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?
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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