A=B=60° then verify: Cos(A -B) = CosA CosB + SinA SinB

1.	A=B=60° then verify: Cos(A -B) = CosA CosB + SinA SinB
Answer» Answer: given A=B=60 verification COS (A-B) = cosA COSB + sinA sinB cos(0 -0 ) = COS60 cos60 + sin60 sin60 $cos( {0}^{0} - {0}^{0}) = \frac{1}{2} \times \frac{1}{2} + \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2}$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{4} + \frac{3}{4}$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1 + 3}{4} = \frac{4}{4}$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {cos0}^{0} = 1$

Answer»

Answer:

given A=B=60

verification

COS (A-B) = cosA COSB + sinA sinB

cos(0 -0 ) = COS60 cos60 + sin60 sin60

$cos( {0}^{0} - {0}^{0}) = \frac{1}{2} \times \frac{1}{2} + \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2}$