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Uchenic456
@Uchenic456
November 2021
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Помогите решить плз...Желательно с обьяснениями
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sedinalana
Verified answer
1
a)cosacosb+sinasinb-2sinasinb=
cosacosb-sinasinb=cos(a+b)
b)cos
²a-[-cosa*cosa/(ctga*ctga)=cos²a+cos²a*sin²a/cos²a=
=cos²a+sin²a=1
2
sin²68+2sin68*cos38+cos²38+sin²38-2sin38*cos68+cos²68=
(sin²68+cos²68)+(sin²38+cos²38)+2(sin68cos38-cos68sin38)=
=1+1+2sin(68-38)=2+2*sin30=2+2*1/2=2+1=3
3
sina=-√(1-cos²a)=-√(1-25/169)=-√(144/169)=-12/13
sin2a=2sinacosa=2*(-5/13)*(-12/13)=120/169
cos2a=cos²a-sin²a=25/169-144/169=-119/169
4
2*1/2*[cos(37-23)+cos(37+23)]-sin(90-14)=cos14+cos60-cos14=1/2
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Answers & Comments
Verified answer
1a)cosacosb+sinasinb-2sinasinb=cosacosb-sinasinb=cos(a+b)
b)cos²a-[-cosa*cosa/(ctga*ctga)=cos²a+cos²a*sin²a/cos²a=
=cos²a+sin²a=1
2
sin²68+2sin68*cos38+cos²38+sin²38-2sin38*cos68+cos²68=
(sin²68+cos²68)+(sin²38+cos²38)+2(sin68cos38-cos68sin38)=
=1+1+2sin(68-38)=2+2*sin30=2+2*1/2=2+1=3
3
sina=-√(1-cos²a)=-√(1-25/169)=-√(144/169)=-12/13
sin2a=2sinacosa=2*(-5/13)*(-12/13)=120/169
cos2a=cos²a-sin²a=25/169-144/169=-119/169
4
2*1/2*[cos(37-23)+cos(37+23)]-sin(90-14)=cos14+cos60-cos14=1/2