Results 1 to 12 of 12

Math Help - Help graphing using simple trig

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    131

    Help graphing using simple trig

    I'm a little frustrated because in my class we never used calculators, and now I have to prepare for a state math test that requires extensive calculator usage. I'm using the TI-83 Plus... My review book says to graph 2sinx+cos2x in Y1 and 2sin^2x-1 in Y2 and observe their solutions (x-coordinates of intersection points) (1.57, 3.67, 5.76).

    Every time I graph these I go put the cursor on where I see them intersect and they never intersect at these points. I think I am not entering the second graph in correctly... I am entering it in as 2sin(x-1)^2.

    Also, any tips about graphing in general would be great since I am nearly a complete noob with graphing calculators. Help is appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    Re: Help graphing using simple trig

    As you've written it I suspect the calculator is doing 2\sin[(x-1)^2] - ie: it thinks it's squaring the argument rather than the sin function

    Try 2*(\sin(x-1))^2.

    I'm afraid I'd be no good at plotting it on the calculator, graphing calculators were rare and we had to do it using pencil and paper and transforming functions

    You can always use Wolfram Alpha to get an idea of what it should look like
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Help graphing using simple trig

    For the second graph. Is it: 2\sin^2(x-1)? Then it's not correct to enter 2\sin(x-1)^2, enter: 2(\sin(x-1))^2.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2011
    Posts
    15

    Re: Help graphing using simple trig

    The original equation is 2sin^2x-1, make sure that it is; sin(x-1) and not sinx - 1 which is totally different. If it is 2 sin ^2 (x - 1) then enter in the calculator as 2[sin(x-1)]^2 and if it is sin^2x - 1 then enter 2 (sinx)^2 - 1 to get it right.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448

    Re: Help graphing using simple trig

    Quote Originally Posted by benny92000 View Post
    I'm a little frustrated because in my class we never used calculators, and now I have to prepare for a state math test that requires extensive calculator usage. I'm using the TI-83 Plus... My review book says to graph 2sinx+cos2x in Y1 and 2sin^2x-1 in Y2 and observe their solutions (x-coordinates of intersection points) (1.57, 3.67, 5.76).

    Every time I graph these I go put the cursor on where I see them intersect and they never intersect at these points. I think I am not entering the second graph in correctly... I am entering it in as 2sin(x-1)^2.

    Also, any tips about graphing in general would be great since I am nearly a complete noob with graphing calculators. Help is appreciated.
    first (and most important) ... are you graphing in radian mode?

    second ... sometimes it's a good idea to subtract one expression from the other (Y1 - Y2) when solving an equation, then one can find where Y1 - Y2 = 0 (same as Y1 = Y2, only easier to find the x-values where the equation works.)



    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2008
    Posts
    131

    Re: Help graphing using simple trig

    Quote Originally Posted by e^(i*pi) View Post
    As you've written it I suspect the calculator is doing 2\sin[(x-1)^2] - ie: it thinks it's squaring the argument rather than the sin function

    Try 2*(\sin(x-1))^2.

    I'm afraid I'd be no good at plotting it on the calculator, graphing calculators were rare and we had to do it using pencil and paper and transforming functions

    You can always use Wolfram Alpha to get an idea of what it should look like
    You're right. It was squaring the argument. The graph on the wolfalpha is the correct graph. However, when I entered it in as you initially suggested, it came up incorrect. This is because I did not communicate what the argument was effectively. My book says 2sin^2 x-1 (with no parenthesis around anything, so the -1 is separate I believe, and not part of the argument??). I'm just a tad confused because the wolfalpha graph is correct with the parenthesis around the x and not the -1. Does that mean the argument is just the x? Sorry if I am confusing you...

    Quote Originally Posted by Siron View Post
    For the second graph. Is it: 2\sin^2(x-1)? Then it's not correct to enter 2\sin(x-1)^2, enter: 2(\sin(x-1))^2.
    Thanks. I think the argument is just the x since there are no parenthesis.

    Quote Originally Posted by mathjeet View Post
    The original equation is 2sin^2x-1, make sure that it is; sin(x-1) and not sinx - 1 which is totally different. If it is 2 sin ^2 (x - 1) then enter in the calculator as 2[sin(x-1)]^2 and if it is sin^2x - 1 then enter 2 (sinx)^2 - 1 to get it right.
    Thank you... I think it is the latter graph since the book has no parenthesis. I initially thought that the x-1 was the argument (I don't know why.) However, when I graphed it the latter way, I came up with the correct graph.
    Quote Originally Posted by skeeter View Post
    first (and most important) ... are you graphing in radian mode?

    second ... sometimes it's a good idea to subtract one expression from the other (Y1 - Y2) when solving an equation, then one can find where Y1 - Y2 = 0 (same as Y1 = Y2, only easier to find the x-values where the equation works.)



    Thank you, sir. And yes, I am graphing in radians. Could you elaborate a little more on why we would subtract Y2 from Y1, and how that would give us the points of intersection?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448

    Re: Help graphing using simple trig

    when two expressions are equal and you subtract them, what value do you get?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Oct 2008
    Posts
    131

    Re: Help graphing using simple trig

    0. But Y1 and Y2 are not equal to each other, are they??
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448

    Re: Help graphing using simple trig

    Quote Originally Posted by benny92000 View Post
    0. But Y1 and Y2 are not equal to each other, are they??
    at certain values of x they are ... where Y1 - Y2 = 0 ... see the last graph posted.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Oct 2008
    Posts
    131

    Re: Help graphing using simple trig

    Ok, and I have a quick question about TI-83 Plus.. How do you make it graph immediately rather than draw it out? It saves me crucial seconds.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448

    Re: Help graphing using simple trig

    Quote Originally Posted by benny92000 View Post
    Ok, and I have a quick question about TI-83 Plus.. How do you make it graph immediately rather than draw it out? It saves me crucial seconds.
    the TI-83 graphs at a rate dependent on the complexity of the graph ... why are seconds crucial?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Oct 2008
    Posts
    131

    Re: Help graphing using simple trig

    I am taking a timed test and sometimes I have to enter several graphs.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 21st 2010, 11:44 PM
  2. Trig graphing(I think)
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 26th 2009, 11:09 PM
  3. simple graphing question relating to optics
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: September 20th 2009, 11:59 AM
  4. Simple Graphing Question
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 19th 2009, 11:16 PM
  5. Simple graphing problem giving me trouble
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: July 15th 2007, 12:16 PM

Search Tags


/mathhelpforum @mathhelpforum