Решите пожалуйста какие сможете
Тема Тригонометрические преобразования и уравнения
Тема Тригонометрические преобразования и уравнения
More Questions From This User See All
Copyright © 2025 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Verified answer
1) a) cos164°cos106°-sin164°sin106°=cos(164°+106°)=cos270°=0б)
sin67°sin53°-cos67°sin37°=sin67°cos37°-cos67°sin37°=sin(67°-37°)=1/2
2) cos8φ +2sin²4φ=cos²4φ- sin²4φ+2sin²4φ=cos²4φ+sin²4φ=1
3) какое задание???
4) [sin(a+t) -sin(a-t) ]/[cos(a+t) -cos(a-t)]=
[sin(a)cos(t)+ sin(t)cos(a)-sin(a)cos(t)+ sin(t)cos(a)] :
[cos(a)cos(t)-sin(a)sin(t)-cos(a)cos(t)-sin(a)sin(t)]=
2sin(t)cos(a)/[-2sin(a)sin(t)]=-cos(a)/sin(a)=-ctg(a)
5) tg(α)=3√3
√3tg(11π/6-2α)=√3 tg(12π/6-π/6-2α)=-√3 tg(π/6+2α)=
=-√3[tg(π/6)+tg(2α)]/[1-tg(π/6)·tg(2α)]=(///)
tg(2α)=2tg(α)/[1-tg²(α)]=2·3√3 /[1-(3√3 )²]=2·3√3/[1-27]=-3√3/13
(///)=-√3[1/(√3)-3√3/13]/[1+(1/(√3))·3√3/13]=-√3[√3/(3)-3√3/13]/[1+3/13]=
=-√3·√3[1/3-3/13]/[(13+3)/13]=-3[(13-9)/39]/(16/13)=-3/4=-0,75