Question

Prove that: ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1-2\cos A\cos B\cos C$.

Answer 1

We write ${\cos }^{2}A=1-{\sin }^{2}A$

and as in $\Delta ABC$ $A+B+C=180$

$⟹\cos C=\cos (180-A-B)=-\cos (A+B)$

$L.H.S.=1-{\sin }^{2}A+{\cos }^{2}B+{\cos }^{2}C$

$=1+({\cos }^{2}B-{\sin }^{2}A)+{\cos }^{2}C$

$=1+\cos (B+A)\cos (B-A)+{\cos }^{2}C$ ..... $({\cos }^{2}C-{\sin }^{2}D=\cos (C+D\left)\cos \right(C-D\left)\right)$

$=1-\cos C\cos (B-A)+{\cos }^{2}C$

$=1-\cos C\left[\cos \right(B-A)-\cos C]$

$=1-\cos C\left[\cos \right(B-A)+\cos (B+A\left)\right]$ .

$=1-\cos C\left(2\cos A\cos B\right)$ ....... By compound angles formula

$=1-2\cos A\cos B\cos C$.