Question

Помогите!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

$frac{3+4cos4alpha+cos8alpha}{3-4cos4alpha+cos8alpha}=frac{2+4cos4alpha+2cos^24alpha}{2-4cos4alpha+2cos^24alpha}=frac{(1+cos4alpha)^2}{(1-cos4alpha)^2}=(frac{1+2cos^22alpha-1}{1-2cos^22alpha+1})^2$=$(frac{cos^22alpha}{1-cos^22alpha})^2=(frac{cos^22alpha}{sin^22alpha})^2=ctg^42alpha$

Answer 2

(3+4cos4a+cos8a)/(3-4cos4a+cos8a)=ctg⁴2a;

(3+4cos4a+cos8a)/(3-4cos4a+cos8a)=

(2+4cos4a+2cos²4a)/(2-4cos4a+2cos²4a)=

2(1+2cos4a+cos²4a)/2(1-2cos4a+cos²4a)=

(1+cos4a)²/(1-cos4a)²=

(2cos²2a/2sin²2a)²=

(ctg²2a)²=

ctg⁴2a.

 

ctg⁴2a=ctg⁴2a.