Question

if A=60° and B=30° , prove that cos(A-B)=cosA cosB - sinA sinB​

Answer 1

$\cos(a + b) = \cos(a) \cos(b) - \sin(a) \sin(b) \\ a = 60° \: and \: b = 30° \\ \cos(60° + 30°) = \cos(60°) \cos(30°) - \sin(60°) \sin(30° ) \\ 0 = \frac{1}{2} \times \frac{ \sqrt{3} }{2} - \frac{ \sqrt{3} }{2} \times \frac {1}{2} \\ 0 = \frac{ \sqrt{3} }{4} - \frac{ \sqrt{3} }{4} \\ 0 = 0$