Question

The principle amplitude of ${(\sin 40^\circ + i\cos 40^\circ )^5}$ is
A.$70^\circ $
B.$ - 110^\circ $
C.${70^{110}}$
D.${70^{ - 70}}$

Answer 1

Hint: Here, we will use the De Moivre’s theorem and the concept of All STC rule. And the value of the imaginary number and its equivalent value and substitute its value using the theorem and simplify the equation accordingly.

Complete step-by-step answer:
The complex numbers, $i = \sqrt { - 1} $
Squaring both the sides-
$
  {i^2} = 1 \\
   \Rightarrow {i^5} = 1.i \\
   \Rightarrow {i^5} = i \\
 $
${(\sin 40^\circ + i\cos 40^\circ )^5}$
Take common from the above equation-
=${(\sin 40^\circ + i\cos 40^\circ )^5} = {i^5}{(\cos 40^\circ - i\sin 40^\circ )^5}$
By using De Moivre’s theorem in the above equation -
=$r{(\cos \theta + i\sin \theta )^n} = {r^n}(\cos n\theta + i\sin n\theta )$
=${(\sin 40^\circ + i\cos 40^\circ )^5} = i(\cos 200^\circ - i\sin 200^\circ )$
Convert the above angle in order to use the All STC rule-
=${(\sin 40^\circ + i\cos 40^\circ )^5} = i[\cos (180^\circ + 20^\circ ) - i\sin (180^\circ + 20^\circ )]$
Use, ALL STC rule in the above equation- since the above angle is in the third quadrant-
=${(\sin 40^\circ + i\cos 40^\circ )^5} = i[\cos ( - 20^\circ ) + i\sin (20^\circ )]$
=${(\sin 40^\circ + i\cos 40^\circ )^5} = - i\cos 20^\circ - i\sin 20^\circ $
The above equation can be re-written as –
$
   - \cos 20^\circ = \cos ( - 110^\circ ) \\
  i\sin 20^\circ = + i\sin ( - 110^\circ ) \\
 $
=${(\sin 40^\circ + i\cos 40^\circ )^5} = \cos ( - 110^\circ ) + i\sin ( - 110^\circ )$
Hence, the required answer - the principal Amplitude is $ = 110^\circ $
Hence, from the given multiple choices – the option B is the correct answer.


Note: Remember the All STC rule, it is also known as ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ($0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ($90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ($180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ($270^\circ {\text{ to 360}}^\circ $ ). Simplify the equivalent angle accordingly wisely.