Getting the sines and cosines right

Graphic - sincos
http://www.flickr.com/photos/colleague/3244321695/

Are you unsure about which of the components of the launch velocity force is v0sinθ and which is v0cosθ? If so

To ‘fix’ it you do need to know a few basic things about the sine and the cosine; specifically:

\[ \sin {0} = \cos {(\pi/2)} =0 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \cos {0} = \sin {(\pi/2)} =1 \]

Now just appeal to the ‘special cases’ trick (Guideline 0.10). You can see that the horizontal (x) component of the velocity has to be vx=v0cosθ because

Please learn this now...it will save you (and us) much pain.