anonymous
  • anonymous
Describe the graph of the cosine function.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
@nincompoop @Preetha
anonymous
  • anonymous
@campbell_st

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campbell_st
  • campbell_st
so do you know what the graph looks like...?
anonymous
  • anonymous
Hehe, not really
anonymous
  • anonymous
Let me check online
anonymous
  • anonymous
Pretty similar to the sine graph if you ask me...
campbell_st
  • campbell_st
here is a site that will graph it for you on the left side just enter y = cos(x) and you'll see the graph https://www.desmos.com/calculator
anonymous
  • anonymous
ok, so now, just to make sure, the domain is all real numbers, the range is -1≤y≤1, the y intercept is 1, but what would the x- intercepts be??
anonymous
  • anonymous
@campbell_st
campbell_st
  • campbell_st
so where does it cut the y-axis, are you working in radians or degrees..?
anonymous
  • anonymous
Degrees
anonymous
  • anonymous
@campbell_st
anonymous
  • anonymous
Sorry to bother u, but I'm pretty lost...
campbell_st
  • campbell_st
so so cos(90) =0 and cos(270) = 0 so the x- intercepts are at x = 90 and x = 270 then repeat every 180 degrees
anonymous
  • anonymous
Oooh, ok, so, as an answer, I put "starting at 90°, every 180°"?
campbell_st
  • campbell_st
you could...
anonymous
  • anonymous
would it be right?
anonymous
  • anonymous
@ganeshie8 could u help me out pleease??
SolomonZelman
  • SolomonZelman
There is a pattern to x-intercepts. For what values is cos(x)=0, the solutions to this are the x-intercepts. For example at x=90 the cos(x) is 0 and so it is for x=270, for x=450 and on adding 180º each time.... \(\large\color{black}{ \displaystyle x=90^\circ }\) is your first y-intercept then, \(\large\color{black}{ \displaystyle x=90^\circ+180^\circ=270^\circ }\) is another x-intercept then, \(\large\color{black}{ \displaystyle x=90^\circ +180^\circ +180^\circ+.... }\) Each time you add 180º you get an x-intercept. So, you can generate the pattern: \(\large\color{black}{ \displaystyle x=90^\circ +(180^\circ\times {\rm k})}\) (for all positive integer values of k, and for 0 as well) You can also go -180º , subtract 180º, to get the x-intercept. \(\large\color{black}{ \displaystyle x=90^\circ-180^\circ=-90^\circ }\) is an x-intercept \(\large\color{black}{ \displaystyle x=90^\circ-180^\circ-180^\circ=-270^\circ }\) \(\large\color{black}{ \displaystyle x=90^\circ -180^\circ -180^\circ-180^\circ-.... }\) So, you can generate the pattern: \(\large\color{black}{ \displaystyle x=90^\circ -(180^\circ\times {\rm k})}\) (for all negative integer values of k) -------------------------------- it follows that x-intercepts all go by the pattern \(\large\color{black}{ \displaystyle x=90^\circ -(180^\circ\times {\rm k});~~~\color{blue}{\rm \forall~~k\in{\bf Z}}}\)
SolomonZelman
  • SolomonZelman
The blue notation here means "for all integer values of k"
SolomonZelman
  • SolomonZelman
saying that k can be equal to ..... \(-5\), \(-4\), \(-3\), \(-2\), \(-1\), \(0\), \(1\), \(2\), \(3\), \(4\), \(5\), .....
anonymous
  • anonymous
Hehe, sorry @SolomonZelman , I'm not quite following... XD
SolomonZelman
  • SolomonZelman
Its alright....
anonymous
  • anonymous
So what you're saying is that the x-intercepts are all multiples of 90?
anonymous
  • anonymous
@SolomonZelman
SolomonZelman
  • SolomonZelman
the x-intercepts start from 90º. and they are all the following: x=90-180 x=90-180-180 x=90-180-180-180 x=90-180-180-180-180 x=90-180-180-180-180-180 x=90-180-180-180-180-180..... and so on.... x=90+(k•180) where K IS NEGATIVE integer or 0. And ALSO, x-intercepts are from 90 and +180 x=90+180 x=90+180+180 x=90+180+180+180 x=90+180+180+180+180 x=90+180+180+180+180+180 x=90+180+180+180+180+180...... and so on.... x=90+(k•180) where K IS POSITIVE integer or 0.
SolomonZelman
  • SolomonZelman
this why, you can generate a pattern with k (where k is both positive integers and negative integers and 0) and this pattern is: x=90+(k•180)
anonymous
  • anonymous
Is 180 an x-intercept? No right?
anonymous
  • anonymous
Aren't the x-intercepts the same as the sine and tangent graphs?
SolomonZelman
  • SolomonZelman
no 180 is not, but 90 plus 180 (add or subtract 180 from 90 any number of times)
SolomonZelman
  • SolomonZelman
lets look at it simple. (we are talking about the cos(x)=y graph) the x-intercepts are the values of x that make the y=0. these are as I posted: `~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~` the x-intercepts start from 90º. and they are all the following: x=90-180 x=90-180-180 x=90-180-180-180 x=90-180-180-180-180 x=90-180-180-180-180-180 x=90-180-180-180-180-180..... and so on.... x=90+(k•180) where K IS NEGATIVE integer or 0. And ALSO, x-intercepts are from 90 and +180 x=90+180 x=90+180+180 x=90+180+180+180 x=90+180+180+180+180 x=90+180+180+180+180+180 x=90+180+180+180+180+180...... and so on.... x=90+(k•180) where K IS POSITIVE integer or 0. `~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
SolomonZelman
  • SolomonZelman
so they are ....... -450º, -270º, -90º, 90º, 270º, 450º, .... (adding or subtracting 180 each time)
anonymous
  • anonymous
Yes, I understand now! So examples of intercepts would be 90, 270, 450, 630, 810, etc?
SolomonZelman
  • SolomonZelman
yes, or you can subtract 180 from 90 many times 90 90-180=-90 -90-180=-270 -270-180=-450 and on....
SolomonZelman
  • SolomonZelman
So, for this reason they can be all given using a pattern x = 90 + k•180 for all integers of k. (is this still unclear?)
SolomonZelman
  • SolomonZelman
(all integers = ..... −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, ..... )
anonymous
  • anonymous
Great, and just to make sure, the rest of the cosine graph would be: y-intercept is 1, the domain are all real numbers and the range is -1≤x≤1
anonymous
  • anonymous
right?
SolomonZelman
  • SolomonZelman
the rest of the cosine graphs?
anonymous
  • anonymous
No, I mean the rest of the characteristics of the cosine graph
SolomonZelman
  • SolomonZelman
yes, yes. But provided they are not shifted sideways. they can have any angle of cx, such that y=cos(c•x)
SolomonZelman
  • SolomonZelman
and the range will change if you shift it up/down
SolomonZelman
  • SolomonZelman
so y=cos(x), has a range [-1,1] y-intercept 1
SolomonZelman
  • SolomonZelman
but, y=cos(x+c) has a range of [-1,1] y-intercept 1-c
SolomonZelman
  • SolomonZelman
and y=cos(x)+c has a range of [-1+c,1+c] y-intercept of 1
SolomonZelman
  • SolomonZelman
this way y=cos(x+a)+b has a range of [-1+b,1+b] y-intercept 1-a
anonymous
  • anonymous
I just started learning the trig graphs, so I don't think the curves will shift. thanks anyways! One more thing, is it fair to say that the x-intercepts are all odd multiples of 90?
SolomonZelman
  • SolomonZelman
Yes, negative or positive multiples of 90
SolomonZelman
  • SolomonZelman
I mean negative or positive, odd multiples of 90
anonymous
  • anonymous
Thank you soooo much!!!
SolomonZelman
  • SolomonZelman
in this case of y=cos(x), which is not the case if you shift the graph sideways such that y=cos(x+b), and which is not [neccessarily, but for most values in fact not] the case if you multiply the angle or the function times a scale factor, such that: y=cos(bx), or y=bcos(x).
SolomonZelman
  • SolomonZelman
this is just graph shifts..... the rule you can get in a book or online..... my PC is about to resart because I am scanning and updating. cu
anonymous
  • anonymous
Thanks!

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