How would I find the standard form of the complex number $9 (cos (\frac{7 π}{6}) = i sin (\frac{7 π}{6}))$ $9(cos((7pi)/6)=isin((7pi)/6))$ ?

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How would I find the standard form of the complex number $9 (cos (\frac{7 π}{6}) = i sin (\frac{7 π}{6}))$ $9(cos((7pi)/6)=isin((7pi)/6))$ ?

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Answer 1 · 2015-10-09T03:02:48

Just evaluate the trigonometric functions and expand to get $- \frac{9 \sqrt{3}}{2} - \frac{9}{2} i$ $-(9sqrt(3))/2-9/2 i$ (standard form for a complex number is $a + b i$ $a+bi$ , where $a,b\in RR$ )

Explanation:

Note that $cos((7pi)/6)=-sqrt(3)/2$ and $sin((7pi)/6)=-1/2$ . Therefore

$9(cos((7pi)/6)+i sin((7pi)/6))=9(-sqrt(3)/2-i 1/2)$

$=-(9sqrt(3))/2-9/2 i$

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How would I find the standard form of the complex number $9 (cos (\frac{7 π}{6}) = i sin (\frac{7 π}{6}))$ $9(cos((7pi)/6)=isin((7pi)/6))$ ?

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Explanation:

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How would I find the standard form of the complex number $9 (cos (\frac{7 π}{6}) = i sin (\frac{7 π}{6}))$ $9(cos((7pi)/6)=isin((7pi)/6))$ ?