cos(15°) =_____________ A. (1 – \(\sqrt{3}\))/2\(\sqrt{2}\) B.

cos-15deg-a-1-sqrt-3-2-sqrt-2-b

cos(15°) =_____________

A. (1 – \(\sqrt{3}\))/2\(\sqrt{2}\)

B. (\(\sqrt{3}\) + 1)/2\(\sqrt{2}\)

C. (\(\sqrt{3}\) – 1)/2\(\sqrt{2}\)

D. (-\(\sqrt{3}\) – 1)/2\(\sqrt{2}\)

I got this question at a job interview.

This is a very interesting question from Trigonometric Functions of Sum and Difference of Two Angles-1 topic in portion Trigonometric Functions of Mathematics – Class 11

Right answer is B. (\(\sqrt{3}\) + 1)/2\(\sqrt{2}\)

Easiest explanation: cos(15°) = cos (45°-30°) = cos45° cos30° + sin45° sin30°

= (1/\(\sqrt{2}\) * \(\sqrt{3}\)/2) + (1/\(\sqrt{2}\) * 1/2) {cos(x – y)=cos x cos y + sin x sin y}

= (\(\sqrt{3}\) +1)/2\(\sqrt{2}\).