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anonymous

  • one year ago

Why does Sin[x -c] Simplify to -Sin[c-x] ? I mean what would be the reason to not just leave it as x-c ?

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  1. anonymous
    • one year ago
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    First sin function is an odd, so f(x) = -f(-x)

  2. anonymous
    • one year ago
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    and you ca use trig identities to prove that also

  3. jdoe0001
    • one year ago
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    well.. or one could use the "symmetry identities" and notice that sin(-a) = -sin(a) thus sin([x-c]) is also equals to sin(- [c -x]) which also would equal -sin([c-x]) or just -sin(c-x)

  4. jdoe0001
    • one year ago
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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

  5. jdoe0001
    • one year ago
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    hmm actuallly

  6. jdoe0001
    • one year ago
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    x+c would equal -(-c-x) <--- expanding that would give x+c

  7. anonymous
    • one year ago
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    Thanks guys.. now I understand the conversion process a bit better. I'm still a bit curious when mathematica is asked to simplify sin[x-c] why it chooses the -sin[c-x] 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?

  8. anonymous
    • one year ago
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    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

  9. anonymous
    • one year ago
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    It's because it is easier to understand \(-\sin(x-c)\) as a function that is shifted and then reflected.

  10. anonymous
    • one year ago
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    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.

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