已知sin(a+π/4)=1/3,a∈(π/2,π)则sin2a=
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![已知sin(a+π/4)=1/3,a∈(π/2,π)则sin2a=](/uploads/image/z/956030-14-0.jpg?t=%E5%B7%B2%E7%9F%A5sin%28a%2B%CF%80%2F4%29%3D1%2F3%2Ca%E2%88%88%28%CF%80%2F2%2C%CF%80%29%E5%88%99sin2a%3D)
已知sin(a+π/4)=1/3,a∈(π/2,π)则sin2a=
已知sin(a+π/4)=1/3,a∈(π/2,π)则sin2a=
已知sin(a+π/4)=1/3,a∈(π/2,π)则sin2a=
由sin(a+π/4)=1/3,a∈(π/2,π)即在第二象限,
sinacosπ/4+cosasinπ/4=1/3
(√2/2)sina+(√2/2)cosa=1/3
两边平方得:
(1/2)sin²a+sinacosa+(1/2)cos²a=1/9,
sinacosa=1/9-1/2(∵sin²a+cos²a=1)
2sinacosa=2/9-1=-7/9,
∴sin2a=-7/9.