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

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x)=\sin(x)+\cos(x) \\ \text{ plug \in } x \\ f(x)=? \\ \text{ if you get } f(x)=f(x) \text{ then } f \text{ is even } \\ \text{ if you get } f(x)=f(x) \text{ then } f \text{ is odd }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0f(x)=sin(x)+cos(x) now what? @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.1use the fact that sin is an odd function and that cos is an even function

freckles
 one year ago
Best ResponseYou've already chosen the best response.1that is you are using that sin(x)=sin(x) and that cos(x)=cos(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay i got cos(x)=f(x)+sin(x) @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.1did you add sin(x) on both sides?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x)=\sin(x)+\cos(x) \\ f(x)=\sin(x)+\cos(x) \] is sin(x)+cos(x) the original function or the opposite of the original function?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles sin(x)+cos(x) is the opposite of the original function

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\sin(x)+\cos(x) \neq \sin(x)+\cos(x) \text{ the signs on } \sin \text{ differ } \\ \sin(x)+\cos(x) \neq (\sin(x)+\cos(x)) \text{ both the signs would have to be differ}\] 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1another 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1but 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1if we had: \[f(x)=x^2+\cos(x) \\ \text{ and we plug in } x \\ f(x)=(x)^2+\cos(x) \\ f(x)=x^2+\cos(x) \\ \text{ this would mean } f \text{ is even } \\ \text{ if we had } f(x)=x^3+\sin(x) \\ \text{ and we plug in } x \\ \text{ we would have } \\ f(x)=(x)^3+\sin(x) \\ f(x)=x^3\sin(x) \\ f(x)=(x^3+\sin(x)) \text{ this is the opposite of } f \\ \text{ so this } f \text{ would be odd }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1anyways was just giving examples of an odd and even function just in case you were confused or whatever

freckles
 one year ago
Best ResponseYou've already chosen the best response.1let me know if you still don't understand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no i think i understand now. thank you so much @freckles
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