# TEST TOMORROW: Need help simplifying Trig Expressions

• Apr 7th 2013, 06:14 PM
srob22
TEST TOMORROW: Need help simplifying Trig Expressions
Sin x + 1- cos x
1-cos x sin x

cos x sec2x-cos x
sin x tan x + cos x

How would you simplify these two equations? (step-by-step)

Thank you so much for help,
Sean (Nod)
• Apr 7th 2013, 06:33 PM
zhandele
Re: TEST TOMORROW: Need help simplifying Trig Expressions
Check the pdf, I hope it helps.

In your first expression (not equation, expression), I'm not sure I understood what you wrote. The second one I'm pretty sure of.
• Apr 7th 2013, 07:38 PM
ibdutt
Re: TEST TOMORROW: Need help simplifying Trig Expressions
• Apr 8th 2013, 08:30 AM
Soroban
Re: TEST TOMORROW: Need help simplifying Trig Expressions
Hello, srob22!

Quote:

$\displaystyle \text{Simplify: }\:\frac{\sin x}{1-\cos x} + \frac{1-\cos x}{\sin x}$

Multiply the first fraction by $\displaystyle \frac{1+\cos x}{1+\cos x}:$

. . $\displaystyle \frac{\sin x}{1-\cos x}\cdot{\color{blue}\frac{1+\cos x}{1+\cos x}} + \frac{1-\cos x}{\sin x} \;=\;\frac{\sin x(1+\cos x)}{1-\cos^2\!x} + \frac{1-\cos x}{\sin x}$

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

. . $\displaystyle =\;\frac{1+\cos x + 1 - \cos x}{\sin x} \;=\;\frac{2}{\sin x} \;=\;2\csc x$

Quote:

$\displaystyle \text{Simplify: }\:\frac{\cos x\sec^2\!x - \cos x}{\sin x\tan x + \cos x}$

We have: .$\displaystyle \dfrac{\cos x(\sec^2\!x - 1)}{\sin x\!\cdot\!\frac{\sin x}{\cos x} + \cos x} \;=\;\dfrac{\cos x\tan^2\!\!x}{\frac{\sin^2\!x}{\cos x} + \cos x} \;=\;\dfrac{\cos x\frac{\sin^2\!x}{\cos^2\!x}}{\frac{\sin^2\!x+\cos ^2\!x}{\cos x}}$

. . . . . . . $\displaystyle =\;\dfrac{\frac{\sin^2\!x}{\cos x}}{\frac{1}{\cos x}} \;=\; \dfrac{\sin^2\!x}{\cos x}\cdot\dfrac{\cos x}{1} \;=\;\sin^2\!x$