Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. \ \

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Consider  S (n) : (cosx+isinx)n= cos(nx) +isin(nx) \ \

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It can be established by mathematical induction for natural numbers, and extended to all integers from there. \ \

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For n ≥ 0, we proceed by mathematical induction. \ \

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n = 0 \ \

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S (0) = cos (0x) + i sin(0x) = 1 +i 0 = 1. \ \

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S (0) is clearly true since \ \

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n = 1 \ \

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S (1) :(cosx+i sinx)1= cos(1x)+isin(1x) = cosx+isinx \ \

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S (1) is clearly true \ \

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n = 2 \ \

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S (2) :(cosx+i sinx)2= cos2x+(isinx)2+2cosx(isinx ) \ \

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= (cos2x - sin2x)+i(2cosxsinx) \ \

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= cos(2x)+isin2x \ \

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S (2) is clearly true \ \

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Now, considering S (n+1): \ \

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S (n+1) = (cosx+isinx)n+1 = (cosx+isinx)n(cosx+isinx) \ \
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= [cos(nx) +isin(nx)](cosx+isinx) \ \
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= cos(nx)cosx+icos(nx)sinx+isin(nx)cosx+i2sin(nx)sinx \ \
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= [cos(nx)cosx−sin(nx)sinx] +i[cos(nx)sinx+ sin(nx)cosx] \ \
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= cos(nx+x) +isin(nx+x) \ \
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= cos[(n+ 1)x] +isin[(n+ 1)x]. \ \
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We deduce that S(n) implies S(n+1). \ \
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By the principle of mathematical induction it follows that the result is true for all natural numbers n. \ \
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(cosx+isinx)n= cos(nx) +isin(nx) is proved by mathematical induction \ \