已知函数f(x)=asin(ωχ+π/3),g(x)=btan(ωχ-π/3)(ω>0)的最小正周期之和为3π/2,且f(π/2)=g(π/2f(π/4)+√3g(π/4)=1,求f(x)g(x)的解析式求f(x)和g(x)的解析式
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![已知函数f(x)=asin(ωχ+π/3),g(x)=btan(ωχ-π/3)(ω>0)的最小正周期之和为3π/2,且f(π/2)=g(π/2f(π/4)+√3g(π/4)=1,求f(x)g(x)的解析式求f(x)和g(x)的解析式](/uploads/image/z/7309760-32-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dasin%28%CF%89%CF%87%2B%CF%80%2F3%29%2Cg%28x%29%3Dbtan%28%CF%89%CF%87-%CF%80%2F3%EF%BC%89%EF%BC%88%CF%89%EF%BC%9E0%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E4%B9%8B%E5%92%8C%E4%B8%BA3%CF%80%2F2%2C%E4%B8%94f%28%CF%80%2F2%29%3Dg%28%CF%80%2F2f%28%CF%80%2F4%29%2B%E2%88%9A3g%28%CF%80%2F4%EF%BC%89%3D1%2C%E6%B1%82f%28x%29g%28x%29%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%E6%B1%82f%28x%29%E5%92%8Cg%28x%29%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F)
已知函数f(x)=asin(ωχ+π/3),g(x)=btan(ωχ-π/3)(ω>0)的最小正周期之和为3π/2,且f(π/2)=g(π/2f(π/4)+√3g(π/4)=1,求f(x)g(x)的解析式求f(x)和g(x)的解析式
已知函数f(x)=asin(ωχ+π/3),g(x)=btan(ωχ-π/3)(ω>0)的最小正周期之和为3π/2,且f(π/2)=g(π/2
f(π/4)+√3g(π/4)=1,求f(x)g(x)的解析式
求f(x)和g(x)的解析式
已知函数f(x)=asin(ωχ+π/3),g(x)=btan(ωχ-π/3)(ω>0)的最小正周期之和为3π/2,且f(π/2)=g(π/2f(π/4)+√3g(π/4)=1,求f(x)g(x)的解析式求f(x)和g(x)的解析式
由f(x)和g(x)的最小正周期之和为3π/2得
2π/ω +π/ω =3π/2
得ω =2
由f(π/2)=g(π/2)得
-asin(π/3)=-btan(π/3)
得a=2b
f(x)=2bsin(2χ+π/3),g(x)=btan(2χ-π/3)(ω>0)
由f(π/4)+√3g(π/4)=1
得2bCosπ/3+√3bCotπ/3=1
得b=1/2 a=1
f(x)=sin(2χ+π/3),g(x)=1/2tan(2χ-π/3)
f(x)g(x)=1/2sin(2χ+π/3)tan(2χ-π/3)
因为f(x)和g(x)的最小正周期之和为3π/2,
所以ω=2
所以f(x)=asin(2χ+π/3),g(x)=btan(2χ-π/3)
因为f(π/2)=g(π/2)代入得a=2b,
因为f(π/4)+√3g(π/4)=1,代入得a+2b=2,
所以a=1,b=1/2
所以f(x)=sin(2χ+π/3),g(x)=0.5tan(2χ-π/3)<...
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因为f(x)和g(x)的最小正周期之和为3π/2,
所以ω=2
所以f(x)=asin(2χ+π/3),g(x)=btan(2χ-π/3)
因为f(π/2)=g(π/2)代入得a=2b,
因为f(π/4)+√3g(π/4)=1,代入得a+2b=2,
所以a=1,b=1/2
所以f(x)=sin(2χ+π/3),g(x)=0.5tan(2χ-π/3)
所以f(x)g(x)=0.5tan(2χ-π/3)sin(2χ+π/3)
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