Calculating Vector Components: X & Y

[seang]

For the vectors in the picture, we're supposed to break down each vector into its x and y components. I don't understand why the x component is given by cos(theta). It seems like it should be sin(theta) to me

 

[Astronuc]

While waiting for approval, let's assume a Cartesian (x,y) coordinate system with x-axis horizontal and y-axis vertical with positive coordinate in upper right quadrant.

Take F to be in the right half, either above or below. If the angle \theta between F and the x-axis, then the component F_x would be given by F cos \theta. If however, the angle was taken from the y-axis, then F_x would be given by sin \theta.

With respect to F, F_x, F_y, think of F as the hypotenuse of a triangle and F_x and F_y as the legs, and then apply the Pythagorean theorem, i.e. appropriate trigonometric relationship.

 

[Tide]

The x components are gven by the sine of the respective angles in the diagram you showed. Who said otherwise?

 

[HallsofIvy]

The x component of either vector cannot be sin(\theta) or cos(\theta). There is no \theta in the picture!

If, as is often done- but not in this picture, \theta is measured from the positive x-axis, then the x component of the vector would be given by the length of the vector times cos(\theta).

 

[Astronuc]

Well, v_1x, would be given by v₁ sin \theta_1, and v_2x, would be given by v₂ sin \theta_2, that is assuming the vertical axis is the y-axis, and x-axis is horizontal. These give the magnitudes, but realize that v_2x is in the +x direction, and v_1x is in the -x direction.

For future reference - http://en.wikipedia.org/wiki/Trigonometric_function
and
http://en.wikipedia.org/wiki/Trigonometric_identities