Question

Prove that: ${\cos }^2\alpha +{\cos }^2(\alpha +\beta )-2\cos \alpha \cos \beta \cos (\alpha +\beta )={\sin }^2\beta$

Answer 1

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

L.H.S=

${\cos }^2\alpha +{\cos }^2(\alpha +\beta )-2\cos \alpha \cos \beta \cos (\alpha +\beta )$

$={\cos }^2\alpha +{[\cos \alpha \cos \beta -\sin \alpha \sin \beta ]}^2-2\cos \alpha \cos \beta (\cos \alpha \cos \beta -\sin \alpha \sin \beta )$

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

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

$={\cos }^2\alpha (1-{\cos }^2\beta )+{\sin }^2\alpha {\sin }^2\beta$

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

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

$={\sin }^2\beta$

$L.H.S=R.H.S$, Hence proved.