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Elementary Trig Question

  1. May 4, 2008 #1
    Say I have an acute angle A in standard position in the first quadrant with a rotating arm of length r , terminating at coordinate P(a,b) . Now say I rotate it 90 degrees further from that position to the second quadrant , this ends up at coordinate Q(-a,b) . I draw perpendiculars from those two positions to the x axis .

    for the first triangle , sin A = b/r and cos A = a/r

    as for the second triangle , why is sin (90+A) = a/r = cos A
    and why is cos(90+A) = -b/r = -sin A

    Im confused ... how can you compute the trig ratio for an angle greater than 90 degress ? Why not take the ratio for the acute angle (180 - (90+a))

    Pls Help .
     
  2. jcsd
  3. May 4, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi ! Welcome to PF! :smile:

    No … (-a,b) is the reflection of (a,b) in the y-axis.

    You want Q(b,-a).

    Then everythihng will make sense! :smile:
     
  4. May 4, 2008 #3
    hmm ... but that doesn't solve my problem ... I need to know why the so called "reference angle" or the "corresponding acute angle" actually works for angles greater than 90 degrees , when taking sines and cosines or tangents . Is there a geometric argument for it ???
     
  5. May 4, 2008 #4

    cristo

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    Im not entirely sure what your question is; but if it boils down to "why is cos(x)=sin(x+90) ?", then you can answer this by looking at the graphs of the sine and cosine functions.
     
  6. May 4, 2008 #5
    Heres my problem ...

    For the triangle in the second quadrant , why does my book state

    sin(90+theta) = a/r and cos(90+theta) = -b/r

    when clearly (90+theta) isn't even one of the angles in the triangle ... ??? This was what I meant when I said why not calculate (180-90-theta) ....

    I dont get this part .... !!! Ive looked up 2/3 trig books by native authors (I live in Nepal) and they don't seem to explain why this is so ... and I have yet to find a good trig book on the internet .... Thanks a mil .
     
    Last edited: May 4, 2008
  7. May 4, 2008 #6

    tiny-tim

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    Hi pokemonDoom! :smile:
    No … the triangle is not in the second quadrant … its base is still in the first quadrant, and its tip is in the second quadrant.

    For an easy example, consider the two triangles OAB and OAC, where O = (0,0), A = (1,0), B = (1,1), and C = (-1,1). So OA = 1, and OB = OC = √2. Put D = (-1,0).

    Then angle AOB = 45º, and angle AOC = 135º.

    And cosAOB = OA/OB = 1/√2, cosAOC = AD/OC = -1/√2.

    But sinAOB = AB/OB = 1/√2, sinAOC = DC/OC = 1/√2. :smile:
     
  8. May 5, 2008 #7
    Hello Tim ,

    Im confused ...

    how can the cosine of <AOC be AD/OC ... if I know better , its probably a typo .

    and in triangle-DOC (which lies squarely on the second quadrant of my graph) how is it that the sine of <AOC is equivalent to the sine of <DOC ... This is the question in the first place .

    Many thanks .
     
  9. May 5, 2008 #8

    tiny-tim

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    Hi pokemonDoom! :smile:

    Yes … sorry … it should be cosAOC = OD/OC = -1/√2. :redface:
    Because sin = opposite/hypotenuse.

    The side opposite angle AOB is AB, and the side opposite angle AOC is DC.

    And AB = DC, so sinAOB = sinAOC. :smile:

    (and of course, the diagram works for any angle, not just 45º)
     
  10. May 5, 2008 #9
    ... I get it !!!

    But isnt the angle opposite side DC <DOC ... ? Im not trying to sound dumb but what youre trying to say is exactly what the question is .

    Why should sineAOC =CD/OC , when its perfectly legible that sineDOC = CD/OC .

    In essence what youre trying to say is for any angle greater than 90 degrees , the ratio of that angle is equal to the ratio of the acute angle formed with the nearest x axis ... ?? but why? is there a geometric proof of some kind ?? instead of just guessing that the ratios are equal ?
     
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