# Proving Identities

• Jun 6th 2011, 09:30 AM
starshine84
Proving Identities
I need help with the following question:

Prove the identity.

(sinx + cosx) ^2 = 1 + 2sinx cosx

I know that I have to split the equation into two sides, left hand side and right hand side, and then reduce the most difficult side. However, I'm not sure which side is the most difficult side or how to start reducing.

I know that using the double angle formula sin2x = 2sinx cosx, that I can make the right side 1 + sin2x but I'm not sure if that helps at all.

Help??
• Jun 6th 2011, 09:34 AM
Ackbeet
Why not just expand out the LHS, and use a familiar identity to simplify? What do you get?
• Jun 6th 2011, 09:39 AM
starshine84
Oh!!

Is this right?

(sinx+cosx)^2
= (sinx + cosx) (sinx + cosx)
= sin^2x + sinx cosx + cosx sinx + cos^2x
= sin^2x + cos^2x + 2sinx cosx
= 1 + 2sinx cosx

Which then equals the right side...
• Jun 6th 2011, 09:40 AM
e^(i*pi)
Quote:

Originally Posted by starshine84
Oh!!

Is this right?

(sinx+cosx)^2
= (sinx + cosx) (sinx + cosx)
= sin^2x + sinx cosx + cosx sinx + cos^2x
= sin^2x + cos^2x + 2sinx cosx
= 1 + 2sinx cosx

Which then equals the right side...

Yes
• Jun 6th 2011, 09:42 AM
starshine84
Awesome, thank you so much!! :)
• Jun 6th 2011, 09:43 AM
Ackbeet
You're welcome for my contribution!
• Jun 6th 2011, 09:50 AM
e^(i*pi)
Quote:

Originally Posted by Ackbeet
You're welcome for my contribution!

And I did bugger all, I really don't deserve that thanks
• Jun 6th 2011, 09:52 AM
Ackbeet
Quote:

Originally Posted by e^(i*pi)
And I did bugger all, I really don't deserve that thanks

I disagree: confirming that the OP'er's proof is correct is useful.