Question

sin (60^o + θ) - cos (30^o - θ) is equal to (where (60^o + θ) and (30^o - θ) are both acute angles):

• A. 2 cos θ
• B. 0
• C. 2 sin θ
• D. 1

Answer 1

The correct option is B

0

sin (60^o + θ) - cos (30^o - θ)

= sin (60^o + θ) - cos (90^o - (60^o + θ)

= sin (60^o + θ) - sin (60^o + θ)

= 0