Tuesday, 27 May 2014

trigonometry - How to prove by induction that |sin(nx)|leqn|sinx|?



Here n belongs to natural numbers. Firstly, I proved the relation by putting n=1 . Then, taking |sin(mx)|m|sinx|

true, I had to prove |sin(m+1)x|(m+1)|sinx|
Now, here I got stuck. How to prove it?? Please help.


Answer



The inductive step: Using the triangle inequality and the fact that sin and cos function are bounded by 1 we get



|sin((n+1)x)|=|sin(nx)cosx+cos(nx)sin(x)||sin(nx)|cos(x)|+|cos(nx)||sin(x)||sin(nx)|+|sin(x)|n|sin(x)|+|sin(x)|=(n+1)|sin(x)|


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