Trig identities

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February 20th 2008, 02:04 AM
gracey

Trig identities

1 / cos^4ф - 1 / cos^2ф = tan^4ф + tan^2ф
February 20th 2008, 02:25 AM
mr fantastic

Quote:

Originally Posted by gracey

cos^4 ф - sin^4 ф + 1 = 2cos^2ф

Mr F says: Note that $\cos^4 \phi - \sin^4 \phi = (\cos^2 \phi + \sin^2 \phi)(\cos^2 \phi - \sin^2 \phi) = \cos^2 \phi - \sin^2 \phi$
[snip]

..
February 20th 2008, 02:36 AM
mr fantastic

Quote:

Originally Posted by gracey

[snip]
1 / cos^4ф - 1 / cos^2ф = tan^4ф + tan^2ф

Right Hand Side $\, =\tan^4 \phi + \tan^2 \phi = \frac{\sin^4 \phi}{\cos^4 \phi} + \frac{\sin^2 \phi}{\cos^2 \phi}$ .

Now note that:

$\frac{\sin^4 \phi}{\cos^4 \phi} = \frac{(\sin^2 \phi)^2}{\cos^4 \phi} = \frac{(1 - \cos^2 \phi)^2}{\cos^4 \phi} = \frac{1 - 2 \cos^2 \phi + \cos^4 \phi}{\cos^4 \phi} = \frac{1}{\cos^4 \phi} - \frac{2}{\cos^2 \phi} + 1$ .

$\frac{\sin^2 \phi}{\cos^2 \phi} = \frac{1 - \cos^2 \phi}{\cos^2 \phi} = \frac{1}{\cos^2 \phi} - 1$ .

Therefore the right hand side becomes .......
March 6th 2008, 11:36 PM
mrbuttersworth

2 Attachment(s)

I did this one differently. I went like...

left side
step one multiply $\frac{1}{cos^2x}$ by $\frac{cos^2x}{cos^2x}$ and then add because we've achieved a common denominator.
Then using the pythagorean identity, we can change $1-cos^2x$ into $sin^2x$

Right side
convert the tan's into their proper sin and cosine values.
multiply ${sin^2x}{cos^2x}$ by $\frac{cos^2x}{cos^2x}$ and add because we've achieved a common denominator.
factor out a $sin^2x$
use the pythagorean identity to turn $sin^2x+cos^2x$ into 1
finally we end up with $\frac{sin^2x}{cos^4x}$

L=R
March 6th 2008, 11:37 PM
mrbuttersworth

Also, I would like to know how to write equations neater and more aesthetically pleasing than the method i'm using now. How do you guys do it?
March 6th 2008, 11:59 PM
mr fantastic

Quote:

Originally Posted by mrbuttersworth

Also, I would like to know how to write equations neater and more aesthetically pleasing than the method i'm using now. How do you guys do it?

Read this.