Consider the function f(x)=sin⁡(x)sin⁡((2k+1)x) with k a positive integer. It seems that f is strictly increasing in [0,π2(2k+1)].Is there some easy proof of this monotonicity property that does not invole differentiation (through suitable inequalities)?

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Consider the function  f(x)=sin⁡(x)sin⁡((2k+1)x)  with k a positive integer. It seems that f is strictly increasing in [0,π2(2k+1)].Is there some easy proof of this monotonicity property that does not invole differentiation (through suitable inequalities)?

Addisyn Thompson · Accepted Answer

sin⁡(x)sin⁡((k+1)x)=sin⁡(x)sin⁡(kx)cos⁡(x)+cos⁡(kx)sin⁡(x)=1sin⁡(kx)sin⁡(x)cos⁡(x)+cos⁡(kx)Consequently:sin⁡(x)sin⁡(kx)&amp;nbsp;&amp;nbsp;increasing⇔sin⁡(kx)sin⁡(x)&amp;nbsp;&amp;nbsp;decreasing⇒sin⁡(kx)sin⁡(x)cos⁡(x)&amp;nbsp;&amp;nbsp;decreasing⇔sin⁡(kx)sin⁡(x)cos⁡(x)+cos⁡(kx)&amp;nbsp;&amp;nbsp;decreasing⇔1sin⁡(kx)sin⁡(x)cos⁡(x)+cos⁡(kx)&amp;nbsp;&amp;nbsp;increasing⇔sin⁡(x)sin⁡((k+1)x)&amp;nbsp;&amp;nbsp;increasingLet P(k) denote the proposition that&amp;nbsp;sin⁡(x)sin⁡(kx)&amp;nbsp;is increasing. So, we&#039;ve demonstrated that&amp;nbsp;P(k)→P(k+1). So, we may use&amp;nbsp;P(1)(based on the second response) to inductively prove that&amp;nbsp;sin⁡(x)sin⁡((2k+1)x)&amp;nbsp;is increasing over&amp;nbsp;[0,π2(2k+1)]&amp;nbsp;for all&amp;nbsp;k∈N

kumewekwah0 · Accepted Answer

You have&amp;nbsp;sin⁡2x=2sin⁡xcos⁡x&amp;nbsp;and&amp;nbsp;sin⁡3x=sin⁡xcos⁡2x+cos⁡xsin⁡2x=sin⁡x(cos⁡2x+2cos2x)&amp;nbsp;and similar formulas for higher&amp;nbsp;sin⁡(kx). Then&amp;nbsp;sin⁡(3x)sin⁡x=cos⁡2x+2cos2x&amp;nbsp;If the RHS is positive and decreasing in x&amp;nbsp;cos⁡2x&amp;gt;0. That gives you an interval where&amp;nbsp;sin⁡xsin⁡3x&amp;nbsp;is increasing.&amp;nbsp;It is undoubtedly much more difficult than distinguishing to make this into a proof for universal k, but it ought to be achievable.

Consider the function f(x)=\frac{\sin(x)}{\sin((2k+1)x)} with k a positive integer. It seems that

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