Question

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

Answer 1

$1+\frac{cos4\alpha}{tg(\frac{3\pi}{4}-2\alpha)}=1+\frac{2cos4\alpha (1-tg^2(\frac{3\pi}{4}-2\alpha))}{2tg(\frac{3\pi}{4}-2\alpha)(1-tg^2(\frac{3\pi}{4}-2\alpha))}=1+\frac{2cos4\alpha}{tg(\frac{3\pi}{2}-4\alpha)(1-tg^2(\frac{3\pi}{4}-2\alpha))}=1+\frac{2cos4\alpha}{ctg4\alpha(1-tg^2(\frac{3\pi}{4}-2\alpha))}=1+\frac{2sin4\alpha}{1-tg^2(\frac{3\pi}{4}-2\alpha)}=1+\frac{2sin4\alpha cos^2(\frac{3\pi}{4}-2\alpha)}{cos^2(\frac{3\pi}{4}-2\alpha)-sin^2(\frac{3\pi}{4}-2\alpha)}=$$1+\frac{2sin4\alpha cos^2(\frac{3\pi}{4}-2\alpha)}{cos2(\frac{3\pi}{4}-2\alpha)}=1+\frac{sin4\alpha (1+cos2(\frac{3\pi}{4}-2\alpha))}{cos(\frac{3\pi}{2}-4\alpha)}=1+\frac{sin4\alpha (1+cos2(\frac{3\pi}{4}-2\alpha))}{-sin(4\alpha)}=1-(1+cos(\frac{3\pi}{2}-4\alpha))=1- (1-sin4\alpha)=1- 1+sin4\alpha=sin4\alpha$

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