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Daca ctg x= √ 2 pe 3
Calculati √3 ori sin2x +√2 ori cos2x


Răspuns :

[tex] \sqrt{3}sin 2x + \sqrt{2}cos2x=? ; ctg x = \frac{ \sqrt{2} }{3} \\ \\ tg x = \frac{1}{ctg x}= \frac{ \frac{1}{ \sqrt{2} } }{3} } = \frac{3}{ \sqrt{2}}= \frac{3 \sqrt{2} }{2} \\ \\ tg^{2}(x)= \frac{sin^{2}(x) }{cos^{2}(x)}} \\ \\ \frac{sin^{2}(x) }{cos^{2}(x)}}= (\frac{3 \sqrt{2} }{2} )^{2} \\ \\\frac{sin^{2}(x) }{cos^{2}(x)}}= \frac{9}{2} \\ \\ sin^{2}(x) +cos^{2}(x)=1; sin^{2}(x) =1-cos^{2}(x) [/tex]

[tex] \frac{1-cos^{2}(x)}{cos^{2}(x)}= \frac{9}{2} \\ \\2-2cos^{2}(x)=9 cos^{2} (x) \\ \\ 2=11cos^{2}(x) \\ \\cos x = \sqrt{ \frac{11}{2}}= \frac{ \sqrt{22}}{2} \\ \\ sin^{2}(x)=1- cos^{2}(x)=(1- \frac{ \sqrt{22}}{2}) ^{2} =- \frac{9}{2} \\ \\sin (x)=- \frac{ \sqrt{10}}{2} \\ \\ \sqrt{3}\cdot sin (x)cos(x) + \sqrt{2}cos (2x)=? \\ \\ sin (2x) =2sin(x)cos (x)=2\cdot-\frac{ \sqrt{10}}{2}\cdot \frac{ \sqrt{22}}{2}} =- \sqrt{55}} [/tex]
[tex]cos2x =cos^{2}(x)-sin^{2} (x)=\frac{7=-5}{2} -\frac{7}{2}=-6 [/tex]

[tex]\sqrt{3}sin 2x + \sqrt{2}cos2x= \sqrt{3} \cdot- \sqrt{55}+ \sqrt{2}\cdot (-6)= \sqrt{165}-2 \sqrt{3} [/tex]