# How do you verify the identity (sin^3beta+cos^3beta)/(sinbeta+cosbeta)=1-sinbetacosbeta?

Jan 24, 2017

see explanation.

#### Explanation:

There are 3 approaches to verifying the identity.

• Manipulate the left side into the form of the right side

• Manipulate the right side into the form of the left side

• Manipulate both sides until a point is reached where they are
$\textcolor{w h i t e}{x} \text{both equal}$

Using the first approach.

$\text{left side } = \frac{{\sin}^{3} \beta + {\cos}^{3} \beta}{\sin \beta + \cos \beta}$

The numerator is a $\textcolor{b l u e}{\text{sum of cubes}}$ and can be factorised in general as follows.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} + a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{here " a=sinbeta" and } b = \cos \beta$

$\Rightarrow {\sin}^{3} \beta + {\cos}^{3} \beta =$

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

The left side can now be simplified.

$\frac{\left(\cancel{\sin \beta + \cos \beta}\right) \left({\sin}^{2} \beta - \sin \beta \cos \beta + {\cos}^{2} \beta\right)}{\cancel{\sin \beta + \cos \beta}}$

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

$\text{Using the identity } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\sin}^{2} \beta + {\cos}^{2} \beta = 1} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{Then} {\sin}^{2} \beta - \sin \beta \cos \beta + {\cos}^{2} \beta = 1 - \sin \beta \cos \beta$

$\text{Thus left side = right side "rArr"verified}$