Adding Trigonomic Functions

  • Thread starter erok81
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  • #1
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I am stuck with my trig homework. My teacher isn't the best at teaching this stuff and the power point slides from my school stop before they get to this. Also in the answer book (and text) it doesn't really explain this either.

The problem is:

[tex]sin(-\pi)+cos(5\pi)[/tex]

I have no idea what to do next. I know how the radian measure works but [tex]\pi[/tex] and [tex]5\pi[/tex] don't mean anything (to me) in regular degree's.

If anyone could give me a quick explanation on how to do these I'd really appreciate it. :approve:
 

Answers and Replies

  • #2
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You don't have to go to degrees. If you imagine the unit circle, [itex] \pi \ and \ -\pi [/tex] are coincidental points (180 deg), because you are just adding revolutions and returning to the same point. [itex]\sin(\pi) = \sin(-\pi) = 0 [/tex]. [itex] 5\pi [/tex] is just [itex] \pi + 4pi [/tex], or just [itex]\pi[/tex] with two added revolutions. [itex]\cos(5\pi) = \cos(\pi) = -1 [/tex].
 
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  • #3
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erok81 said:
I am stuck with my trig homework. My teacher isn't the best at teaching this stuff and the power point slides from my school stop before they get to this. Also in the answer book (and text) it doesn't really explain this either.

The problem is:

[tex]sin(-\pi)+cos(5\pi)[/tex]

I have no idea what to do next. I know how the radian measure works but [tex]\pi[/tex] and [tex]5\pi[/tex] don't mean anything (to me) in regular degree's.

If anyone could give me a quick explanation on how to do these I'd really appreciate it. :approve:
convert them if you have trouble with radians. pi would be 180 degress, so what would negative pi be? it just means that you're going counter clockwise rather than clockwise. as far as five times pi, one complete revolution is two pi, in that case two complete revolutions would be 4 pi -- so, where would you be for 5 pi?
 
  • #4
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For some reason I always get hung up on these easy ones. I tend to make them a lot harder than they really are. The more complex ones I understand right away, however that works. :rofl:

I hate to sound stupid, but I still don't get it.

The [tex]cos(5\pi)[/tex] I understand how that goes around 2.5 times. I get how the [tex]sin(-\pi)[/tex] goes half way backwards.

For some stupid reason I don't get how [tex]sin(-\pi)[/tex] equals zero, or how the [tex]cos(5\pi)[/tex] equals -1.

I have read the section in my math book over and over again, but can't find anything about these. :confused:
 
  • #5
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Ok. on the unit circle, sin x measure the displacement in the Y direction. When your on the x axis, your displacmeent in the Y direction is 0, and the only two points you are on the x axis on the unit circle are at x = 0 and x = Pi .

Cos is the same thing but in the X direction, so it is zero at x = Pi/2 and 3 Pi/2
 
  • #6
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Oh duh, see what I mean. I don't know how I didn't see/realize that in the first place.


Thanks for the help, I appreciate it. :biggrin:
 

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