Identities

**anna_sims** · October 26th 2006, 05:54 PM

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.

**ThePerfectHacker** · October 26th 2006, 06:11 PM

Originally Posted by anna_sims

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

Express everything in terms of sine and cosine.

$\frac{\frac{1}{\sin x}-\sin x}{\frac{\cos^2 x}{\sin^2 x}}$
Add the numerator,
$\frac{\frac{1-\sin^2 x}{\sin x}}{\frac{\cos^2 x}{\sin^2 x}}$
Pythagoren identities,
$\frac{\frac{\cos^2 x}{\sin x}}{\frac{\cos^2 x}{\sin^2 x}}$
Fractions means divison thus,
$\frac{\cos^2 x}{\sin x}\div \frac{\cos^2 x}{\sin^2 x}$
Thus,
$\frac{\cos^2 x}{\sin x}\cdot \frac{\sin^2 x}{\cos^2 x}$
Cancel,
$\frac{\sin x}{1}=\sin x$

**Soroban** · October 26th 2006, 10:49 PM

Hello, Anna!

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

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

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

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

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

**earboth** · October 26th 2006, 11:12 PM

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:

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

EB

**Soroban** · October 27th 2006, 10:45 AM

Nice job, Earboth!

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

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

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

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

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

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

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