What is the product of [2cis 33˚][3cis 22˚]

Step-by-step explanation:

Given the complex numbers in polar cordinates

[2cis 33˚][3cis 22˚]

Generally, if z = rcis, then z = r(cos)

2cis33° = 2(cos33°+isin33°)

3cis22° =  3(cos22°+isin22°)

Taking tthe product of both complex numbers;

[2cis 33˚][3cis 22˚] = [2(cos33°+isin33°)][3(cos22°+isin22°)]

= 6{cos33cos22+icos33sin22+isin33cos22+i²sin33sin22}

since i² = 1

= 6{cos33cos22+icos33sin22+isin33cos22-sin33sin22}

= 6{(cos33cos22-sin33sin22)+i(cos33sin22+sin33cos22)}

Applying the trigonometric identity as shown;

cos(A+B)= cosAcosB-sinAsinB and;

sin(A+B)= sinAcosB + cosAsinB

[2cis 33˚][3cis 22˚] = 6{(cos(33+22)+ i(sin(33+22)}

[2cis 33˚][3cis 22˚] = 6(cos55°+isin55°)

[2cis 33˚][3cis 22˚] = 6cis55°

The product of the complex numbers is 6cis55°

i think it is 70

step-by-step explanation:

step-by-step explanation:

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