# Proving a Couple of Trig Identities

• Nov 7th 2011, 03:52 PM
Dante
Proving a Couple of Trig Identities
sin^2x - tan^2x = -sin^2x tan^2x

I kept working on tan on both sides, but cannot seem to get them to equal.

also this one

sin^4x + 2sin^2x.cos^2x + cos^4x = 1

I went :
sin^2x.sin^2x + 2sin^2x.cos^2x + cos^2x.cos^2x

I tried to make middle one sin2x didn't seem to get me anywhere.
Also tried changing one of the left sin^2x into 1-cos^2x and
one of the right cos^2x into 1-sin^2x

but they didn't cancel each other out.
• Nov 7th 2011, 04:03 PM
skeeter
re: Proving a Couple of Trig Identities
Quote:

Originally Posted by Dante
sin^2x - tan^2x = -sin^2x tan^2x

I kept working on tan on both sides, but cannot seem to get them to equal.

also this one

sin^4x + 2sin^2x.cos^2x + cos^4x = 1

I went :
sin^2x.sin^2x + 2sin^2x.cos^2x + cos^2x.cos^2x

I tried to make middle one sin2x didn't seem to get me anywhere.
Also tried changing one of the left sin^2x into 1-cos^2x and
one of the right cos^2x into 1-sin^2x

but they didn't cancel each other out.

$\sin^2{x} - \tan^2{x} =$

$\sin^2{x}\left(1 - \frac{1}{\cos^2{x}}\right) =$

$\sin^2{x}\left(1 - \sec^2{x}\right) =$

$\sin^2{x}\left(-\tan^2{x}\right) = -\sin^2{x}\tan^2{x}$

$\sin^4{x} +2\sin^2{x}\cos^2{x} + \cos^4{x} =$

$\left(\sin^2{x} + \cos^2{x}\right)^2 = \, ?$
• Nov 7th 2011, 04:08 PM
Plato
re: Proving a Couple of Trig Identities
Quote:

Originally Posted by Dante
[B]sin^2x - tan^2x = -sin^2x tan^2x[/B

$\sin^2(x)-\tan^2(x)=\frac{\sin^2(x)(\cos^2(x)-1)}{\cos^2(x)}$
• Nov 7th 2011, 06:11 PM
Dante
re: Proving a Couple of Trig Identities
Quote:

Originally Posted by skeeter
$\sin^2{x} - \tan^2{x} =$

$\sin^2{x}\left(1 - \frac{1}{\cos^2{x}}\right) =$

$\sin^2{x}\left(1 - \sec^2{x}\right) =$

$\sin^2{x}\left(-\tan^2{x}\right) = -\sin^2{x}\tan^2{x}$

$\sin^4{x} +2\sin^2{x}\cos^2{x} + \cos^4{x} =$

$\left(\sin^2{x} + \cos^2{x}\right)^2 = \, ?$

Woah, that is intense ... I some what understand these steps but it's a lot more advanced than the identities which we are currently being taught.

Is it possible if you can explain to me how you went from

http://latex.codecogs.com/png.latex?...5E4%7Bx%7D%20=

to

http://latex.codecogs.com/png.latex?...20=%20%5C,%20?

Thanks :)
• Nov 8th 2011, 03:11 AM
sbhatnagar
re: Proving a Couple of Trig Identities
Quote:

Originally Posted by Dante
Is it possible if you can explain to me how you went from

http://latex.codecogs.com/png.latex?...5E4%7Bx%7D%20=

to

http://latex.codecogs.com/png.latex?...20=%20%5C,%20?

Thanks :)

$a^2+b^2+2ab=(a+b)^2$