Prove that, $cos(\alpha + \beta ) + sin(\alpha – \beta )$ = $2sin\left(\frac{\pi }{4} + \alpha \right)\cos \left(\frac{\pi }{4} + \beta \right)$
L.H.S= cos (α+β) +sin(α-β)
= cos α cosβ – sinα sinβ + sinα cosβ – cosα sinβ
= cos α cosβ – cosα sinβ + sinα cosβ – sinα sinβ
= cosα (cosβ – sinβ) + sinα (cosβ – sinβ)
= (cosβ – sinβ) (cosα +sinα)
=2($\frac{1}{\sqrt{2}}$cosβ- $\frac{1}{\sqrt{2}}$sinβ)( $\frac{1}{\sqrt{2}}$cosα+$ \frac{1}{\sqrt{2}}$sinα)
= 2(cos $\frac{{\mathrm{\pi}}}{4}$cosβ – sin$\frac{{\mathrm{\pi}}}{4}$ sinβ) (sin $\frac{{\mathrm{\pi}}}{4}$ cosα + cos $\frac{{\mathrm{\pi}}}{4}$ sinα)
= 2 cos($\frac{{\mathrm{\pi}}}{4}$+β) sin ($\frac{{\mathrm{\pi}}}{4}$+α)
= 2 sin ($\frac{{\mathrm{\pi}}}{4}$+α) cos($\frac{{\mathrm{\pi}}}{4}$+β) =R.H.S[Proved]