设函数f(x)=3sin(wx+π/6),w>0,x属于(负无穷大,正无穷大),且以π/2为最小正周期若f(a/4+π/12)=9/5,则sina的值为什么为正负4/5?
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![设函数f(x)=3sin(wx+π/6),w>0,x属于(负无穷大,正无穷大),且以π/2为最小正周期若f(a/4+π/12)=9/5,则sina的值为什么为正负4/5?](/uploads/image/z/6877757-29-7.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D3sin%28wx%2B%CF%80%2F6%29%2Cw%3E0%2Cx%E5%B1%9E%E4%BA%8E%EF%BC%88%E8%B4%9F%E6%97%A0%E7%A9%B7%E5%A4%A7%2C%E6%AD%A3%E6%97%A0%E7%A9%B7%E5%A4%A7%EF%BC%89%2C%E4%B8%94%E4%BB%A5%CF%80%2F2%E4%B8%BA%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E8%8B%A5f%28a%2F4%2B%CF%80%2F12%29%3D9%2F5%2C%E5%88%99sina%E7%9A%84%E5%80%BC%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%BA%E6%AD%A3%E8%B4%9F4%2F5%3F)
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设函数f(x)=3sin(wx+π/6),w>0,x属于(负无穷大,正无穷大),且以π/2为最小正周期若f(a/4+π/12)=9/5,则sina的值为什么为正负4/5?
设函数f(x)=3sin(wx+π/6),w>0,x属于(负无穷大,正无穷大),且以π/2为最小正周期
若f(a/4+π/12)=9/5,则sina的值为什么为正负4/5?
设函数f(x)=3sin(wx+π/6),w>0,x属于(负无穷大,正无穷大),且以π/2为最小正周期若f(a/4+π/12)=9/5,则sina的值为什么为正负4/5?
f(x)的最小正周期是T=2π/w=π/2
得到w=4
所以f(x)=3sin(4x+π/6)
所以f(a/4+π/12)=3sin(a+π/2)=3cosa=9/5
所以cosa=3/5
故sina=4/5或-4/5
由f(x)最小正周期为π/2,w>0得: w=(2π)/(π/2)=4
f(a/4+π/12)=3sin(a+π/3+π/6)=3sin(π/2 +a)=3cos(-a)=3cosa=9/5
cosa=3/5
所以sina=±√(1-9/25=±4/5
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