I mean what would be the reason to not just leave it as x-c ?
Stacey Warren - Expert brainly.com
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First sin function is an odd, so f(x) = -f(-x)
and you ca use trig identities to prove that also
well.. or one could use the "symmetry identities"
and notice that sin(-a) = -sin(a)
sin([x-c]) is also equals to sin(- [c -x])
which also would equal -sin([c-x]) or just -sin(c-x)
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you could expand it if you wish anyhow, using the sum identity, and you'd end up with sin(x+c)
keeping in mind that x+c is also equals to -(c-x) <--- expand that and you'd get x+c
x+c would equal -(-c-x) <--- expanding that would give x+c
now I understand the conversion process a bit better.
I'm still a bit curious when mathematica is asked to simplify
why it chooses the
form as the preferred form, this doesn't seem simplified to me. Is there a general rule that gives one form priority over the other?
It depends on what will you do with that expression or neighborhood expressions to make them look identical and factor out or something like that.
anyway this example "for me" is meaningless as itself and the word simplify is ambiguous.
So i would go with making it sin x cos c -cos x sin c
It's because it is easier to understand \(-\sin(x-c)\) as a function that is shifted and then reflected.
Basically, \(-\sin(x)\) is generally easier to evaluate than \(\sin(-x)\) because the table of \(\sin(x)\) values likely did not have negative \(x\) values as it would be redundant.