设函数f(x)=3sin(wx+π/6),w>0,x属于R,且以π/2为最小正周期 (1)求f(0)? (2)求f(x)的解析式?
设函数f(x)=3sin(wx+π/6),w>0,x属于R,且以π/2为最小正周期 (1)求f(0)? (2)求f(x)的解析式?
(3)已知f(a/4+π/12)=9/5,求sina的值
a为α
(3)已知f(a/4+π/12)=9/5,求sina的值
a为α
数学人气:775 ℃时间:2019-08-18 16:47:24
优质解答
f(0)=3sin(w0+π/6),=3sin(π/6)=3/2,π/2为最小正周期 ∴根据公式T=2π/w ∴W=4∴f(x)=3sin(4x+π/6)第三问f(a/4+π/12)=9/5∴f(a/4+π/12)=3sin(4(a/4+π/12)+π/6)=3sin(a+π/3+π/6)=3cos(a)=9/5所以cosa=3/5根...
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