Question

Докажите тождество:
1+(cos 4α/tg(3/4П-2α))=sin 4α

Answer 1

$1+frac{cos4alpha}{tg(frac{3pi}{4}-2alpha)}=1+frac{2cos4alpha (1-tg^2(frac{3pi}{4}-2alpha))}{2tg(frac{3pi}{4}-2alpha)(1-tg^2(frac{3pi}{4}-2alpha))}=1+frac{2cos4alpha}{tg(frac{3pi}{2}-4alpha)(1-tg^2(frac{3pi}{4}-2alpha))}=1+frac{2cos4alpha}{ctg4alpha(1-tg^2(frac{3pi}{4}-2alpha))}=1+frac{2sin4alpha}{1-tg^2(frac{3pi}{4}-2alpha)}=1+frac{2sin4alpha cos^2(frac{3pi}{4}-2alpha)}{cos^2(frac{3pi}{4}-2alpha)-sin^2(frac{3pi}{4}-2alpha)}=$$1+frac{2sin4alpha cos^2(frac{3pi}{4}-2alpha)}{cos2(frac{3pi}{4}-2alpha)}=1+frac{sin4alpha (1+cos2(frac{3pi}{4}-2alpha))}{cos(frac{3pi}{2}-4alpha)}=1+frac{sin4alpha (1+cos2(frac{3pi}{4}-2alpha))}{-sin(4alpha)}=1-(1+cos(frac{3pi}{2}-4alpha))=1- (1-sin4alpha)=1- 1+sin4alpha=sin4alpha$

Answer 2

1+(cos4a/tg(3π/4-2a))=sin4a

1)tg(3π/4-2a)=tg(π-π/4-2a)=

tg((π-(π/4+2a))=-tg(π/4+2a)=

-(1+tg2a)/(1-tg2a)=

-(cos2a+sin2a)/(cos2a-sin2a)

2)1-cos4a(cos2a-sin2a)/(cos2a+sin2a)=

cos2a+sin2a-(cos²2a-sin²2a)(cos2a-sin2a)/
(cos2a+sin2a)=
(cos2a+sin2a)(1-(cos2a-sin2a)²/(cos2a+sin2a)=
1-(cos2a-sin2a)²=
1-cos²2a+2sin2a•cos2a-
sin²a=1-1+2sin2a•cos2a
=sin4a

sin4a=sin4a