Question

Prove that cos 55° + cos 65° + cos 175° = 0.

Answer 1

Answer:

0

Step-by-step explanation:

Formula used:

cosC + cosD

= 2 cos((C+D)/2) cos((C-D)/2)

cos 55° + cos 65° + cos 175°

= cos 55° + (cos 65° + cos 175°)

= cos 55° +

2 cos((65° + 175° )/2) cos((65° -175° )/2)

= cos 55° + 2 cos(120° ) cos(-55°)

= cos 55° + 2 (-1/2 ) cos(-55°)

= cos 55° - cos55°

= 0

Answer 2

cos 55° + cos 65° + cos 175° = 0

${cos55}^{0} + 2.cos( \frac{65 + 175}{2} ).cos( \frac{175 - 65}{2})$

$cos {55}^{0} + 2.cos( \frac{240}{2} ).cos( \frac{110}{2} )$

$cos {55}^{0} + 2.cos {120}^{0}. cos {55}^{0} \\\\\bf{By \: taking \: (cos {55}^{0} )as \: common}$

$cos {55}^{0} (1 + 2.cos {120}^{0})$

$cos {55}^{0} (1 + 2( - \frac{ 1}{2}))$

$cos {55}^{0} (0) = 0$

= RHS

$\bf{Identities\:used:-}\\\\CosC+CosD=2.Cos\frac{C+D}{2}.Cos\frac{C-D}{2}\\\\Cos{120}^{0}=-\frac{1}{2}$