The range of f(x)=cos2n(x)+sin2n(x) I have f(x)=cos2n(x)+sin2n(x), n∈N,n≥2,x∈R I need to find the range of the function. I took n=2 and I got f(x)=1−12sin2(2x) and the range of this is [1/2,1] Also, for n=3 I got [1/4,1]. How to find the range for n?

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The range of f(x)=cos2n(x)+sin2n(x)  I have f(x)=cos2n(x)+sin2n(x),  n∈N,n≥2,x∈R I need to find the range of the function. I took n=2 and I got f(x)=1−12sin2(2x) and the range of this is [1/2,1]   Also, for n=3 I got [1/4,1].   How to find the range for n?

tainiaadjouctlw · Accepted Answer

With t=cos2(x) such that 0≤t≤1, you bracket tn+(1−t)n The stationary points are the roots of tn−1−(1−t)n−1=0 or t=12 Hence, f(x)∈[21−n,1] Without derivatives: tn+(1−t)n is obviously symmetric around t=12,  and with  s=t+12, (12+s)n+(12−s)n If we develop using the binomial formula, only even degree terms will remain so that the polynomial is monotonic for s&amp;gt;0. Hence the minimum is achieved for s=0, and the maximum for s=12

The range of f(x)=\cos^{2n}(x)+\sin^{2n}(x) I have f(x)=\cos^{2n}(x)+\sin^{2n}(x), \ \ n \in

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