Question

Объясните решение.
3-4cos2B+cos4B / 3+4cos2B+cos4B = tg^4B

Answer 1

tg%4b=sin^4b/cos^4b=(1-cos2b)²/(1+cos2b)²=
=(1-2cos2b+cos²2b)/(1+2cos2b+cos²2b)=
=[1-cos2b+(1+cos4b)/2]/[1+cos2b+(1+cos4b)/2]=
=(2-2cos2b+1+cos4b)/(2+2cos2b+1+cos4b)=
=(3-cos2b+cos4b)/(3+cos2b+cos4b)
чтд