Question

How do you multiply `e^(( 9 pi )/ 8 i) * e^( pi/2 i )` in trigonometric form?

Answer 1

`e^((9pi)/8i) * e^(pi/2i)=color(green)(sin(pi/8)-icos(pi/8))`

Explanation:

Using Euler's formula
`color(white)("XXX")e^((9pi)/8i) = cos((9pi)/8)+i * sin((9pi)/8)`

`color(white)("XXXXX")=cos(pi+pi/8)+i * sin(pi+pi/8)`

`color(white)("XXXXX")=-cos(pi/8)+i * (-sin(pi/8))`

`color(white)("XXXXX")=-cos(pi/8)- i * sin(pi/8)`

and
`color(white)("XXX")e^(pi/2i)=cos(pi/2)+ i * sin(pi/2)`

`color(white)("XXXXX")=0 + i * 1`

`color(white)("XXXXX")=i`

Therefore
`color(white)("XXX")e^((9pi)/8i) * e^(pi/2i) = [-cos(pi/8)- i * sin(pi/8)] * i`

`color(white)("XXXXXXX")=- i * cos(pi/8) - (-1) * sin(pi/8)`

`color(white)("XXXXXXX")=sin(pi/8)-icos(pi/8)`