已知sin(5π-α)=√2cos(7/2π+β)和√3cos(-α)=-√2cos(π+β),且0<α<π,0<β<π
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已知sin(5π-α)=√2cos(7/2π+β)和√3cos(-α)=-√2cos(π+β),且0<α<π,0<β<π
求α和β的值.
求α和β的值.
![已知sin(5π-α)=√2cos(7/2π+β)和√3cos(-α)=-√2cos(π+β),且0<α<π,0<β<π](/uploads/image/z/18701832-48-2.jpg?t=%E5%B7%B2%E7%9F%A5sin%EF%BC%885%CF%80-%CE%B1%EF%BC%89%3D%E2%88%9A2cos%EF%BC%887%2F2%CF%80%2B%CE%B2%EF%BC%89%E5%92%8C%E2%88%9A3cos%EF%BC%88-%CE%B1%EF%BC%89%3D-%E2%88%9A2cos%EF%BC%88%CF%80%2B%CE%B2%EF%BC%89%2C%E4%B8%940%EF%BC%9C%CE%B1%EF%BC%9C%CF%80%2C0%EF%BC%9C%CE%B2%EF%BC%9C%CF%80)
此题关键是化简,我用a代替α,b代替β
sin(5π-a-4π)=√2cos(7/2π+b-4π)
sin(π-a)=√2cos(π/2-b)
sina=√2sinb(1)
√3cos(-a)=-√2cos(π+b)
√3cosa=√2cosb
cosa=√(2/3)cosb(2)
(1)²+(2)²
sin²a+cos²a=2sin²b+2/3cos²b
1=4/3sin²b+2/3
4/3sin²b=1/3
sin²b=1/4
0<b<π
所以
sinb>0
sinb=1/2
b=π/6或5π/6
sinb=1/2代入(1)
sina=√2/2
因为0<a<π
所以a=π/4或3π/4
sin(5π-a-4π)=√2cos(7/2π+b-4π)
sin(π-a)=√2cos(π/2-b)
sina=√2sinb(1)
√3cos(-a)=-√2cos(π+b)
√3cosa=√2cosb
cosa=√(2/3)cosb(2)
(1)²+(2)²
sin²a+cos²a=2sin²b+2/3cos²b
1=4/3sin²b+2/3
4/3sin²b=1/3
sin²b=1/4
0<b<π
所以
sinb>0
sinb=1/2
b=π/6或5π/6
sinb=1/2代入(1)
sina=√2/2
因为0<a<π
所以a=π/4或3π/4
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