Given:
The expression is:
[tex]\dfrac{1}{2}(\cos (72)+i\sin (72))^5[/tex]
To find:
The [tex]a+bi[/tex] form for the given expression.
Solution:
According to De Moivre's theorem,
[tex](\cos \theta+i\sin \theta)^n=\cos (n\theta )+i\sin (n\theta )[/tex]
We have,
[tex]\dfrac{1}{2}(\cos (72)+i\sin (72))^5[/tex]
Using De Moivre's theorem, we get
[tex]=\dfrac{1}{2}(\cos (72\times 5)+i\sin (72\times 5))[/tex]
[tex]=\dfrac{1}{2}(\cos (360)+i\sin (360))[/tex]
[tex]=\dfrac{1}{2}(1+i(0))[/tex]
[tex]=\dfrac{1}{2}+0i[/tex]
It is the [tex]a+bi[/tex] form of the given expression. Here, [tex]a=\dfrac{1}{2},\ b=0[/tex].
Therefore, the required expression is [tex]\dfrac{1}{2}+0i[/tex].