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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
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
Check out this site...
http://www.ping.be/~ping1339/index.html#Main-Purpose-=-MATH-
marlon
Thanks a lot for the answer and the great link!!!
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