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Trigonometry proofs needed

  1. Jun 26, 2013 #1
    I don't need unit circle proof or proof using trignometric equality. I know those. I am attaching a image. Please read it. It is from the book Plane Trigonometry by S L Loney. See what it says.
    I know that angle MOP = angle M'P'O and both triangles formed are congruent, OM' = MP and M'P" = OM and OP' = OP.

    Angle P'OM' + angle MOP = 90.
    If angle MOP is theta then angle M'P'O is also theta.
    Angle OPM = angle P'OM' = 90 - theta because 3 angles of triangle = 180. ( 90 + theta + (90 - theta))

    What I don't understand is how is andle MOP = 90 - angle P'OM' and how it is = to angle OP'M'. ( as OP'M' = theta)

    http://www.edaboard.com/attachments/92503d1371565911-trigproof.png

    The point A in the diagram is wrong. That is my mistake. Point A is to the left of M on x-axis.

    There is no error. angle MOP or AOP is theta. AOP' and MOP' are both same. Point A is just to the right of M and A' to the left of M'. My question is how did he calculate
    sin(90 + theta) = sin(AOP') = M'P'/OP' = OM/OP. This is giving a problem. M'P'/OP' is cos(theta) in second quardrant.
    If we take angle MOP as theta and angle P'OM' as (90 - theta) as sum of the angles should be 90. then angle P'OM' is 90 - theta.
    Again if we consider just the triangle in 2nd quadrant then the angle M'P'O is theta and angle P'M'O is 90 and hence the other angle of the triangle is (90 - theta).

    M'P'/OP' is sin(90 - theta) of triangle in 2nd quadrant or cos(theta) of triangle in first quadrant.

    Then how does he say that M'P'/OP' is sin(90 + theta)?

    So many people refer the book by S L Loney. I am attaching a new image.

    http://www.edaboard.com/attachments/92532d1371636717-trigproof2.png
     
  2. jcsd
  3. Jun 26, 2013 #2

    Mark44

    Staff: Mentor

    Your images don't show.
     
  4. Jun 26, 2013 #3

    verty

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    Homework Helper

    @Jayanthd: You get sin(90 - θ) but the author gets sin(90 + θ)? By the CAST rule, those are the same.
     
  5. Jun 26, 2013 #4
    Can you explain in detail. Please explain how the author considers sin(90 - θ) as sin(90 + θ).
     
  6. Jun 26, 2013 #5

    verty

    User Avatar
    Homework Helper

    Notice that I gave a link, click on the words "the CAST rule".
     
  7. Jun 26, 2013 #6

    I have seen that link. It doesn't tell how sin(90 - θ) becomes sin(90 + θ).

    Can you please explain. This is the only thing in Trigonometry that I am not able to understand.
     
  8. Jun 26, 2013 #7

    Mark44

    Staff: Mentor

    It's easy enough to show using the identities for the sine of a sum or difference.
    sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
    sin(A - B) = sin(A)cos(B) - cos(A)sin(B)
     
  9. Jun 26, 2013 #8
    I know that solution. What I need is geometrical proof using right angled triangle.
     
  10. Jun 26, 2013 #9

    Mark44

    Staff: Mentor

    Why?

    Also, assuming θ is an acute angle (0 ≤ θ < 90°), it's not possible to have a right triangle where one angle is 90° + θ.
     
  11. Jun 26, 2013 #10

    Because I can't memorize sin(90 + theta), sin(90 - theta), sin(180 - theta), sin(180 + theta) and other ratios. If explained geometrically then it will be easy to derive whenever needed. I know the Unit Circle approach.
     
  12. Jun 26, 2013 #11

    Mark44

    Staff: Mentor

    That's kind of a silly reason. One formula will serve for all of those:
    sin(A + B) = sin(A)cos(B) + cos(A)sin(B).

    You don't need a separate formula for sin(A - B). Just use the formula above with sin(A + (-B)).
    As already noted, it isn't possible to draw a right triangle with one angle equal to 90° + θ, where θ is an acute angle.
     
  13. Jun 27, 2013 #12

    I know that a right angled triangle can have angles 90, θ, and (90 - θ) but how did S L Loney proves sin(90 + θ) = cos(θ) as mentioned in the first post.

    Is it that nobody here knows geometrical proof for sin(90 + θ) and sin(180 - θ)?
     
  14. Jun 27, 2013 #13

    Mark44

    Staff: Mentor

    It would be helpful if you posted the images that were supposed to be in your first post. I mentioned that the links were broken in my first reply.
    What do you mean by a "geometrical proof"? You have already been told how you can prove that sin(90° + θ) = sin(90° - θ) using the sum identity. The proof is also very simple using the unit circle. Your objection to the first method seems to be only that you don't want to memorize the identity. I don't know what your objection to using the unit circle is.
     
  15. Jun 27, 2013 #14
    Trigonometry Proofs

    Here are the images. Consider the 2nd image for the drawing.

    attachment.php?attachmentid=59917&d=1372354550.jpg

    attachment.php?attachmentid=59918&stc=1&d=1372354604.png
     

    Attached Files:

    Last edited: Jun 27, 2013
  16. Jun 27, 2013 #15

    Mark44

    Staff: Mentor

    The two triangles are congruent, with angle M'P'O being equal to θ. This means that
    Angle MOP' = 90° + θ and angle M'OP' = 180° - (90° + θ) = 90° - θ.
    These angles add to 180°, so are supplementary angles.
    It follows that
    sin(90° + θ) = sin (180° - (90° + θ)), because the sines of supplementary angles are equal.
    So sin(90° + θ) = sin(90° - θ)
    = cos(θ), because 90° - θ and θ are complementary angles (i.e., they add to 90°).
     
  17. Jun 27, 2013 #16

    Thank you very much. Now it is clear to me. I have similar problem with sin(180 - θ).


    Edit: Now another question is regarding 180 - θ.

    See images. Now tell me how sin(AOP') = sin θ.

    Is this right? sin(180 - θ) = sin(180 - (180 - θ)) = sin(θ) ??
     

    Attached Files:

    Last edited by a moderator: Jun 27, 2013
  18. Jun 27, 2013 #17

    Mark44

    Staff: Mentor

    When you insert images, adjust their sizes so that they aren't so large. 900 X 600 pixels is the maximum size.
    Yes. AOP (θ) and AOP' (180° - θ) are supplementary angles, so their sines are equal. This is easy to show using either the unit circle or the sum formula.
     
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