Thread: TEST TOMORROW: Need help simplifying Trig Expressions

1. 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

2. 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.

4. Re: TEST TOMORROW: Need help simplifying Trig Expressions

Hello, srob22!

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

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

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

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

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

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

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

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