Thread: 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??

Why not just expand out the LHS, and use a familiar identity to simplify? What do you get?

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...

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

Awesome, thank you so much!!

You're welcome for my contribution!

Originally Posted by Ackbeet

You're welcome for my contribution!

And I did bugger all, I really don't deserve that thanks

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.