Convert 2 e^{\pi i/6} to rectangular form.
Well, converting a number in polar form to rectangular form is like trying to convert a polar bear into a rectangular bear. It just doesn't quite work. But don't worry, let me explain how we can handle it!
First, let's remember Euler's formula: e^(ix) = cos(x) + i*sin(x). In this case, we have e^(πi/6).
So, plugging it into Euler's formula, we get:
e^(πi/6) = cos(π/6) + i*sin(π/6)
Cos(π/6) is equal to √3/2, and sin(π/6) is equal to 1/2. Therefore, we can rewrite it as:
e^(πi/6) = √3/2 + i/2
And there you have it! The rectangular form of 2*e^(πi/6) is (√3/2 + i/2)*2, which simplifies to √3 + i.
To convert \(2e^{\frac{\pi i}{6}}\) to rectangular form, we can use Euler's formula which states that \(e^{ix} = \cos(x) + i\sin(x)\).
Let's break down the given expression:
\(2e^{\frac{\pi i}{6}} = 2(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right))\)
Using the values of sine and cosine for \(\frac{\pi}{6}\), which are \(\frac{\sqrt{3}}{2}\) and \(\frac{1}{2}\) respectively, we get:
\(2(\frac{1}{2} + \frac{\sqrt{3}}{2}i)\)
Simplifying this expression, we have:
\(1 + \sqrt{3}i\)
To convert 2*e^(πi/6) to rectangular form, we will use Euler's formula, which states that e^(ix) = cos(x) + i*sin(x).
First, let's break down the given expression:
2*e^(πi/6)
According to Euler's formula, e^(πi/6) can be written as cos(π/6) + i*sin(π/6).
Now let's calculate the values of cos(π/6) and sin(π/6):
cos(π/6) = √3/2
sin(π/6) = 1/2
So, e^(πi/6) can be written as (√3/2) + (1/2)i.
Now we can multiply this expression by 2:
2 * ((√3/2) + (1/2)i)
Distributing the 2, we get:
(2 * √3/2) + (2 * 1/2)i
Simplifying, we get:
√3 + i
Therefore, 2*e^(πi/6) in rectangular form is √3 + i.
recall that e^(iθ) = cosθ + i sinθ
So plug in your numbers.