# A quick Trig Identity Question.

by logan233
Tags: identity, trig
 Share this thread:
 P: 11 Hopefully this will make sense... We have the trig. identities shown below: sin(u)cos(v) = 0.5[sin(u+v) + sin(u-v)] cos(u)sin(v) = 0.5[sin(u+v) - sin(u-v)] How are these different? I realize u and v switched between the sine and cosine functions, but what is the difference between u and v? I recognize that there is a difference between taking sine of a number u and sine of a different number v, and same with taking the cosine of a those numbers, I just don't see how we differentiate between u and v. Like say we have... x(t) = sin(2πt)cos(2π10t) and we choose u = 2πt and v = 2π10t so that x1(t) = 0.5[sin(2π11t) + sin(2π9t)] but what if we choose u = 2π10t and v = 2πt then x2(t) = 0.5[sin(2π11t) - sin(2π9t)], which is different than the original x(t) even though we simply chose u and v to be different?
Sci Advisor
HW Helper
PF Gold
P: 3,288
 Quote by logan233 x(t) = sin(2πt)cos(2π10t) and we choose u = 2πt and v = 2π10t so that x1(t) = 0.5[sin(2π11t) + sin(2π9t)]
Your mistake is here. The formula is ##0.5[\sin(u+v) + \sin(u-v)]##, so the argument to the second ##\sin## should be ##-2\pi 9 t##, not ##2\pi 9 t##. Then use the fact that ##\sin(-x) = -\sin(x)## to see that the two answers are in fact the same.
 P: 11 Oh I see now. Thank you.

 Related Discussions Precalculus Mathematics Homework 3 Calculus & Beyond Homework 1 Precalculus Mathematics Homework 11 Precalculus Mathematics Homework 4 Calculus 4