Question

If $E={\left(\sin {10}^{\circ }+\cos {10}^{\circ }\right)}^2+{\left(\cos {10}^{\circ }-\sin {10}^{\circ }\right)}^2$, then find ${\log }_{0.5}E$

Answer 1

$E={(\sin {10}^o+\cos {10}^o)}^2+{(\cos {10}^o-\sin {10}^o)}^2$

$E={\sin }^2{10}^o+{\cos }^2{10}^o+2\sin {10}^o\cos {10}^o+{\cos }^2{10}^o+{\sin }^2{10}^o-2\sin {10}^o\cos {10}^o$

$E=1+1$

$E=2$

$\therefore {\log }_{0.5}E={\log }_{1/2}\left(2\right)=-{\log }_{2}2=-1$