How do you prove #cos(2x) = [cos(x)]^2  [sin(x)]^2# and #sin(2x) = 2sin(x)cos(x)# using Euler's Formula: #e^(ix) = cos(x) + isin(x)#?
Use Euler's Formula to evaluate
To get: Equitable real and imaginary parts
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To prove ( \cos(2x) = \cos^2(x)  \sin^2(x) ) and ( \sin(2x) = 2\sin(x)\cos(x) ) using Euler's formula ( e^{ix} = \cos(x) + i\sin(x) ), we start by expressing ( \cos(2x) ) and ( \sin(2x) ) in terms of Euler's formula:

For ( \cos(2x) ): [ \cos(2x) = \frac{e^{i(2x)} + e^{i(2x)}}{2} ]

For ( \sin(2x) ): [ \sin(2x) = \frac{e^{i(2x)}  e^{i(2x)}}{2i} ]
Now, we use Euler's formula to express ( e^{i(2x)} ) and ( e^{i(2x)} ):
[ e^{i(2x)} = \cos(2x) + i\sin(2x) ] [ e^{i(2x)} = \cos(2x) + i\sin(2x) = \cos(2x)  i\sin(2x) ]
Substituting these expressions back into ( \cos(2x) ) and ( \sin(2x) ):

For ( \cos(2x) ): [ \cos(2x) = \frac{\cos(2x) + i\sin(2x) + \cos(2x)  i\sin(2x)}{2} ] [ = \frac{2\cos(2x)}{2} ] [ = \cos^2(x)  \sin^2(x) ]

For ( \sin(2x) ): [ \sin(2x) = \frac{\cos(2x) + i\sin(2x)  (\cos(2x)  i\sin(2x))}{2i} ] [ = \frac{2i\sin(2x)}{2i} ] [ = 2\sin(x)\cos(x) ]
Therefore, using Euler's formula, we have proven that ( \cos(2x) = \cos^2(x)  \sin^2(x) ) and ( \sin(2x) = 2\sin(x)\cos(x) ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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