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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