What letter is missing here?$$(\cos(x)+i\sin(x))^?=(e^{ix})^?=e^{i?x}=\cos(nx)+i\sin(nx).$$

Question

BlackTom AI · Accepted Answer

To analyze this identity, start from the complex exponential form of a complex number on the unit circle, written as cos(x) + i sin(x) = e^{ix}. When you raise this to a power, you apply the exponent to the exponent: (e^{ix})^k = e^{ikx}. Using Euler's formula, e^{ikx} = cos(kx) + i sin(kx). Therefore, (cos x + i sin x)^k becomes cos(kx) + i sin(kx).

Comparing this to the target expression cos(nx) + i sin(nx), you need the inner multiplier of x to be the same, i.e., kx must equal nx for all x. This forces k to be n, a constant independent of x. Consequently, the missing letter that makes the identity hold for all x is the exponent that when multiplied by x yields nx inside the cosine and sine.

A common point of confusion is thinking the exponent should be 1 or some other number; however, unless the exponent scales x to nx, the right-hand side cannot match cos(nx) + i sin(nx) for all x. Remember that de Moivre’s formula generalizes exactly with the exponent corresponding to the multiple of x in the argument of the trigonometric functions.

In summary, the relationship relies on the property (cos x + i sin x)^k = cos(kx) + i sin(kx), which aligns with cos(nx) + i sin(nx) only when k = n. Since the question asks for the missing letter that makes the equality hold, the variable that completes this pattern must be the same n that appears in nx.

What letter is missing here?\[(\cos(x)+i\sin(x))^?=(e^{ix})^?=e^{i?x}=\cos(nx)+i\sin(nx).\]简答题

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