A quick Trig Identity Question.

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The discussion centers on the trigonometric identities: sin(u)cos(v) = 0.5[sin(u+v) + sin(u-v)] and cos(u)sin(v) = 0.5[sin(u+v) - sin(u-v)]. The key difference between variables u and v lies in their roles in the sine and cosine functions, affecting the resulting expressions. A specific example illustrates that choosing different values for u and v can yield different results, but a correction reveals that the outputs are equivalent when properly applying the identities. The clarification emphasizes the importance of correctly interpreting the sine function's arguments.

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logan233
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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?
 
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logan233 said:
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.
 
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Oh I see now. Thank you.
 

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