We are asked to find the standard form of the product of two complex numbers: $\sqrt{2}[\cos(\pi/4)+i\sin(\pi/4)]$ and $\sqrt{2}[\cos(7\pi/4)+i\sin(7\pi/4)]$.

1. Problem Description

We are asked to find the standard form of the product of two complex numbers: $\sqrt{2}[\cos(\pi/4)+i\sin(\pi/4)]$ and $\sqrt{2}[\cos(7\pi/4)+i\sin(7\pi/4)]$.

2. Solution Steps

First, let's write the two complex numbers as $z_1 = \sqrt{2}[\cos(\pi/4)+i\sin(\pi/4)]$ and $z_2 = \sqrt{2}[\cos(7\pi/4)+i\sin(7\pi/4)]$.

We can find the product of $z_1$ and $z_2$ by multiplying their magnitudes and adding their angles.

The magnitude of $z_1$ is $\sqrt{2}$ and the magnitude of $z_2$ is $\sqrt{2}$. Therefore, the magnitude of the product is $\sqrt{2} \cdot \sqrt{2} = 2$.

The angle of $z_1$ is $\pi/4$ and the angle of $z_2$ is $7\pi/4$. Therefore, the angle of the product is $\pi/4 + 7\pi/4 = 8\pi/4 = 2\pi$.

So, the product is $2[\cos(2\pi) + i\sin(2\pi)]$.

Since $\cos(2\pi) = 1$ and $\sin(2\pi) = 0$, the product is $2[1 + i(0)] = 2[1] = 2$.

Alternatively, we can use Euler's formula: $e^{i\theta} = \cos(\theta) + i\sin(\theta)$.

Then $z_1 = \sqrt{2} e^{i\pi/4}$ and $z_2 = \sqrt{2} e^{i7\pi/4}$.

The product $z_1 z_2 = (\sqrt{2} e^{i\pi/4})(\sqrt{2} e^{i7\pi/4}) = 2 e^{i(\pi/4 + 7\pi/4)} = 2 e^{i(8\pi/4)} = 2 e^{i2\pi} = 2(\cos(2\pi) + i\sin(2\pi)) = 2(1 + i(0)) = 2$.

3. Final Answer

The final answer is

2.