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danilv1
@danilv1
July 2022
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Помогите вычислить:
cos510° - sin1200° - tg1005°
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sedinalana
Verified answer
Сos(360+150)-sin(1080+120)-tg(900+105)=cos150-sin120-tg105=
=cos(180-30)-sin(90+30)-tg(60+45)=
=-cos30-cos30-(tg60+tg45)/(1-tg60*tg45)=-2*√3/2-(√3+1)/(1-√3)=
=-√3-(√3+1)(√3+1)/[(1-√3)(1+√3)=-√3-(3+2√3+1)/(1-3)=
=-√3+(4+2√3)/2=-√3+2+√3=2
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Answers & Comments
Verified answer
Сos(360+150)-sin(1080+120)-tg(900+105)=cos150-sin120-tg105==cos(180-30)-sin(90+30)-tg(60+45)=
=-cos30-cos30-(tg60+tg45)/(1-tg60*tg45)=-2*√3/2-(√3+1)/(1-√3)=
=-√3-(√3+1)(√3+1)/[(1-√3)(1+√3)=-√3-(3+2√3+1)/(1-3)=
=-√3+(4+2√3)/2=-√3+2+√3=2