已知函数f(X)=(sin2x+cos2x+1)/2cosx,求函数值域
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先化简
f(x)
=(sin2x+cos2x+1)/2cosx
=(2sinxcosx+2(cosx)^2-1+1)/(2cosx)
=sinx+cosx
=√2(√2/2sinx+√2/2cosx)
=√2(sinxcosπ/4+sinπ/4cosx)
=√2sin(x+π/4)
-1<=sin(x+π/4)<=1
所以-√2<=f(x)<=√2
又因为cosx≠0，即x≠kπ
sin(x+π/4)≠±√2/2
f(x)≠±1
所以-√2<=f(x)<=√2，且f(x)≠±1
f(x)
=(sin2x+cos2x+1)/2cosx
=(2sinxcosx+2(cosx)^2-1+1)/(2cosx)
=sinx+cosx
=√2(√2/2sinx+√2/2cosx)
=√2(sinxcosπ/4+sinπ/4cosx)
=√2sin(x+π/4)
-1<=sin(x+π/4)<=1
所以-√2<=f(x)<=√2
又因为cosx≠0，即x≠kπ
sin(x+π/4)≠±√2/2
f(x)≠±1
所以-√2<=f(x)<=√2，且f(x)≠±1
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