Question

write [2(cos15*+isin15*)]^3 in standard form a+bi

Answer 1

hello : 
by moivre theorem : 
[2(cos15*+isin15*)]^3 = 2^3(cos(3×45*)+isin(3×45*))
                                   =8 (cos45*+isin45*)
                                   = 8 ( √2/2 +i √2/2)
                                   = 4√2 + 4√2 i ..... ( in standard form a+bi)
a = b = 4√2 

Answer 2

Answer:

Hence, the standard form of the given expression is:

       $[2(\cos 15+i\sin 15)]^3=4\sqrt{2}+4\sqrt{2}i$

Step-by-step explanation:

We have to represent the expression:

$[2(\cos 15+i\sin 15)]^3$ in the standard form: $a+bi$

Now, this expression could also be written as:

$[2(\cos 15+i\sin 15)]^3=2^3(\cos 15+i\sin 15)^3$

i.e.

$[2(\cos 15+i\sin 15)]^3=8(\cos 15+i\sin 15)^3$

Now we will use the De Movier's Theorem that:

$(\cos \theta+i\sin \theta)^n=\cos n\theta+i\sin n\theta$

Hence, we have:

$[2(\cos 15+i\sin 15)]^3=8(\cos (15\times 3)+i\sin (15\times 3))$

i.e.

$[2(\cos 15+i\sin 15)]^3=8[\cos 45+i\sin 45]$

i.e.

$[2(\cos 15+i\sin 15)]^3=8[\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}]$

i.e.

$[2(\cos 15+i\sin 15)]^3=\dfrac{8}{\sqrt{2}}+i\dfrac{8}{\sqrt{2}}$

i.e.

$[2(\cos 15+i\sin 15)]^3=4\sqrt{2}+4\sqrt{2}i$