Question

3-4cos2L+cos4L/3+4cos2L+cos4L
Помогите, пожалуйста.

Answer 1

$\frac{3-4Cos2\alpha+Cos4\alpha}{3+4Cos2\alpha+Cos4\alpha}=\frac{3-4(Cos^{2}\alpha-Sin^{2}\alpha)+(8Sin^{4}\alpha-8Sin^{2}\alpha+1)}{3+4(Cos^{2}\alpha-Sin^{2}\alpha)+(8Cos^{4}\alpha-8Cos^{2}\alpha+1)}=\frac{3-4Cos^{2}\alpha+4Sin^{2}\alpha+8Sin^{4}\alpha-8Sin^{2}\alpha+1}{3+4Cos^{2}\alpha-4Sin^{2}\alpha+8Cos^{4}\alpha-8Cos^{2}\alpha+1} =\\\\=\frac{4-4(Sin^{2}\alpha+Cos^{2}\alpha)+8Sin^{4}\alpha}{4-4(Sin^{2}\alpha+Cos^{2}\alpha)+8Cos^{4} \alpha}= \frac{4-4+8Sin^{4}\alpha}{4-4+8Cos^{4}\alpha}=tg^{4}\alpha$