Question

What is cos $ 80{}^\circ $ + cos $ 40{}^\circ $ − cos $ 20{}^\circ $ equal to?
A. 2
B. 1
C. 0
D. −19

Answer 1

Hint:Use the sum to product identity cos 2A + cos 2B = 2 cos (A + B) cos (A − B) to arrive at an expression which has angles whose trigonometric ratios we know, or they have the same angles. Here, $ 80{}^\circ $ =

2 × $ 40{}^\circ $ and $ 40{}^\circ $ = 2 × $ 20{}^\circ $ , and $ 40{}^\circ $ + $ 20{}^\circ $ = $ 60{}^\circ
$ and $ 40{}^\circ $ − $ 20{}^\circ $ = $ 20{}^\circ $ . Convert all the terms into trigonometric ratios of 20˚ and simplify.
Recall that cos $ 60{}^\circ $ = $ \dfrac{1}{2} $ .

Complete step by step solution:
The given expression is cos $ 80{}^\circ $ + cos $ 40{}^\circ $ − cos $ 20{}^\circ $ .

It can be written as:
= cos (2 × $ 40{}^\circ $ ) + cos (2 × $ 20{}^\circ $ ) − cos $ 20{}^\circ $
Using cos 2A + cos 2B = 2 cos (A + B) cos (A − B), we can write it as:
= 2 cos ( $ 40{}^\circ $ + $ 20{}^\circ $ ) cos ( $ 40{}^\circ $ − $ 20{}^\circ $ ) − cos $ 20{}^\circ $
= 2 cos $ 60{}^\circ $ cos $ 20{}^\circ $ − cos $ 20{}^\circ $
Substituting the value of cos $ 60{}^\circ $ = $ \dfrac{1}{2} $ , we get:
= $ 2\left( \dfrac{1}{2} \right) $ cos $ 20{}^\circ $ − cos $ 20{}^\circ $
= cos $ 20{}^\circ $ − cos $ 20{}^\circ $
= 0

Hence, the correct answer is C.

Note: All the trigonometric ratios are positive in the interval 0 < θ < $ 90{}^\circ $ .

Useful trigonometric identities:
$ {{\sin }^{2}}\theta $ + $ {{\cos }^{2}}\theta $ = 1
Sum-Product formula:
sin 2A + sin 2B = 2 sin (A + B) cos (A − B)
sin 2A − sin 2B = 2 cos (A + B) sin (A − B)
cos 2A + cos 2B = 2 cos (A + B) cos (A − B)
cos 2A + cos 2B = −2 sin (A + B) sin (A − B)