My attempt:
First I converted this expression to sum notation:
sin(1∘)+sin(3∘)+sin(5∘)+...+sin(175∘)+sin(177∘)+sin(179∘) = ∑90n=1sin(2n−1)∘
Next, I attempted to use Euler's formula for the sum, since I needed this huge expression to be simplified in exponential form:
∑90n=1sin(2n−1)∘ = Im(∑90n=1cis(2n−1)∘)
Im(∑90n=1cis(2n−1)∘) = Im(∑90n=1ei(2n−1)∘)
Im(∑90n=1ei(2n−1)∘) = Im(ei+e3i+e5i+...+e175i+e177i+e179i)
Next, I used the sum of the finite geometric series formula on this expression:
Im(ei+e3i+e5i+...+e175i+e177i+e179i) = Im(ei(1−e180i)1−e2i)
Im(ei(1−e180i)1−e2i) = Im(2ei1−e2i)
Now I'm stuck in here;
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