Question

Решите тождество и 2 примера , пожалуйста .

Answer 1

1) tg 55 - tg 35 = 2tg 20
sin (55 - 35) : (cos 55 · cos 35) = 2tg 20
sin 20 : [0.5(cos (55 -35) + cos (55 + 35)] = 2tg 20
2sin 20 : (cos 20 + cos 90) = 2tg 20
2sin 20 : cos 20 = 2tg 20
2tg 20 ≡ 2tg 20

2) (2sin α - sin 2α) : (2sin α + sin 2α) =
= (2sin α - 2sin α ·cos α) : (2sin α + 2sin α · cos α) =
= [2sin α (1 - cos α)] : [2sin α (1 + cos α) =
= (1 - cos α) : (1 + cos α) =
= 2sin² 0.5α : 2cos² 0.5α =
= 2tg² 0.5α

3) 1 - tg π/16 · tg 3π/16) : (tg π/16 + tg 3π/16) =
= (tg π/16 + tg 3π/16) : [(tg π/16 + tg 3π/16) · tg (π/16 + 3π/16) =
= 1 : tg 4π/16 = 1 : tg π/4 = 1 : 1 = 1

Answer 2

Tg55° - tg 35° = $\frac{sin(55 ^{o}+35 ^{o} ) }{cos55 ^{o}cos35 ^{o} } = \frac{sin20 ^{o} }{ \frac{1}{2}(cos20 ^{o}+cos90 ^{o}) } = \frac{2sin20 ^{o} }{cos20 ^{o} } =2tg20 ^{o}$


$\frac{2sin \alpha -sin2 \alpha }{2sin \alpha +sin2 \alpha } = \frac{2sin \alpha -2sin \alpha cos \alpha }{sin \alpha +2sin \alpha cos \alpha } = \frac{2sin \alpha (1-cos \alpha )}{2sin \alpha (1+cos \alpha )} = \frac{1-cos \alpha }{1+cos \alpha } = \frac{2sin ^{2} \frac{ \alpha }{2} }{2cos ^{2} \frac{ \alpha }{2} } =$

$=\frac{sin ^{2} \frac{ \alpha }{2} }{cos ^{2} \frac{ \alpha }{2} } =tg ^{2} \frac{ \alpha }{2}$


$\frac{1-tg \frac{ \pi }{16}tg \frac{3 \pi }{16} }{tg \frac{ \pi }{16}+ tg\frac{3 \pi }{16} } =( \frac{tg \frac{ \pi }{16}+ tg\frac{3 \pi }{16}}{1-tg \frac{ \pi }{16}tg \frac{3 \pi }{16}} ) ^{-1} =(tg( \frac{ \pi }{16} + \frac{3 \pi }{16} )) ^{-1} =(tg \frac{ \pi }{4} ) ^{-1} =1 ^{-1} =1$