Question

Prove that: $\cos ({36}^0-A)\cos ({36}^0+A)+\cos ({54}^0+A)\cos ({54}^0-A)=\cos 2A$

Answer 1

Let $Z=\cos ({36}^0-A)\cos ({36}^0+A)+\cos ({54}^0+A)\cos ({54}^0-A)$

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

$\therefore Z={\cos }^{2}A-{\sin }^2{36}^0+{\cos }^{2}A-{\sin }^2{54}^0$

$Z=2{\cos }^{2}A-({\cos }^2{54}^0+{\sin }^2{54}^0)$

$Z=2{\cos }^{2}A-1$

$Z=\cos 2A$