Question

Prove that $\cos {40}^o+\cos {50}^o+\cos {70}^o+\cos {80}^o=\cos {20}^o+\cos {10}^o$

Answer 1

Consider the given L.H.S.

$=\cos {40}^o+\cos {50}^o+\cos {70}^o+\cos {80}^o$

$=\cos {70}^o+\cos {50}^o+\cos {80}^o+\cos {40}^o$

We know that

$\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cdot \cos \left(\frac{C-D}{2}\right)$

Therefore,

$=2\cos \left(\frac{{70}^o+{50}^o}{2}\right)\cdot \cos \left(\frac{{70}^o-{50}^o}{2}\right)+2\cos \left(\frac{{80}^o+{40}^o}{2}\right)\cdot \cos \left(\frac{{80}^o-{40}^o}{2}\right)$

$=2[\cos \left(\frac{{120}^o}{2}\right)\cdot \cos \left(\frac{{20}^o}{2}\right)+\cos \left(\frac{{120}^o}{2}\right)\cdot \cos \left(\frac{{40}^o}{2}\right)]$

$=2[\cos \left({60}^o\right)\cdot \cos \left({10}^o\right)+\cos \left({60}^o\right)\cdot \cos \left({20}^o\right)]$

$=2[\frac{1}{2}\cdot \cos \left({10}^o\right)+\frac{1}{2}\cdot \cos \left({20}^o\right)]$

$=\cos \left({10}^o\right)+\cos \left({20}^o\right)$

R.H.S

Hence, proved.