The principal amplitude of
`(sin40^(@)+icos40^(@))^(5)`, is

Options

A. `70^(@)`
B. `-110^(@)`
C. `110^(@)`
D. `-70^(@)`

Explanation

To find the principal amplitude of the expression \((\sin 40^\circ + i \cos 40^\circ)^5\), we will follow these steps:

### Step 1: Rewrite the expression
We start with:
\[ z = (\sin 40^\circ + i \cos 40^\circ)^5 \]

### Step 2: Use trigonometric identities
We can use the identities:
\[ \sin 40^\circ = \cos(90^\circ - 40^\circ) = \cos 50^\circ \]
\[ \cos 40^\circ = \sin(90^\circ - 40^\circ) = \sin 50^\circ \]
Thus, we can rewrite \(z\) as:
\[ z = (\cos 50^\circ + i \sin 50^\circ)^5 \]

### Step 3: Use Euler's formula
According to Euler's formula:
\[ \cos \theta + i \sin \theta = e^{i\theta} \]
Applying this, we have:
\[ z = (e^{i 50^\circ})^5 = e^{i(5 \times 50^\circ)} = e^{i 250^\circ} \]

### Step 4: Rewrite in trigonometric form
Using the identity again, we can express this in terms of sine and cosine:
\[ z = \cos 250^\circ + i \sin 250^\circ \]

### Step 5: Find the principal amplitude
The principal amplitude is the angle in the exponential form. The angle \(250^\circ\) can be expressed in a more standard form:
\[ 250^\circ = 360^\circ - 110^\circ \]
Thus, we can write:
\[ \cos 250^\circ = \cos(360^\circ - 110^\circ) = \cos 110^\circ \]
\[ \sin 250^\circ = \sin(360^\circ - 110^\circ) = -\sin 110^\circ \]
This confirms that the principal amplitude is:
\[ \text{Principal Amplitude} = -110^\circ \]

### Final Answer
The principal amplitude of \((\sin 40^\circ + i \cos 40^\circ)^5\) is \(-110^\circ\).

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