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anonymous

  • one year ago

Does the graph of y = sin x + cos x have even symmetry, odd symmetry, or neither?

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  1. freckles
    • one year ago
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    \[f(x)=\sin(x)+\cos(x) \\ \text{ plug \in } -x \\ f(-x)=? \\ \text{ if you get } f(-x)=f(x) \text{ then } f \text{ is even } \\ \text{ if you get } f(-x)=-f(x) \text{ then } f \text{ is odd }\]

  2. anonymous
    • one year ago
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    f(-x)=sin(-x)+cos(-x) now what? @freckles

  3. freckles
    • one year ago
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    use the fact that sin is an odd function and that cos is an even function

  4. freckles
    • one year ago
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    that is you are using that sin(-x)=-sin(x) and that cos(-x)=cos(x)

  5. anonymous
    • one year ago
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    okay i got cos(x)=f(-x)+sin(x) @freckles

  6. freckles
    • one year ago
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    did you add sin(x) on both sides?

  7. freckles
    • one year ago
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    \[f(-x)=\sin(-x)+\cos(-x) \\ f(-x)=-\sin(x)+\cos(x) \] is -sin(x)+cos(x) the original function or the opposite of the original function?

  8. freckles
    • one year ago
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    or neither?

  9. anonymous
    • one year ago
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    @freckles -sin(x)+cos(x) is the opposite of the original function

  10. freckles
    • one year ago
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    \[-\sin(x)+\cos(x) \neq \sin(x)+\cos(x) \text{ the signs on } \sin \text{ differ } \\ -\sin(x)+\cos(x) \neq -(\sin(x)+\cos(x)) \text{ both the signs would have to be differ}\] what I'm saying is that the opposite of sin(x)+cos(x) is -(sin(x)+cos(x)) but we did not get this so it is not odd and also we did not get sin(x)+cos(x) back so we know this is not even

  11. freckles
    • one year ago
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    another way to say -(sin(x)+cos(x)) is -sin(x)-cos(x) this is what I meant by both of the signs needed to be different

  12. freckles
    • one year ago
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    but anyways in summary -sin(x)+cos(x) is neither the same as or the opposite of sin(x)+cos(x) so sin(x)+cos(x) is neither even or odd

  13. freckles
    • one year ago
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    if we had: \[f(x)=x^2+\cos(x) \\ \text{ and we plug in } -x \\ f(-x)=(-x)^2+\cos(-x) \\ f(-x)=x^2+\cos(x) \\ \text{ this would mean } f \text{ is even } \\ \text{ if we had } f(x)=x^3+\sin(x) \\ \text{ and we plug in } -x \\ \text{ we would have } \\ f(-x)=(-x)^3+\sin(-x) \\ f(-x)=-x^3-\sin(x) \\ f(-x)=-(x^3+\sin(x)) \text{ this is the opposite of } f \\ \text{ so this } f \text{ would be odd }\]

  14. freckles
    • one year ago
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    anyways was just giving examples of an odd and even function just in case you were confused or whatever

  15. freckles
    • one year ago
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    let me know if you still don't understand

  16. anonymous
    • one year ago
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    no i think i understand now. thank you so much @freckles

  17. freckles
    • one year ago
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    np

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