(Cos70 Cos40)+(Sin70 Sin40)=??

1 Answer
Feb 24, 2018

\qquad \qquad \qquad \quad \quad cos70 cos40 + sin 70 sin 40 \ = \ \sqrt{3}/2.

Explanation:

"Nice question !!"

"The pattern of the question matches the pattern of the rule for"
"the cosine of a sum/difference -- "

\qquad \qquad \qquad \qquad \quad cos( x - y ) \ = \ cosx cosy + sin x sin y.

"So:"

\qquad \qquad \qquad \qquad \quad cosx cosy + sin x sin y \ = \ cos( x - y ).

"So, with" \ x = 70, \quad y = 40, "we have:"

\qquad \qquad \qquad \qquad \quad cos70 cos40 + sin 70 sin 40 \ = \ cos( 70 - 40 ).

"Thus:"

\qquad \qquad \qquad \qquad \quad cos70 cos40 + sin 70 sin 40 \ = \ cos( 30 ). \qquad \qquad \qquad \qquad (I)

"But, recalling the 30-60-90 right triangle, we have:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad cos( 30 ) \ = \ \sqrt{3}/2.

"So, by (I):"

\qquad \qquad \qquad \qquad \quad cos70 cos40 + sin 70 sin 40 \ = \ \sqrt{3}/2.

"This is our answer -- nice question !!"