Question

Show that - $\cos {40}^{\circ }+\cos {50}^{\circ }+\cos {70}^{\circ }+\cos {80}^{\circ }=\cos {20}^{\circ }+\cos {10}^{\circ }$.

Answer 1

Taking L.H.S

$\left(\cos {40}^{\circ }+\cos {80}^{\circ }\right)+\left(\cos {50}^{\circ }+\cos {70}^{\circ }\right)$

As we know,

$\cos c+\cos d=2cos\frac{\left(c+d\right)}{2}.\cos \frac{\left(c-d\right)}{2}$

So,

$=\left[2\cos \frac{\left(40+80\right)}{2}.\cos \frac{\left(40-80\right)}{2}\right]+\left[2\cos \frac{\left(50+70\right)}{2}.\cos \frac{\left(50-70\right)}{2}\right]$

$=2\cos \frac{{120}^{\circ }}{2}.cos\left(\frac{-{40}^{\circ }}{2}\right)+2\cos \frac{{120}^{\circ }}{2}\cos \left(\frac{-{20}^{\circ }}{2}\right)$

$=2\cos {60}^{\circ }.\cos \left(-{20}^{\circ }\right)+2\cos {60}^{\circ }\cos \left(-{10}^{\circ }\right)$

$=2\times \frac{1}{2}.\cos {20}^{\circ }+2\times \frac{1}{2}.\cos {10}^{\circ }$

$=\cos {20}^{\circ }+\cos {10}^{\circ }$

which is equal to R.H.S