- #1

- Thread starter zack7
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- #1

- #2

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[itex]\cos(x\pi)=(-1)^x[/itex] if [itex]x\in\mathbb{Z}[/itex]

Does that help?

Does that help?

- #3

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[itex]\frac{-(cos(n*pi+pi)}{(n+1)}[/itex]+[itex]\frac{-(cos(n*pi-pi)}{(n-1)}[/itex]

to

[itex]\frac{(cos(n*pi)}{(n+1)}[/itex]-[itex]\frac{(cos(n*pi)}{(n-1)}[/itex]

to

[itex]\frac{-(2 cos(pi n)}{(n^2-1)}[/itex]

- #4

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How would you simplify cos(θ+π) and cos(θ-π)?What I don't get is how to I simplify from

[itex]\frac{-(cos(n*π+π)}{(n+1)}[/itex]+[itex]\frac{-(cos(n*π-π)}{(n-1)}[/itex]

to

[itex]\frac{(cos(n*π)}{(n+1)}[/itex]-[itex]\frac{(cos(n*π)}{(n-1)}[/itex]

How would combine a/(n+1) - a/(n-1) into a single fraction?to

[itex]\frac{-(2 cos(π n)}{(n^2-1)}[/itex]

- #5

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Yup got it, totally forgot the trig identityYou may be looking for something more complicated than necessary.

How would you simplify cos(θ+π) and cos(θ-π)?

How would combine a/(n+1) - a/(n-1) into a single fraction?

cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)

Thank you

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