已知cos(30°+α)=3/5,α∈(0,π) 求cos(2α+105°)的值
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已知cos(30°+α)=3/5,α∈(0,π) 求cos(2α+105°)的值
![已知cos(30°+α)=3/5,α∈(0,π) 求cos(2α+105°)的值](/uploads/image/z/17140188-12-8.jpg?t=%E5%B7%B2%E7%9F%A5cos%2830%C2%B0%2B%CE%B1%29%3D3%2F5%2C%CE%B1%E2%88%88%280%2C%CF%80%29+%E6%B1%82cos%282%CE%B1%2B105%C2%B0%EF%BC%89%E7%9A%84%E5%80%BC)
cos(30°+α)=3/5>0 ,α∈(0,180°) .
(30°+α)∈(0,180°) sin(30°+α)=4/5
cos(60°+2α)=(cos(30°+α))^2-1=-7/25
sin(60°+2α)=2sin(30°+α)*cos(30°+α)=12/25
cos(2α+105°)=cos(2α+60°+45°)=(√2/2)cos(60°+2α)-(√2/2)sin(60°+2α)
=(√2/2)[cos(60°+2α)-sin(60°+2α)]=(√2/2)[-7/25-12/25]=-19√2/50
本题主要就是对倍角公式的运用!
(30°+α)∈(0,180°) sin(30°+α)=4/5
cos(60°+2α)=(cos(30°+α))^2-1=-7/25
sin(60°+2α)=2sin(30°+α)*cos(30°+α)=12/25
cos(2α+105°)=cos(2α+60°+45°)=(√2/2)cos(60°+2α)-(√2/2)sin(60°+2α)
=(√2/2)[cos(60°+2α)-sin(60°+2α)]=(√2/2)[-7/25-12/25]=-19√2/50
本题主要就是对倍角公式的运用!
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