Finding the Real Part of a Complex Number Power

The problem asks for the real part of the complex number expression $\left(\sin x + i \cos x\right)^3$. We can solve this using methods involving complex numbers and trigonometry, such as converting to polar form and using De Moivre's Theorem, or by expanding the expression using the binomial theorem.

Method 1: Using De Moivre's Theorem

De Moivre's Theorem is useful for raising complex numbers in polar form to a power. The polar form of a complex number is $r\left(\cos \theta + i \sin \theta\right)$, where $r$ is the modulus and $\theta$ is the argument.

The given expression is $\sin x + i \cos x$. This is not exactly in the standard polar form $\cos \theta + i \sin \theta$. However, we can rewrite it using trigonometric identities:

• We know that $\cos\left(\frac{\pi}{2} - x\right) = \sin x$
• And $\sin\left(\frac{\pi}{2} - x\right) = \cos x$

So, we can write $\sin x + i \cos x$ as $\cos\left(\frac{\pi}{2} - x\right) + i \sin\left(\frac{\pi}{2} - x\right)$.

Now the complex number is in the form $\cos \theta + i \sin \theta$, where the modulus $r=1$ and the argument $\theta = \frac{\pi}{2} - x$.

According to De Moivre's Theorem, for a complex number $r\left(\cos \theta + i \sin \theta\right)$ raised to the power $n$, the result is $r^n\left(\cos n\theta + i \sin n\theta\right)$.

In our case, $r=1$, $\theta = \frac{\pi}{2} - x$, and $n=3$. Applying De Moivre's Theorem:

$\left(\sin x + i \cos x\right)^3 = \left(\cos\left(\frac{\pi}{2} - x\right) + i \sin\left(\frac{\pi}{2} - x\right)\right)^3$

$= 1^3 \left(\cos\left(3\left(\frac{\pi}{2} - x\right)\right) + i \sin\left(3\left(\frac{\pi}{2} - x\right)\right)\right)$

$= \cos\left(\frac{3\pi}{2} - 3x\right) + i \sin\left(\frac{3\pi}{2} - 3x\right)$

Now, we use angle subtraction identities for cosine and sine:

• $\cos\left(\frac{3\pi}{2} - \alpha\right) = \cos\left(\frac{3\pi}{2}\right)\cos\alpha + \sin\left(\frac{3\pi}{2}\right)\sin\alpha$
• $\sin\left(\frac{3\pi}{2} - \alpha\right) = \sin\left(\frac{3\pi}{2}\right)\cos\alpha - \cos\left(\frac{3\pi}{2}\right)\sin\alpha$

We know that $\cos\left(\frac{3\pi}{2}\right) = 0$ and $\sin\left(\frac{3\pi}{2}\right) = -1$. Substituting these values with $\alpha = 3x$:

• $\cos\left(\frac{3\pi}{2} - 3x\right) = 0 \cdot \cos\left(3x\right) + (-1) \cdot \sin\left(3x\right) = -\sin\left(3x\right)$
• $\sin\left(\frac{3\pi}{2} - 3x\right) = (-1) \cdot \cos\left(3x\right) - 0 \cdot \sin\left(3x\right) = -\cos\left(3x\right)$

So, $\left(\sin x + i \cos x\right)^3 = -\sin\left(3x\right) + i \left(-\cos\left(3x\right)\right)$.

The real part of this expression is the term without $i$, which is $-\sin\left(3x\right)$.

Method 2: Using Binomial Expansion

Alternatively, we can expand the expression $\left(\sin x + i \cos x\right)^3$ using the binomial theorem $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Here, $a = \sin x$ and $b = i \cos x$.

$\left(\sin x + i \cos x\right)^3 = \left(\sin x\right)^3 + 3\left(\sin x\right)^2 \left(i \cos x\right) + 3\left(\sin x\right) \left(i \cos x\right)^2 + \left(i \cos x\right)^3$

$= \sin^3 x + 3i \sin^2 x \cos x + 3\sin x \left(i^2 \cos^2 x\right) + \left(i^3 \cos^3 x\right)$

Substitute $i^2 = -1$ and $i^3 = i \cdot i^2 = i \cdot (-1) = -i$:

$= \sin^3 x + 3i \sin^2 x \cos x + 3\sin x \left(-1 \cdot \cos^2 x\right) + \left(-i \cos^3 x\right)$

$= \sin^3 x + 3i \sin^2 x \cos x - 3\sin x \cos^2 x - i \cos^3 x$

Group the real terms (terms without $i$) and the imaginary terms (terms with $i$):

Real Part: $\sin^3 x - 3\sin x \cos^2 x$

Imaginary Part: $3i \sin^2 x \cos x - i \cos^3 x = i \left(3\sin^2 x \cos x - \cos^3 x\right)$

We are interested in the real part: $\sin^3 x - 3\sin x \cos^2 x$.

Let's simplify this expression using the identity $\cos^2 x = 1 - \sin^2 x$:

Real Part $= \sin^3 x - 3\sin x \left(1 - \sin^2 x\right)$

$= \sin^3 x - 3\sin x + 3\sin x \sin^2 x$

$= \sin^3 x - 3\sin x + 3\sin^3 x$

$= 4\sin^3 x - 3\sin x$

Recall the triple angle identity for sine: $\sin 3x = 3\sin x - 4\sin^3 x$.

Comparing our real part $4\sin^3 x - 3\sin x$ with the formula for $\sin 3x$, we see that:

$4\sin^3 x - 3\sin x = -(3\sin x - 4\sin^3 x) = -\sin 3x$.

Both methods yield the same result for the real part.

Conclusion on the Real Part

The real part of $\left(\sin x + i \cos x\right)^3$ is $-\sin 3x$.

Revision Table: Complex Numbers and Trigonometry

Additional Information: Complex Number Manipulation

Working with complex numbers often involves converting between rectangular form ($a+bi$) and polar form ($r(\cos \theta + i \sin \theta)$). The choice of method for problems involving powers usually depends on the starting form of the complex number and the power involved.

• When the complex number is already in or easily convertible to polar form $r(\cos \theta + i \sin \theta)$, De Moivre's Theorem is typically the most straightforward method for finding powers. It avoids lengthy expansions.
• When dealing with lower powers or needing the expanded form for other purposes, the binomial theorem can be used. However, this requires careful handling of powers of $i$ ($i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, and the pattern repeats).
• Trigonometric identities are crucial for simplifying expressions obtained from both methods, especially when converting between forms or simplifying arguments like $\frac{3\pi}{2} - 3x$.

Understanding how to manipulate complex numbers and use trigonometric identities is key to solving problems like finding the real or imaginary parts of complex expressions raised to powers.