Question

Solve :$\cos {24}^o+\cos {55}^o+\cos {125}^o+\cos {204}^o+\cos {300}^o=\frac{1}{2}$

Answer 1

$\cos {24}^o+\cos {55}^o+\cos {125}^o+\cos {204}^o+\cos {300}^o....\left(1\right)$

write the above equation as

$=\cos { { 24 }^{ o } } +\sin { \left( { 90 }^{ 0 }-{ 55 }^{ 0 } \right) } +\cos { \left( { 90 }^{ 0 }-{ 35 }^{ 0 } \right) } +\cos { \left( { 180 }^{ 0 }-{ 24 }^{ 0 } \right) } +\cos { \left( { 360 }^{ 0 }-{ 60 }^{ 0 } \right) } ....(2)$

We know that $\cos { \theta } =\sin { \left( 90-\theta \right) } $

From the properties of trignometry

$\cos { \left( 180+\theta \right) } =\cos { \theta } ...(3)$

$\cos { \left( 90+\theta \right) } =\sin { \theta } ....(4)$

$\cos { \left( 360-\theta \right) } =\cos { \theta } ...(5)$

by using 3,4,5 and substitute then in (2) we get

$\cos {24}^o+\sin {35}^o+(-\sin {35}^o)+(-\cos {24}^o)+\cos {60}^o$

$=\cos {24}^o-\cos {24}^o+\sin {35}^o-\sin {35}^o+\cos {60}^o=\frac{1}{2}$