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.

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.

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

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

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

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

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

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

re: Proving a Couple of Trig Identities

Quote:

Originally Posted by

**Dante** [B]sin^2x - tan^2x = -sin^2x tan^2x[/B

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

re: Proving a Couple of Trig Identities

Quote:

Originally Posted by

**skeeter** $\displaystyle \sin^2{x} - \tan^2{x} =$

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

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

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

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

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

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 :)

re: Proving a Couple of Trig Identities

Quote:

Originally Posted by

**Dante**

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