Trigonometric identities and complex numbers

PedroB
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Homework Statement



Show, using complex numbers, that sin(x)+cos(x)=(√2)cos(x-∏/4)

Homework Equations



cos(x)=(e^(ix)+e^(-ix))/2
sin(x)=(e^(ix)-e^(-ix))/2i

e^ix=cos(x)+isin(x)

The Attempt at a Solution



I was given the hint that sin(x)=Re(-ie^(ix)), but have thus far not been able to determine its usefullness. I have tried squaring the expression, which (after simplification) yields:

1+sin(2x)

but cannot seem to go further. I assume that the crux of the solution lies in fully expressing sin(x)+cos(x) as purely cos in terms of complex exponentials, but everything I try just brings me back to the original expression. Am I missing a fundamental equivalence between either of these trigonometric functions and a complex number? Any help would be greatly appreciated, thank-you in advance.

(Obviously all work done in trying to solve this problem involves the assumption that I do not know the final answer)
 
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PedroB said:

Homework Statement



Show, using complex numbers, that sin(x)+cos(x)=(√2)cos(x-∏/4)

Homework Equations



cos(x)=(e^(ix)+e^(-ix))/2
sin(x)=(e^(ix)-e^(-ix))/2i

e^ix=cos(x)+isin(x)

The Attempt at a Solution



I was given the hint that sin(x)=Re(-ie^(ix)), but have thus far not been able to determine its usefullness. I have tried squaring the expression, which (after simplification) yields:

1+sin(2x)

but cannot seem to go further. I assume that the crux of the solution lies in fully expressing sin(x)+cos(x) as purely cos in terms of complex exponentials, but everything I try just brings me back to the original expression. Am I missing a fundamental equivalence between either of these trigonometric functions and a complex number? Any help would be greatly appreciated, thank-you in advance.

(Obviously all work done in trying to solve this problem involves the assumption that I do not know the final answer)

In addition to your hint, another identity is cosx = Re(eix).

Start with the right side of your identity, √2 cos(x - ##\pi/4##). The identity I show just above can be used to rewrite √2 cos(x - ##\pi/4##) in its exponential form. The rest is fairly straightforward but does take a few steps.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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