Results 1 to 6 of 6

Math Help - Trig indentity

  1. #1
    Member
    Joined
    Apr 2006
    Posts
    201
    Awards
    1

    Trig indentity

    Hi All,

    How about do I go to answer this trig. identity question. I have attached it as a picture as wasn't to sure on how to insert the symbols.

    Also if it is not possible to plot the sketch just explain it in words what I have to do, i'm sure I would be able to understand.

    Thanks.
    Attached Thumbnails Attached Thumbnails Trig indentity-trigidentity.gif  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by dadon
    Hi All,

    How about do I go to answer this trig. identity question. I have attached it as a picture as wasn't to sure on how to insert the symbols.

    Also if it is not possible to plot the sketch just explain it in words what I have to do, i'm sure I would be able to understand.

    Thanks.
    I can't show the graph, but I can help with the rest.

    Via the identity:
    R cos(\theta - \beta)=R cos\theta cos\beta + R sin\theta sin\beta

    By comparison then:
    a cos\theta + b sin\theta = R cos\theta cos\beta + R sin\theta sin\beta
    So
    a = R cos\beta
    b = R sin\beta

    Dividing these two equations gives us:
    \frac{R sin\beta}{R cos\beta}=tan\beta = \frac{b}{a}
    or
    \beta = tan^{-1} \left ( \frac{b}{a} \right )

    and
    R^2 sin^2 \beta + R^2 cos^2 \beta = R^2 (1) = a^2+b^2
    so
    R = \sqrt{a^2+b^2}

    Now, in order to graph 2 cos\theta + 3 sin\theta you are essentially graphing:
    \sqrt{2^2+3^2}cos \left ( \theta - \left [ tan^{-1} \left ( \frac{3}{2} \right ) \right ] \right )

    or roughly \sqrt{13} \, cos(\theta - 0.983) vs. \theta.
    This graph is simply a cosine graph, with amplitude \sqrt{13} and translated 0.983 radians (or so) to the right.

    -Dan
    Last edited by topsquark; May 2nd 2006 at 10:16 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by dadon
    Hi All,

    How about do I go to answer this trig. identity question. I have attached it as a picture as wasn't to sure on how to insert the symbols.

    Also if it is not possible to plot the sketch just explain it in words what I have to do, i'm sure I would be able to understand.

    Thanks.
    Here is the plot

    RonL
    Attached Thumbnails Attached Thumbnails Trig indentity-wire.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2006
    Posts
    201
    Awards
    1

    Post re:

    Hey Cheers for that guys!

    I get want you have done throughout but how do you get <br />
a = R cos\beta<br />
and <br />
b = R sin\beta<br />
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by topsquark
    a cos\theta + b sin\theta = R cos\theta cos\beta + R sin\theta sin\beta
    You get them by comparing forms. Let me write it out a bit more explicitly:
    a cos\theta + b sin\theta = R cos\theta cos\beta + R sin\theta sin\beta

    Now, take a look at the cos\theta term. They have to be the same on both sides of the equation so we can write:
    a cos\theta = R cos\theta cos\beta
    Dividing both sides by cos\theta we obtain:
    a = R cos\beta

    A similar argument with the sine function gives the second relation.

    Technically, I am assuming that cos\theta \neq 0 so some care needs to be taken as far as interpretation for all values of \theta. However, since \theta is being taken as an independent variable and not a constant I feel we can more or less disregard this problem. (Theoretically I suppose we should be plotting an open circle at any point that cos\theta = 0.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Apr 2006
    Posts
    201
    Awards
    1

    Post re:

    Thanks for sharing you time to solve my problem and explainly in a way that I understand fully.

    Cheers

    dadon
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Trig Indentity Limits
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: June 2nd 2010, 11:20 PM
  2. prov trig indentity
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: January 12th 2010, 03:21 PM
  3. trig indentity proof help,
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: December 15th 2009, 01:58 PM
  4. please check my working, trig indentity proof
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: December 14th 2009, 09:22 AM
  5. Trig Indentity Help (Proof)
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: March 5th 2008, 04:21 PM

Search Tags


/mathhelpforum @mathhelpforum