# Identities

• Oct 26th 2006, 05:54 PM
anna_sims
Identities
Could anybody give me some help with these identities that I need to complete for homework? I need to have them correct with all work for tomorrow, and they need to be proofs. If I could at least get help with starting them, that would be great. Thanks.

1) (csc X - sin X)/(cot^2 X) = sin X

2) (tan X - sin X)/(sin^3 X) = (sec X)/(1+cos X)

3) ((cos X)/(1- sin X)+(1+ sin X)/(cos X)) = 2(sec X + tan X)

Again, any help will be appreciated.
• Oct 26th 2006, 06:11 PM
ThePerfectHacker
Quote:

Originally Posted by anna_sims

1) (csc X - sin X)/(cot^2 X) = sin X

Express everything in terms of sine and cosine.

$\displaystyle \frac{\frac{1}{\sin x}-\sin x}{\frac{\cos^2 x}{\sin^2 x}}$
$\displaystyle \frac{\frac{1-\sin^2 x}{\sin x}}{\frac{\cos^2 x}{\sin^2 x}}$
Pythagoren identities,
$\displaystyle \frac{\frac{\cos^2 x}{\sin x}}{\frac{\cos^2 x}{\sin^2 x}}$
Fractions means divison thus,
$\displaystyle \frac{\cos^2 x}{\sin x}\div \frac{\cos^2 x}{\sin^2 x}$
Thus,
$\displaystyle \frac{\cos^2 x}{\sin x}\cdot \frac{\sin^2 x}{\cos^2 x}$
Cancel,
$\displaystyle \frac{\sin x}{1}=\sin x$
• Oct 26th 2006, 10:49 PM
Soroban
Hello, Anna!

Quote:

$\displaystyle 3)\;\frac{\cos x}{1- \sin x} + \frac{1 + \sin x}{\cos x} \:=\:2(\sec x + \tan x)$

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

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

The problem becomes: .$\displaystyle \frac{1 + \sin x}{\cos x} + \frac{1 + \sin x}{\cos x} \:=\:2\left(\frac{1 + \sin x}{\cos x}\right)$

. . $\displaystyle = \:2\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right) \:=\:2(\sec x + \tan x)$

• Oct 26th 2006, 11:12 PM
earboth
Quote:

Originally Posted by anna_sims
Could anybody give me some help with ...
2) (tan X - sin X)/(sin^3 X) = (sec X)/(1+cos X)
...

Hi,

I'll take the RHS of the equation and show that it is equal to the LHS:

$\displaystyle \frac{\tan(x)-\sin(x)}{\sin^3(x)}=\frac{\sec(x)}{1+\cos(x)}$
$\displaystyle =\frac{\frac{1}{\cos(x)}}{1+\cos(x)}\cdot \frac{1-\cos(x)}{1-\cos(x)}$
$\displaystyle =\frac{\frac{1}{\cos(x)}-1}{1-\cos^2(x)}\cdot \frac{\sin(x)}{\sin(x)}$
$\displaystyle =\frac{\frac{\sin(x)}{\cos(x)}-\sin(x)}{\sin^3(x)}$
$\displaystyle =\frac{\tan(x)-\sin(x)}{\sin^3(x)}$

EB
• Oct 27th 2006, 10:45 AM
Soroban
Nice job, Earboth!

I tried going from left to right . . . not worth the effort!

Quote:

$\displaystyle 2)\;\frac{\tan x - \sin x}{\sin^3x} \:= \:\frac{\sec x}{1 + \cos x}$

We have: .$\displaystyle \frac{\frac{\sin x}{\cos x} - \sin x}{\sin^3x}\;=\;\frac{\sin x\left(\frac{1}{\cos x} - 1\right)} {\sin^3x}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}:\;\;\frac{\cos x}{\cos x} \cdot \frac{\sin x\left(\frac{1}{\cos x} - 1\right)}{\sin^3x}\;=\;\frac{\sin x(1 - \cos x)}{\sin^3x\cos x} \;=\;\frac{1-\cos x}{\sin^2x\cos x}$

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

Therefore: .$\displaystyle \frac{1}{1+\cos x}\cdot\frac{1}{\cos x} \;=\;\frac{\sec x}{1 + \cos x}$