Question

Show that cos42°+cos78°+cos162°=0.

Answer 1

Answer:

Using cosA + cosB = 2 cos(½(A+B)) cos(½(A-B)), we can write

cos42° + cos78° = 2 cos60° cos18°

= 2 . ½ . cos18° (since cos60° = ½)

= cos18°

So, we have

cos42° + cos78° = cos18° --------(1)

Also, cos(180°-A) = -cosA.

Therefore,

cos162° = cos(180°-18°) = -cos18°-----(2)

Using (1) and (2),

cos42° + cos78° + cos162° = cos18° + (-cos18°) = 0

Hence proved.