Question

Упростите выражение: а) sin⁡(π+α)+cos⁡(3π/2-α) ; б) tg⁡(π/2+α)-ctg⁡(2π-α);

Answer 1

$a)Sin(\pi+\alpha)+Cos(\frac{3\pi }{2}-\alpha)=-Sin\alpha -Sin\alpha=-2Sin\alpha\\\\b)tg(\frac{\pi }{2}+\alpha)-Ctg(2\pi-\alpha)=-Ctg\alpha +Ctg\alpha=0$

Answer 2

a) -sina-sina=-2sina

b) ctga-ctga=0

Формулы приведения

Answer 3

а)

sin(π+α) = sin πcos α + cos πsin α = 0*cos α + (-1)*sin α = -sin α

cos(3π/2 - α) = cos 3π/2cos α + sin 3π/2sin α = 0*cos α + (-1)*sin α = -sin α

Таким образом,

sin(π+α)+cos(3π/2-α) = -sin α - sin α = -2sin α

б)

tg(π/2+α) = cot α (т.к. tg(π/2+α) = 1/tgα = ctgα)

ctg(2π-α) = -tan α (т.к. ctg(2π-α) = 1/tg(π-α) = 1/(-tan α) = -cot α)

Таким образом,

tg(π/2+α)-ctg(2π-α) = cot α - (-cot α) = 2cot α