How do you perform the operation in trigonometric form $(0.45(cos(310)+isin(310)))(0.6(cos(200)+isin(200)))$ ?

Question

2105 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-05T08:20:26

$(0.45(cos310^@+isin310^@))(0.6(cos200^@+isin200^@))=0.27(-sqrt3/2+1/2i)$

Two complex numbers in polar form $z_1=r_1(cosalpha+isinalpha)$ and $z_2=r_2(cosbeta+isinbeta)$ can be multiplied as under,

$z_1*z_2=r_1*r_2[cosalphacosbeta+icosalphasinbeta+isinalphacosbeta+i^2sinalphasinbeta]$

= $r_1*r_2[(cosalphacosbeta-sinalphasinbeta)+i(cosalphasinbeta+sinalphacosbeta)]$

= $r_1*r_2[cos(alpha+beta)+isin(alpha+beta)]$

Hence $(0.45(cos310^@+isin310^@))(0.6(cos200^@+isin200^@))$

= $0.45xx0.6(cos(310^@+200^@)+isin(310^@+200^@))$

= $0.27(cos510^@+isin510^@)$

= $0.27(cos(360^@+150^@)+isin(360^@+150^@))$

= $0.27(cos150^@+isin150^@)$

= $0.27(-sqrt3/2+1/2i)$

How do you perform the operation in trigonometric form (0.45(cos(310)+isin(310)))(0.6(cos(200)+isin(200)))(0.45(cos(310)+isin(310)))(0.6(cos(200)+isin(200)))?