How do I prove this identity?Earn 10 points?

Question

(sinx+cosx)^2=1+sin2x

Answer 1

L.H.S.
= (sinx+cosx)^2
= (sinx)^2 + 2(sinx)(cosx) + (cosx)^2 ----- Using (a+b)^2 = a^2 + 2ab + b^2
= (sinx)^2 + (cosx)^2 + 2(sinx)(cosx)
= 1 + 2(sinx)(cosx) ----- Using (sinx)^2 + (cosx)^2 = 1 [Pythagorean identity]
= 1 + sin2x ---- Using sin2x = 2(sinx)(cosx) [Double-angle formula]
= R.H.S. (proven)

Answer 2

Okay.. for this problem, you need to know a couple of formulas. You will have to know that sin^2(x)+cos^2(x)=1 and sin2x=2sin(x)cos(x). Okay.. you first use the FOIL method for the (sinx+cosx)^2. You will get sin^2(x)+2sin(x)cos(x)+cos^2(x). For the sin^2(x) and cos^2(x), I have told you above that if you add those two, you get 1 right? so you have 1+ 2sin(x)cos(x) left right? Okay for the 1+sin2(x).. I told you that sin2(x) equals 2sin(x)cos(x). So both sides of the equation gives you 1+ 2sin(x)cos(x). Yes?? If you have any other questions about this.. email me at [e-mail] good luck.. And yea... I hope you got help from this.

Answer 3

(sin x + cox x)^2 = (sin x)^2 + 2sin x cos x + (cos x )^2 (see note 1 below)

= 1 + 2 sinx cosx = 1 + sin 2x (see note 2 below)

Note 1: Pythag. Identity: (sin x)^2 + (cos x)^2 = 1

Note 2: Double-angle identity: sin 2x = 2sinxcosx