Question

Тригонометрия, хэээлп

Answer 1

1)cos²a=1:(1+tg²a)=1:(1+4)=1/5⇒cosa=1/√5
sina=√1-1/5=2/√5
(4sina-3cosa)/5sina=(8/√5-3/√5):5/√5=5/√5:10/√5=1/2
3)sin²a=1: (1+ctg²a)=1:(1+9)=1/10⇒sina=1/√10
cosa=√1-1/10=3/√10
(4sina-3cosa)/5sina=(4/√10-9/√10):5√10=-5√10:5/√10=-1

Answer 2

$\frac{4sin \alpha -3cos \alpha }{5sin \alpha } = \frac{ \frac{4sin \alpha }{cos \alpha }- \frac{3cos \alpha }{cos \alpha } }{ \frac{5sin \alpha }{cos \alpha } } = \frac{4tg \alpha -3}{5tg \alpha } \\ a)tg \alpha =2;\frac{4tg \alpha -3}{5tg \alpha } =\frac{4*2 -3}{5*2 } = \frac{5}{10} =0,5 \\ \frac{4sin \alpha -3cos \alpha }{5sin \alpha } = \frac{ \frac{4sin \alpha }{sin \alpha }- \frac{3cos \alpha }{sin \alpha } }{ \frac{5sin \alpha }{sin \alpha } } = \frac{4-3ctg \alpha }{5}$
$b)ctg \alpha =3; \frac{4-3ctg \alpha }{5}= \frac{4-3*3 }{5}= \frac{-5}{5}=-1$