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.

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

$\sin2\alpha =2\sin\alpha \cos\alpha =2\cdot\dfrac{1}{3} \cdot\dfrac{2\sqrt{2} }{3} =\boxed{\dfrac{4\sqrt{2} }{9}}$

$\cos2\alpha =\cos^2\alpha -\sin^2\alpha =\left(\dfrac{2\sqrt{2} }{3}\right)^2-\left(\dfrac{1}{3}\right)^2=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}$

$\sin4\alpha =2\sin2\alpha \cos2\alpha =2\cdot\dfrac{4\sqrt{2} }{9} \cdot\dfrac{7}{9}=\dfrac{56\sqrt{2} }{81}$

$\cos4\alpha =\cos^22\alpha -\sin^22\alpha=\left(\dfrac{7}{9}\right)^2-\left(\dfrac{4\sqrt{2} }{9}\right)^2=\dfrac{49}{81}-\dfrac{32}{81}=\boxed{\dfrac{17}{81}}$

$\mathrm{ctg}4\alpha =\dfrac{\cos4\alpha }{\sin4\alpha } =\dfrac{17}{81}:\dfrac{56\sqrt{2} }{81}=\dfrac{17}{56\sqrt{2}}=\dfrac{17\cdot\sqrt{2}}{56\sqrt{2}\cdot\sqrt{2}}=\boxed{\dfrac{17\sqrt{2}}{112}}$

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

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

$\sin2\alpha =2\sin\alpha \cos\alpha =2\cdot\left(-\dfrac{1}{3}\right) \cdot\left(-\dfrac{2\sqrt{2} }{3}\right) =\dfrac{4\sqrt{2} }{9}$

$\cos2\alpha =\cos^2\alpha -\sin^2\alpha =\left(-\dfrac{2\sqrt{2} }{3}\right)^2-\left(-\dfrac{1}{3}\right)^2=\dfrac{8}{9}-\dfrac{1}{9}=\boxed{\dfrac{7}{9}}$

$\mathrm{tg}2\alpha =\dfrac{\sin2\alpha }{\cos2\alpha } =\dfrac{4\sqrt{2} }{9}:\dfrac{7}{9}=\boxed{\dfrac{4\sqrt{2} }{7}}$

$\sin4\alpha =2\sin2\alpha \cos2\alpha =2\cdot\dfrac{4\sqrt{2} }{9} \cdot\dfrac{7}{9}=\boxed{\dfrac{56\sqrt{2} }{81}}$

Answer 2

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

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

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

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