anonymous
  • anonymous
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 ?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
First sin function is an odd, so f(x) = -f(-x)
anonymous
  • anonymous
and you ca use trig identities to prove that also
jdoe0001
  • jdoe0001
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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jdoe0001
  • jdoe0001
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
jdoe0001
  • jdoe0001
hmm actuallly
jdoe0001
  • jdoe0001
x+c would equal -(-c-x) <--- expanding that would give x+c
anonymous
  • anonymous
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
It's because it is easier to understand \(-\sin(x-c)\) as a function that is shifted and then reflected.
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.