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SpyIntel [72]
3 years ago
10

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

Mathematics
2 answers:
Marysya12 [62]3 years ago
8 0
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 ..... ( <span>in standard form a+bi)
</span>a = b = 4√2 
Luda [366]3 years ago
6 0
<h2>Answer:</h2>

Hence, the standard form of the given expression is:

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

<h2>Step-by-step explanation:</h2>

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

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