Упростите выражение cos4a/(sin4a-tg2a). Created by arishka122. algebra-ru

(sin4a-tg2a)...

m11m

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$\frac{cos4a}{sin4a-tg2a}= \frac{cos4a}{2sin2acos2a- \frac{sin2a}{cos2a} }= \\ \\ = \frac{cos4a*cos2a}{2sin2acos2a*cos2a-sin2a}= \frac{cos4a*cos2a}{sin2a(2cos^22a-1)}= \\ \\ 
= \frac{cos4a*cos2a}{sin2a(2cos^22a-(cos^22a+sin^22a))}= \\ \\ 
= \frac{cos2a}{sin2a}* \frac{cos4a}{2cos^22a-cos^22a-sin^22a}= \\ \\ 
=ctg2a* \frac{cos4a}{cos^22a-sin^22a}=ctg2a* \frac{cos4a}{cos(2*2a)}=ctg2a$

7 votes Thanks 9

sedinalana

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Cos4a:(2sin2acos2a-sin2a/cos2a)=cos4a:(2sin2acos²2a-sin2a)/cos2a=
=cos4a*cos2a/[sin2a(2cos²2a-1)]=cos4a8cos2a/(sin2a*cos4a)=ctg2a

3 votes Thanks 4

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