation must be proved. Thanks.

Oh, and this is the equation i need to prove, i havent done any work to it yet. Help please.
• Jun 16th 2008, 03:42 PM
topsquark
Quote:

Originally Posted by FORK
Heres the question, a full solution step by step is great as i already know the process to answer a question like this but have trouble doing so.\

$\displaystyle tanx-sinx/tanxsinx = tanxsinx/tanx+sinx$

$\displaystyle \frac{tan(x) - sin(x)}{tan(x)~sin(x)} = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$
One way is to start on the left hand side and convert everything to sines and cosines:
$\displaystyle = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)}$

Multiply top and bottom by cos(x):
$\displaystyle = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)} \cdot \frac{cos(x)}{cos(x)}$

$\displaystyle = \frac{\frac{sin(x)}{cos(x)} \cdot cos(x) - sin(x)~cos(x)}{\frac{sin(x)}{cos(x)}~sin(x) \cdot cos(x)}$

$\displaystyle = \frac{sin(x) - sin(x)~cos(x)}{sin^2(x)}$

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

$\displaystyle = \frac{sin(x)(1 - cos(x))}{(1 - cos(x))(1 + cos(x))}$

$\displaystyle = \frac{sin(x)}{1 + cos(x)}$

Now multiply the top and bottom by tan(x):
$\displaystyle = \frac{sin(x)}{1 + cos(x)} \cdot \frac{tan(x)}{tan(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + tan(x)~cos(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$

-Dan
• Jun 16th 2008, 03:50 PM
FORK
Quote:

Originally Posted by topsquark
$\displaystyle \frac{tan(x) - sin(x)}{tan(x)~sin(x)} = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$
One way is to start on the left hand side and convert everything to sines and cosines:
$\displaystyle = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)}$

Multiply top and bottom by cos(x):
$\displaystyle = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)} \cdot \frac{cos(x)}{cos(x)}$

$\displaystyle = \frac{\frac{sin(x)}{cos(x)} \cdot cos(x) - sin(x)~cos(x)}{\frac{sin(x)}{cos(x)}~sin(x) \cdot cos(x)}$

$\displaystyle = \frac{sin(x) - sin(x)~cos(x)}{sin^2(x)}$

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

$\displaystyle = \frac{sin(x)(1 - cos(x))}{(1 - cos(x))(1 + cos(x))}$

$\displaystyle = \frac{sin(x)}{1 + cos(x)}$

Now multiply the top and bottom by tan(x):
$\displaystyle = \frac{sin(x)}{1 + cos(x)} \cdot \frac{tan(x)}{tan(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + tan(x)~cos(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$

-Dan

Wow man, thanks alot, I certainly appreciate it. Ill check this with my teacher tommorow, thanks alot again(Clapping)
• Jun 16th 2008, 03:59 PM
FORK
Quote:

Originally Posted by topsquark
$\displaystyle \frac{tan(x) - sin(x)}{tan(x)~sin(x)} = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$
One way is to start on the left hand side and convert everything to sines and cosines:
$\displaystyle = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)}$

Multiply top and bottom by cos(x):
$\displaystyle = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)} \cdot \frac{cos(x)}{cos(x)}$

$\displaystyle = \frac{\frac{sin(x)}{cos(x)} \cdot cos(x) - sin(x)~cos(x)}{\frac{sin(x)}{cos(x)}~sin(x) \cdot cos(x)}$

$\displaystyle = \frac{sin(x) - sin(x)~cos(x)}{sin^2(x)}$

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

$\displaystyle = \frac{sin(x)(1 - cos(x))}{(1 - cos(x))(1 + cos(x))}$

$\displaystyle = \frac{sin(x)}{1 + cos(x)}$

Now multiply the top and bottom by tan(x):
$\displaystyle = \frac{sin(x)}{1 + cos(x)} \cdot \frac{tan(x)}{tan(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + tan(x)~cos(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$

-Dan

COuple questions though, are you allowed to just multiply them by tan if youre not making the denominator common? And how did you get from your second last step to the last one?
• Jun 16th 2008, 04:02 PM
topsquark
Quote:

Originally Posted by FORK
COuple questions though, are you allowed to just multiply them by tan if youre not making the denominator common? And how did you get from your second last step to the last one?

$\displaystyle \frac{a}{b} \cdot \frac{c}{c} = \frac{ac}{bc}$ (with suitable restrictions on b and c.) When you multiply fractions you do not need common denominators.

Finishing the problem is easy! C'mon, you need to try a little harder.

$\displaystyle \frac{tan(x)~sin(x)}{tan(x) + tan(x)~cos(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + \frac{sin(x)}{cos(x)}~cos(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$

-Dan