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skan
Oct22-04, 03:12 PM
I have basic trignometry doubt

how to u get 5cos(2t-53.13degrees) from 3cos2t + 4 sin2t.

Can someone suggest a site from where I can lean basic trinometry

thanks.
skan

marlon
Oct22-04, 05:14 PM
well, nice question...
use cos(x-y)=cos(x)cos(y) + sin(x)sin(y).

3cos2t + 4sin2t = 3(cos2t + (4/3)sin2t). The trick is to define 4/3 = tan(y)=siny/cosy where y = 53,13°. Now you have 3(cos2t + (siny/cosy)sin2t). or this becomes following expression : (3/cosy)(cosy * cos2t + siny * sin2t) =
(3/cosy)cos(2t - 53.13°) and cosy = cos(53.13°) = 0.6

3/cosy = 3/0.6 = 5

problem solved

regards
marlon

marlon
Oct22-04, 05:25 PM
Check out this site...


http://www.ping.be/~ping1339/index.html#Main-Purpose-=-MATH-

marlon

skan
Oct22-04, 06:02 PM
Thanks a lot for the answer and the great link!!!