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 can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (\sin 40^\circ + i \cos 40^\circ)^5 \]
We can rewrite \(\sin 40^\circ\) and \(\cos 40^\circ\) in terms of cosine and sine:
\[ \sin 40^\circ = \cos(90^\circ - 40^\circ) = \cos 50^\circ \]
\[ \cos 40^\circ = \sin(90^\circ - 40^\circ) = \sin 50^\circ \]
Thus, we can rewrite the expression as:
\[ (\cos 50^\circ + i \sin 50^\circ)^5 \]

### Step 2: Apply De Moivre's Theorem
Using De Moivre's theorem, we know that:
\[ (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) \]
Applying this to our expression:
\[ (\cos 50^\circ + i \sin 50^\circ)^5 = \cos(5 \times 50^\circ) + i \sin(5 \times 50^\circ) = \cos 250^\circ + i \sin 250^\circ \]

### Step 3: Simplify the angle
Next, we simplify \(250^\circ\):
\[ 250^\circ = 360^\circ - 110^\circ \]
Using the properties of sine and cosine:
\[ \cos 250^\circ = \cos(360^\circ - 110^\circ) = \cos 110^\circ \]
\[ \sin 250^\circ = \sin(360^\circ - 110^\circ) = -\sin 110^\circ \]
Thus, we can write:
\[ \cos 250^\circ + i \sin 250^\circ = \cos 110^\circ - i \sin 110^\circ \]

### Step 4: Find the principal amplitude
The principal amplitude (or argument) of a complex number \(z = x + iy\) is given by:
\[ \text{arg}(z) = \tan^{-1}\left(\frac{y}{x}\right) \]
In our case, since we have:
\[ z = \cos 110^\circ - i \sin 110^\circ \]
The argument can be expressed as:
\[ \text{arg}(z) = -110^\circ \]

### Conclusion
Thus, the principal amplitude of \((\sin 40^\circ + i \cos 40^\circ)^5\) is:
\[ -110^\circ \]