Question

Срочно!!! Решите номер 2,4,5 (2 и 4 ОБЯЗАТЕЛЬНО 5если сможете)

Answer 1

$8Sin2 \beta Cos2 \beta Cos4 \beta =4*(2Sin2 \beta Cos2 \beta )Cos4 \beta =4Sin4 \beta Cos4 \beta =$$2*(2Sin4 \beta Cos4 \beta) = 2Sin8 \beta \\\\ \frac{Sin(2 \alpha +3 \beta )-Sin(2 \alpha -3 \beta )}{Cos(2 \alpha -3 \beta )-Cos(2 \alpha +3 \beta )}= \frac{Sin2 \alpha Cos3 \beta +Cos2 \alpha Sin3 \beta -Sin2 \alpha Cos3 \beta +Cos2 \alpha Sin3 \beta }{Cos2 \alpha Cos3 \beta +Sin2 \alpha Sin3 \beta -Cos2 \alpha Cos3 \beta +Sin2 \alpha Sin3 \beta }$$= \frac{2Cos2 \alpha Sin3 \beta }{2Sin2 \alpha Sin3 \beta } =Ctg2 \alpha$

$tgx = \frac{1}{2} \\\\tg ^{2} x+1= \frac{1}{Cos ^{2}x } \\\\Cos ^{2}x= \frac{1}{tg ^{2}x+1 }\\\\Cos2x=2Cos ^{2}x-1=2* \frac{1}{( \frac{1}{2}) ^{2} +1 }-1= 2* \frac{4}{5} -1= \frac{3}{5} \\\\Cos4x=2Cos ^{2}2x-1 =[tex]2*( \frac{3}{5}) ^{2}-1=2* \frac{9}{25}-1= \frac{18}{25}-1=- \frac{7}{25}\\\\Sin4x= \sqrt{1-Cos ^{2}4x } = \sqrt{1- \frac{49}{625} }= \sqrt{ \frac{576}{625} } = \frac{24}{25}$

Answer 2

2
8sin2bcos2bcos4b=4sin4bcos4b=2sin8b
3
sina=√(1-cos²a)=√(1-25/169)=√144/169)=12/13
tga=sina/cosa=12/13:(-5/13)=-12/13*13/5=-12/5
ctga=1/tga=1:(-12/5)=-5/12
4
(sin2acos3b+cos2asin3b-sin2acos3b+cos2asin3b)/(cos2acos3b+sin2asin3b-
cos2acos3b+sin2asin3b)=2cos2asin3b/2sin2asin3b=cos2a/sin2a=ctg2a
5
tgx=1/2
cos²x=1:(1+tg²x)=1:(1+1/4)=1:5/4=4/5
cos2x=2cos²x-1=2*4/5-1=3/5
sin2x=√(1-cos²2x)=√(1-9/25)=√(16/25)=4/5
sin4x=2sin2x*cos2x=2*3/5*4/5=24/25