# How can verify this equation is an identity? Sin(α+ß)Sin(α-ß)=Sin²α-Sin²ß

Mar 20, 2018

See below.

#### Explanation:

Use the addition and subtraction formulas for sine.

$\sin \left(A + B\right) = \sin A \cos B + \sin B \cos A$
$\sin \left(A - B\right) = \sin A \cos B - \sin B \cos A$

Now

$\left(\sin \alpha \cos \beta + \sin \beta \cos \alpha\right) \left(\sin \alpha \cos \beta - \sin \beta \cos \alpha\right) = {\sin}^{2} \alpha - {\sin}^{2} \beta$

On the left it's clearly a difference of squares.

${\sin}^{2} \alpha {\cos}^{2} \beta - {\sin}^{2} \beta {\cos}^{2} \alpha = {\sin}^{2} \alpha - {\sin}^{2} \beta$

${\sin}^{2} \alpha \left(1 - {\sin}^{2} \beta\right) - {\sin}^{2} \beta \left(1 - {\sin}^{2} \alpha\right) = {\sin}^{2} \alpha - {\sin}^{2} \beta$

${\sin}^{2} \alpha - {\sin}^{2} \alpha {\sin}^{2} \beta - {\sin}^{2} \beta + {\sin}^{2} \alpha {\sin}^{2} \beta = {\sin}^{2} \alpha - {\sin}^{2} \beta$

${\sin}^{2} \alpha - {\sin}^{2} \beta = {\sin}^{2} \alpha - {\sin}^{2} \beta$

As required.

Hopefully this helps!

Mar 20, 2018

The proof of the identity is given below.

#### Explanation:

We know that,

color(red)((1)sin(x+y)=sinxcosy+cosxsiny

color(red)((2)sin(x-y)=sinxcosy-cosxsiny

Using color(red)((1) and (2) ,we get

$L H S = \sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right)$

$= \left(\sin \alpha \cos \beta + \cos \alpha \sin \beta\right) \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right)$

$= {\sin}^{2} \alpha \textcolor{red}{{\cos}^{2} \beta} - \textcolor{b l u e}{{\cos}^{2} \alpha} {\sin}^{2} \beta$

$= {\sin}^{2} \alpha \textcolor{red}{\left(1 - {\sin}^{2} \beta\right)} - \textcolor{b l u e}{\left(1 - {\sin}^{2} \alpha\right)} {\sin}^{2} \beta$

$= {\sin}^{2} \alpha - \cancel{{\sin}^{2} \alpha {\sin}^{2} \beta} - {\sin}^{2} \beta + \cancel{{\sin}^{2} \alpha {\sin}^{2} \beta}$

$= {\sin}^{2} \alpha - {\sin}^{2} \beta$

$= R H S$

Mar 20, 2018

Kindly go through a Proof in the Explanation.

#### Explanation:

The following is another way to prove the assertion :

Recall that, $- 2 \sin x \sin y = \cos \left(x + y\right) - \cos \left(x - y\right)$.

$\therefore \sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right)$,

$= - \frac{1}{2} \left\{- 2 \sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right)\right\}$,

$= - \frac{1}{2} \left\{\cos \left(\left(\alpha + \beta\right) + \left(\alpha - \beta\right)\right) - \cos \left(\left(\alpha + \beta\right) - \left(\alpha - \beta\right)\right)\right\}$,

$= - \frac{1}{2} \left\{\cos 2 \alpha - \cos 2 \beta\right\}$,

$= - \frac{1}{2} \left\{\left(1 - 2 {\sin}^{2} \alpha\right) - \left(1 - 2 {\sin}^{2} \beta\right)\right\}$,

$= {\sin}^{2} \alpha - {\sin}^{2} \beta$.