Question

Какая формула для нахождения sin4a; cos4a и ctg4a?​

Answer 1

1. Так как $0^circ<alpha<90^circ$, то данный угол первой четверти и все тригонометрические функции в ней положительны.

$cosalpha =sqrt{1-sin^2alpha} =sqrt{1-left(dfrac{1}{3}right)^2} =sqrt{1-dfrac{1}{9}} =sqrt{dfrac{8}{9}} =dfrac{2sqrt{2} }{3}$

$sin2alpha =2sinalpha cosalpha =2cdotdfrac{1}{3} cdotdfrac{2sqrt{2} }{3} =boxed{dfrac{4sqrt{2} }{9}}$

$cos2alpha =cos^2alpha -sin^2alpha =left(dfrac{2sqrt{2} }{3}right)^2-left(dfrac{1}{3}right)^2=dfrac{8}{9}-dfrac{1}{9}=dfrac{7}{9}$

$sin4alpha =2sin2alpha cos2alpha =2cdotdfrac{4sqrt{2} }{9} cdotdfrac{7}{9}=dfrac{56sqrt{2} }{81}$

$cos4alpha =cos^22alpha -sin^22alpha=left(dfrac{7}{9}right)^2-left(dfrac{4sqrt{2} }{9}right)^2=dfrac{49}{81}-dfrac{32}{81}=boxed{dfrac{17}{81}}$

$mathrm{ctg}4alpha =dfrac{cos4alpha }{sin4alpha } =dfrac{17}{81}:dfrac{56sqrt{2} }{81}=dfrac{17}{56sqrt{2}}=dfrac{17cdotsqrt{2}}{56sqrt{2}cdotsqrt{2}}=boxed{dfrac{17sqrt{2}}{112}}$

2, Так как $180^circ<alpha<270^circ$, то данный угол третьей четверти, синус и косинус в ней отрицательны.

$sinalpha =-sqrt{1-cos^2alpha} =-sqrt{1-left(-dfrac{2sqrt{2} }{3}right)^2} =-sqrt{1-dfrac{8}{9}} =-sqrt{dfrac{1}{9}} =-dfrac{1}{3}$

$sin2alpha =2sinalpha cosalpha =2cdotleft(-dfrac{1}{3}right) cdotleft(-dfrac{2sqrt{2} }{3}right) =dfrac{4sqrt{2} }{9}$

$cos2alpha =cos^2alpha -sin^2alpha =left(-dfrac{2sqrt{2} }{3}right)^2-left(-dfrac{1}{3}right)^2=dfrac{8}{9}-dfrac{1}{9}=boxed{dfrac{7}{9}}$

$mathrm{tg}2alpha =dfrac{sin2alpha }{cos2alpha } =dfrac{4sqrt{2} }{9}:dfrac{7}{9}=boxed{dfrac{4sqrt{2} }{7}}$

$sin4alpha =2sin2alpha cos2alpha =2cdotdfrac{4sqrt{2} }{9} cdotdfrac{7}{9}=boxed{dfrac{56sqrt{2} }{81}}$

Answer 2

$1); ; sina=frac{1}{3}\\0^circ <a<90^circ ; ; to ; ; cosa>0\\cosa=+sqrt{1-sin^2a}=+sqrt{1-frac{1}{9}}=sqrt{frac{8}{9}}=frac{2sqrt2}{3}\\sin2a=2cdot sinacdot cosa=2cdot frac{1}{3}cdot frac{2sqrt2}{3}=frac{4sqrt2}{9}\\cos2a=cos^2a-sin^2a=frac{8}{9}-frac{1}{9}=frac{7}{9}\\cos4a=cos^22a-sin^22a=(frac{7}{9})^2-(frac{4sqrt2}{9})^2=frac{49-32}{81}=frac{17}{81}\\sin4a=2, sin2acdot cos2a=2cdot frac{4sqrt2}{9}cdot frac{7}{9}=frac{56sqrt2}{81}$

$ctg4a=frac{cos4a}{sin4a}=frac{17}{56sqrt2}=frac{17sqrt2}{112}$

$2); ; cosa=-frac{2sqrt2}{3}\\180^circ <a<270^circ ; ; to ; ; sina<0\\sina=-sqrt{1-cos^2a}=-sqrt{1-frac{8}{9}}=-frac{1}{3}\\cos2a=cos^2a-sin^2a=frac{8}{9}-frac{1}{9}=frac{7}{9}\\sin2a=2, sinacdot cosa=2cdot (-frac{1}{3})cdot (-frac{2sqrt2}{3})=frac{4sqrt2}{9}$

$tg2a=frac{sin2a}{cos2a}=frac{4sqrt2}{7}\\sin4a=2, sin2acdot cos2a=2cdot frac{4sqrt2}{9}cdot frac{7}{9}=frac{56sqrt2}{81}$