Question

Помогите пожалуйста с тригонометрией 10 класс, задание во вложении.

Answer 1

1
а)sin(2π+π/6)=sinπ/6=1/2
б)-tg(2π-π/6)=tgπ/6=√3/3
в)cosπ+ctg(π+π/3)=-1+√3/3
г)tgπ/4*(-ctgπ/4)+cos3π/2*sinπ/2=-1+0*1=-1
д)sin(360+45)+cos(180+45)*tg(180+45)=sin45-cos45*tg45=√2/2-√2/2*1=0
2
sin²t+cos²t/(ctgt*tgt)=sin²t+cos²t=1
3
1)cost=1/2
t=+-π/3+2πn,n∈z
2)cos(π/2+t)=-√3/2
-sint=-√3/2
sint=√3/2
t=(-1)^n*π/3+πn,n∈z
4
cost=-4/5
sint=-√(1-cos²t)=-√(1-16/25)=-√(9/25)=-3/5
tgt=sint/cost=-3/5:(-4/5)=3/5*5/4=3/4
ctgt=1/tgt=1:3/4=4/3

Answer 2

1.
а) sin13π/6 = sin (2π+π/6 ) = sin π/6 =1/2.
б) tq(-11π/6) = - tq11π/6 = - tq(2π -π/6 ) = tq π/6= √3 / 3.
в) cos4π +ctq 4π/3 = 1 +ctq(π +π/3 ) =  1 +ctq π/3  =1 + √3 / 3.
г) tq π/4 *ctq(- π/4) +cos(3π/2)* sinπ/2 =  -tq π/4 *ctq π/4 +0* sinπ/2  = -1.
д) sin405° +cos225°*tq225° = sin(360°+45°) +sin (180°+ 45°)  =sin45° -sin45°=0.
2.
а) sin²t  - cos²t / ( ctqt *tq(-t) )=sin²t  - cos²t /(- ctqt *tqt  ) =.sin²t  + cos²t  =1. 
б)
3.
а) cost   =1 / 2 . ⇒
 t = ± π/3 +2πn , n∈Z. 
б) cos(π/2 +t ) = -√3 /2   ⇒ - sin t  = -√3 /2 ⇔ sin t  = √3 /2 ⇒
 t =(-1)^ n * π/3 +πn , n∈Z. 
4. 
 cost  =- 4/5 ; π < t < 3π/2 .
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 sint - ?  ; tqt ? ;  ctqt - ?
sint  = -√(1-cos²t) = -√(1-(4/5)= -3 /5 ; tqt =  sint  /cost  =(-3/5)/(-4/5) =3/4.
ctqt  =1/tqt = 4/3 .
5.
sin(2π -t) -cos(3π/2 +t)  +1 = 0 ;
-sint  -sint  +1 =0 ⇔sint =1/2⇒   t = (-1)^n  *( π/6) + πn ,n ∈Z.
* * *   { π/6 + 2πn  ;  (π -π/6) +2πn  } ⇔     { π/6 + 2πn  ;  5π /6 +2πn  }  * * *