Question

Prove that $1+\cos {56}^{\circ }+\cos {58}^{\circ }-\cos {66}^{\circ }=4\cos {28}^{\circ }\cos {29}^{\circ }\sin {33}^{\circ }$.

Answer 1

Simplifying the LHS of $1+\cos {56}^{\circ }+\cos {58}^{\circ }-\cos {66}^{\circ }=4\cos {28}^{\circ }\cos {29}^{\circ }\sin {33}^{\circ }$.

$1+\cos {56}^{\circ }+\cos {58}^{\circ }-\cos {66}^{\circ }=\left(1-\cos {66}^{\circ }\right)+\left(\cos {56}^{\circ }+\cos {58}^{\circ }\right)$

$=2{\sin }^2\frac{{66}^{\circ }}{2}+\left(2\cos \frac{{56}^{\circ }+{58}^{\circ }}{2}\cos \frac{{56}^{\circ }-{58}^{\circ }}{2}\right)$

$=2{\sin }^2{33}^{\circ }+2\cos {57}^{\circ }\cos 1^{\circ }$

$=2{\sin }^2{33}^{\circ }+2\cos \left({90}^{\circ }-{33}^{\circ }\right)\cos \left({90}^{\circ }-{89}^{\circ }\right)$

$=2{\sin }^2{33}^{\circ }+2\sin {33}^{\circ }\sin {89}^{\circ }$

$=2\sin {33}^{\circ }\left(\sin {33}^{\circ }+\sin {89}^{\circ }\right)$

$=2\sin {33}^{\circ }\left(2\sin \frac{{33}^{\circ }+{89}^{\circ }}{2}\cos \frac{{33}^{\circ }-{89}^{\circ }}{2}\right)$

$=4\sin {33}^{\circ }\sin {61}^{\circ }\cos {28}^{\circ }$

$=4\sin {33}^{\circ }\sin \left({90}^{\circ }-{29}^{\circ }\right)\cos {28}^{\circ }$

$=4\cos {28}^{\circ }\cos {29}^{\circ }\sin {33}^{\circ }$

$=RHS$