trig desperate help please ASAP 10 points for sure!!?

Question

i am just really stuck.

Find the standard form of the number
1. 9(cos0 + isin0) (note:its zero not theta)

Perform operation and leave in trigonometric form.
1. [(3/2)(cos pi/6 + isin pi/6)][6(cos pi/4 + i sin pi/4]

2. 9(cos20 + isin20)/5*(cos75 + isin75)

Answer 1

9 (cos 0 + i sin 0) = 9 x 1 = 9

9 [√3/2 + i (1/2) ] [1/√2 + i 1/√2 ]
(9/2)[√3/(2√2) + i√3 /(2√2) + i(1/(2√2) - 1/(2√2]
(9/2) [ (√3 - 1) / (2√2)] + i (√3 + 1) / (2√2) ]

(9/5) (cos 20° + i sin 20°) (cos 75° - i sin 75°)
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(cos 75° + i sin75°)(cos75° - i sin 75°)

(9/5) (cos 20° + i sin 20°) (cos 75° - i sin 75°)
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(cos² 75° + sin² 75°)

(9/5) (cos 20° + i sin 20°) (cos 75° - i sin 75°)

Answer 2

1. The sine of zero is zero. The cos of zero is 1. So 9(cos0 + isin0) = 9(1 +0) = 9.