Question

已知cosa=负4/5 a属于(π,3/2π),tanb=-1/3 属于(π/2,π)求cos(a+b)

Answer 1

tanb=-1/3 b∈(π/2,π)

sinb=1/√10

cosb=-3/√10

cosa=-4/5 a∈(π,3π/2)

sina=-3/5

cos(a+b) =cosacosb-sinasinb

=(-4/5)(-3/√10)-(-3/5)(1/√10)

=12/5√10+3/5√10

=15/5√10

=3/√10

=3√10/10