perform the operation and leave the result in trigonometric.

Question

1). 4(cos40° + isin40°)/3(cos160°+isin160°)

 

 

2).[2/3(cosπ/3+isinπ/3)][9(cosπ/6+isinπ/6])]

Answer 1

1).

      4( Cos 40° + i Sin 40° ) / 3 ( Cos 160° + i Sin 160° ) 

 

         Substitute Cos 160° = - Cos20°      Sin 160° = Sin 20°

 

 

       4/3 [ (COS 40° + i Sin 40° ) ( -Cos 20° - i Sin20°) ] /( - Cos 20° +i Sin 20°) (- Cos20° - i Sin 20°)=

 

  4/3 ( - Cos 40° Cos 20° - i cos 20° Sin 40° - i Sin 20° Cos 40°  + Sin20° Sin 40°)/ Cos^2 20° +Sin^2 20°=

   

   - 4/3 ( Cos 40º Cos20°- Sin 40° Sin 20°) +i ( Sin 20 Cos 40 + Sin 40° Cos 20°) = 

 

    -4/3 ( Cos 60 ) + i ( Sin 60) 

Answer 2

Perform the operation and leave the result in trigonometric form.

1). [4(cos(40°)+ i sin(40°))]/[3(cos(160°)+ i sin(160°))]
= 4/3 (cos(40°–160°+360°) + i sin(40°–160°+360°))
= 4/3 (cos(240°) + i sin(240°))

1 rotation was added to angles to bring answer’s angles into usual domain interval [0°,360°).

 

Note that 4/3 is positive because it is the magnitude of the complex number a + bi and defined to be r = √(a^2 + b^2).

2). [2/3 (cos(π/3)+ i sin(π/3))]*[9 (cos(π/6)+ i sin(π/6))]
= 6 (cos(π/3+π/6)+ i sin(π/3+π/6))
= 6 (cos(π/2)+ i sin(π/2))

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Euler's formula, e^(iθ) = cos(θ) + i sin(θ), gives us some remarkable identities:

e^(iα)*e^(iβ) = e^(i(α+β))
=> [cos(α) + i sin(α)]*[cos(β) + i sin(β)]
= cos(α+β) + i sin(α+β)

e^(iα)/e^(iβ) = e^(i(α–β))
=> [cos(α) + i sin(α)]/[cos(β) + i sin(β)]
= cos(α–β) + i sin(α–β)

(e^(iθ))^p = e^(ipθ)
=> [cos(θ) + i sin(θ)]^p = cos(pθ) + i sin(pθ)

Answer 3

4(cos40° + isin40°)/3(cos160°+isin160°)=4(cos(2/9)π+isin(2/9)π)/3(cos(8/9)π+isin(8/9)π)

=4e^(2/9)πi/3e^(8/9)πi=(4/3)e^-(6/9)πi=(4/3)e^-(2/3)πi

=(4/3)(cos(2/3)π-isin(2/3)π)=-4/3(cos(π/3)+isin(π/3))

=-(4/3)(cos60°+isin60°)

 

[2/3(cosπ/3+isinπ/3)][9(cosπ/6+isinπ/6])]

=(2/3)e^iπ/3×(9e^iπ/6)=6e^iπ/2=6(cos(π/2)+isin(π/2))=6i