Explaining the Coefficient of a 45-Degree Rotation Vector

[jeff1evesque]

Homework Statement

A complex vector is written as,
$$\hat{v}(t) = cos(\omega t)\hat{x} + sin(\omega t)\hat{y} = \hat{x} + j\hat{y}$$,

where $$\omega$$ is the angular velocity, and the vector rotates counterclockwise in the x-y plane.If a unit vector is rotated in the x-y plane but is phase shifted by 45degrees, then:

$$\hat{v}(t) = (\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}})\hat{x} + \frac{1}{\sqrt{2}} - j\frac{1}{\sqrt{2}})\hat{x} \Rightarrow (cos(\omega t + 45^{\circ})\hat{x} + (cos(\omega t - 45^{\circ})\hat{y}$$

Can someone explain to me why there are terms $$\frac{1}{\sqrt{2}}$$ in the equation above. I always thought a 45 degree triangle had sides of $$\sqrt{2}, \sqrt{2}, 2$$, but not sure how the coefficient $$\frac{1}{\sqrt{2}}$$ is obtained.thanks,JL

 

[tiny-tim]

Hi JL!

(you've got your x's and y'x mixed up … and you can get LaTeX to write big brackets "to fit" by typing \left( and \right) )

Because cos45º = sin45º = 1/√2 (a 45 degree triangle also has sides of 1, 1/√2, 1/√2)