Deriving a trigonometric identity

[HHermans]

For a homework assignment I'm supposed to prove that sin(x)^2+cos(x)^2=1, using only the following identities (along with algebraic operations):

sin(-x)=-sin(x)
cos(-x)=cos(x)
cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
sin(x+y)=sin(x)cos(y)+cos(x)sin(y)

I can't figure this out, because as far as I know the identity can only be derived from the Pythagorean Theorem.

Any help would be much appreciated.

 

[Hurkyl]

Well, remember that identities such as

$$\cos (\theta + \varphi) = \cos \theta \cos \varphi - \sin \theta \sin \varphi$$

are true no matter what the arguments are: you can plug anything you want in for $\theta$ and $\varphi$.

 

[HHermans]

Well, remember that identities such as

$$\cos (\theta + \varphi) = \cos \theta \cos \varphi - \sin \theta \sin \varphi$$

are true no matter what the arguments are: you can plug anything you want in for $\theta$ and $\varphi$.

I've tried this a few times, and I got nowhere. Is there a specific direction to go?

 

[danago]

$$\cos (a \pm b) = \cos (a)\cos (b) \mp \sin (a)\sin (b)$$

try using some value for a and b that will make the left hand side equal 1.

 

[HHermans]

Thanks very much for everyone's help, I now understand this. cos(x + y) can be written as cos(x + -x), or cos(0).

 

[Hurkyl]

$$\cos (a \pm b) = \cos (a)\cos (b) \mp \sin (a)\sin (b)$$

try using some value for a and b that will make the left hand side equal 1.

Heh, I was thinking more along the lines of looking for an a and b that makes cos^2 and sin^2 appear on the r.h.s. Same thing either way.