Finding the Real Part of a Complex Expression

We are asked to find the real part of the complex number given by the expression \((sin x + i cos x)^4\). This involves raising a complex number to a certain power.

Converting the Complex Number to Polar Form

To simplify raising a complex number to a power, it is often easiest to convert it into polar form, which is typically written as \(r(\cos \theta + i \sin \theta)\). The given expression is \((sin x + i cos x)\). This is not directly in the standard polar form because the real part involves \(\sin x\) and the imaginary part involves \(\cos x\).

However, we can use trigonometric identities to rewrite it in the standard form. We know that:

• \(\sin x = \cos(\frac{\pi}{2} - x)\)
• \(\cos x = \sin(\frac{\pi}{2} - x)\)

Using these identities, we can write the expression as:

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

This is now in the standard polar form with modulus \(r=1\) and argument \(\theta = \frac{\pi}{2} - x\). This makes it easy to apply De Moivre's Theorem.

Applying De Moivre's Theorem

De Moivre's Theorem states that for any real number \(n\),

\[(r(\cos \theta + i \sin \theta))^n = r^n(\cos(n\theta) + i \sin(n\theta))\]

In our case, \(r=1\), \(\theta = \frac{\pi}{2} - x\), and \(n=4\). Applying the theorem:

\[(sin x + i cos x)^4 = \left(1\left(\cos\left(\frac{\pi}{2} - x\right) + i \sin\left(\frac{\pi}{2} - x\right)\right)\right)^4\] \[ = 1^4 \left(\cos\left(4 \times \left(\frac{\pi}{2} - x\right)\right) + i \sin\left(4 \times \left(\frac{\pi}{2} - x\right)\right)\right)\] \[ = 1 \left(\cos\left(2\pi - 4x\right) + i \sin\left(2\pi - 4x\right)\right)\] \[ = \cos(2\pi - 4x) + i \sin(2\pi - 4x)\]

Simplifying and Identifying the Real Part

Now we need to simplify the trigonometric functions using the periodicity of cosine and sine. We know the identities:

• \(\cos(2\pi - \phi) = \cos \phi\)
• \(\sin(2\pi - \phi) = -\sin \phi\)

Letting \(\phi = 4x\), we apply these identities:

\[\cos(2\pi - 4x) = \cos(4x)\] \[\sin(2\pi - 4x) = -\sin(4x)\]

Substituting these back into our expression for \((sin x + i cos x)^4\):

\[(sin x + i cos x)^4 = \cos(4x) + i (-\sin(4x))\] \[ = \cos(4x) - i \sin(4x)\]

The resulting complex number is \(\cos(4x) - i \sin(4x)\). A complex number is written in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. In our result, \(a = \cos(4x)\) and \(b = -\sin(4x)\). Therefore, the real part of \((sin x + i cos x)^4\) is \(\cos(4x)\).

Finding the real part of a complex number expression involving sin x and cos x raised to a power is efficiently done using this method.