# How do you verify the identity (sintheta+costheta)^2-1=sin2theta?

Jan 22, 2017

See proof below

#### Explanation:

We need

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

$\sin 2 \theta = 2 \sin \theta \cos \theta$

Therefore,

$L H S = {\left(\cos \theta + \sin \theta\right)}^{2} - 1$

$= {\cos}^{2} \theta + {\sin}^{2} \theta + 2 \sin \theta \cos \theta - 1$

$= 1 + 2 \sin \theta \cos \theta - 1$

$= 2 \sin \theta \cos \theta$

$= \sin 2 \theta$

$= R H S$

$Q E D$

Jan 22, 2017

$\cancel{1} + 2 \sin \theta \cos \theta \cancel{-} 1 = 2 \sin \theta \cos \theta$
and the identity is verified.

#### Explanation:

Since

$\sin 2 \theta = 2 \sin \theta \cos \theta$,

you can substitute it in the given expression and expand the square of the bynomial:

$\textcolor{red}{{\sin}^{2} \theta + {\cos}^{2} \theta + 2 \sin \theta \cos \theta} - 1 = \textcolor{g r e e n}{2 \sin \theta \cos \theta}$

Then, since

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$, you get:

$\cancel{1} + 2 \sin \theta \cos \theta \cancel{-} 1 = 2 \sin \theta \cos \theta$

and the identity is verified.