已知a属于(派/2,派),且sina/2+cosa/2=2根号3/3(1),求cosa (2)若sin(a+b)=-3/5,b属于(0,派/2),求sinb
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![已知a属于(派/2,派),且sina/2+cosa/2=2根号3/3(1),求cosa (2)若sin(a+b)=-3/5,b属于(0,派/2),求sinb](/uploads/image/z/3775712-32-2.jpg?t=%E5%B7%B2%E7%9F%A5a%E5%B1%9E%E4%BA%8E%28%E6%B4%BE%2F2%2C%E6%B4%BE%29%2C%E4%B8%94sina%2F2%2Bcosa%2F2%3D2%E6%A0%B9%E5%8F%B73%2F3%281%29%2C%E6%B1%82cosa+%282%29%E8%8B%A5sin%28a%2Bb%29%3D-3%2F5%2Cb%E5%B1%9E%E4%BA%8E%280%2C%E6%B4%BE%2F2%29%2C%E6%B1%82sinb)
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已知a属于(派/2,派),且sina/2+cosa/2=2根号3/3(1),求cosa (2)若sin(a+b)=-3/5,b属于(0,派/2),求sinb
已知a属于(派/2,派),且sina/2+cosa/2=2根号3/3
(1),求cosa (2)若sin(a+b)=-3/5,b属于(0,派/2),求sinb
已知a属于(派/2,派),且sina/2+cosa/2=2根号3/3(1),求cosa (2)若sin(a+b)=-3/5,b属于(0,派/2),求sinb
1、sina/2+cosa/2=2√3/3
∴(sina/2+cosa/2)²=(2√3/3)²
(sina/2)²+(cosa/2)²+2sina/2cosa/2=4/3
1+sina=4/3
∴sina=1/3
∴cosa=±√(1-1/9)=±2√2/3
∵a属于( π /2,π ),
∴cosa=-2√2/3
2、∵a属于( π /2,π ),b属于(0,π/2),
∴a+b属于(π/2,3π/2)
∴cos(a+b)=-√(1-9/25)=-4/5
∴sin(a+b)=sinacosb+cosasinb=-3/5
即1/3×cosb-2√2/3×sinb=-3//5……(1)
cos(a+b)=cosacosb-sinasinb=-4/5
即-2√2/3×cosb-1/3×sinb=-4/5……(2)
解sinb=2√2/5