(1.) cos(α+β)=
(2.) sin(α-β)=
(3.) tan(α+β)=
cos(α) = (√5)/5 so tan (α) = -2 in Q4 and α = arg(1-2i)
sin(β) = (√10)/10 so tan(β) = -1/3 in Q2 and β = arg(-3+i)
α+β = arg((1-2i)(-3+i)) = arg(-1+7i), so
(3.) tan(α+β) = -7
(1.) cos(α+β) = -1/√50
α-β = arg((1-2i)/(-3+i)) = arg(-1/2+i/2) = arg(-1+i), so
(2.) sin(α-β) = 1/√2
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Answers & Comments
cos(α) = (√5)/5 so tan (α) = -2 in Q4 and α = arg(1-2i)
sin(β) = (√10)/10 so tan(β) = -1/3 in Q2 and β = arg(-3+i)
α+β = arg((1-2i)(-3+i)) = arg(-1+7i), so
(3.) tan(α+β) = -7
(1.) cos(α+β) = -1/√50
α-β = arg((1-2i)/(-3+i)) = arg(-1/2+i/2) = arg(-1+i), so
(2.) sin(α-β) = 1/√2