1. ## Proving Trigonometric Identities

Hey i need some help

sin^4 (X) - cos^4 (X) = 1 - 2cos^2(X)

2. Originally Posted by oneway1225
Hey i need some help

sin^4 (X) - cos^4 X = 1 - 2cos^2(X)
Note that the left hand side is a difference of two squares: $(\sin^2 x - \cos^2 x) (\sin^2 x + \cos^2 x)$.

3. Hello,

Here is a way of solving this:

$sin^4(x)-cos^4(x)=[sin^2(x)-cos^2(x)][sin^2(x)+cos^2(x)]$ $=sin^2(x)-cos^2(x)=1-cos^2(x)-cos^2(x)=1-2cos^2(x)
$

Have a nice day,
Hush_Hush.

4. Originally Posted by Hush_Hush
Hello,

Here is a way of solving this:

$sin^4(x)-cos^4(x)=[sin^2(x)-cos^2(x)][sin^2(x)+cos^2(x)]$ $=sin^2(x)-cos^2(x)=1-cos^2(x)-cos^2(x)=1-2cos^2(x)
$

Have a nice day,
Hush_Hush.

how did you go from [sin^2(x)-cos^2(x)][sin^2(x)+cos^2(x)] to sin^2(x)-cos^2(x)

5. Originally Posted by oneway1225
how did you go from [sin^2(x)-cos^2(x)][sin^2(x)+cos^2(x)] to sin^2(x)-cos^2(x)
What is always the value of $\sin^2(x)\, +\, \cos^2(x)$?

6. Do you need any more help with this problem?