Question

Prove that $4cos{12}^{\circ }cos{48}^{\circ }cos{72}^{\circ }=1-2si{n}^2{18}^{\circ }$.

Answer 1

We have ,

$LHS=4cos{12}^{\circ }cos{48}^{\circ }cos{72}^{\circ }$

$=2\left(2cos{12}^{\circ }cos{48}^{\circ }\right)cos{72}^{\circ }$

$=2\left[cos\right({12}^{\circ }+{48}^{\circ })+cos({12}^{\circ }-{48}^{\circ }\left)\right]cos{72}^{\circ }$

$[\because 2cosAcosB=cos(A+B)+cos(A-B\left)\right]$

$=2(cos{60}^{\circ }+cos{36}^{\circ })cos{72}^{\circ }$

$=2cos{60}^{\circ }cos{72}^{\circ }+2cos{36}^{\circ }cos{72}^{\circ }$

$=2\times \frac{1}{2}cos{72}^{\circ }+cos({36}^{\circ }+{72}^{\circ })+cos({72}^{\circ }-{36}^{\circ })$

$[\because 2cosAB=cos(A+B)+cos(A-B\left)\right]$

$=cos{72}^{\circ }cos{108}^{\circ }+cos{36}^{\circ }$

$=cos{72}^{\circ }+cos({180}^{\circ }-cos{36}^{\circ }$

$=cos{72}^{\circ }-cos{72}^{\circ }+cos{36}^{\circ }$

$=cos{36}^{\circ }$

$=cos2\times {18}^{\circ }$

$=1-2si{n}^2{18}^{\circ }=RHS$ Hence Proved.