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Trigonometry 6 Online
OpenStudy (anonymous):

which expression is a cube root of 2? 3sqrt2(cos(320) + isin(320)) 3sqrt2(cos280)+isin(280)) 3sqrt2(cos40)+isin(40)) 3sqrt2(cos120)+isin(120))

OpenStudy (anonymous):

Trying to use de moivre's theorem...[r(cosT + i*sinT)]^n = (r^n)(cos(nT) + i*sin(nT))

OpenStudy (anonymous):

YEAH USE TO SOLVE BY DE MOIVER'S THEOREM ( COSX + iSINX)n = COSnX + i SINnX

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