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Simple Trig Question Help!

  1. Feb 6, 2006 #1
    What is the value of cos A if sin A = 0.9063 and angle A is an obtuse angle?

    Does this need the sine or cosine laws?

    What would the obtuse angle thing be hinting at?

    Any help would be appreciated!
    Last edited: Feb 6, 2006
  2. jcsd
  3. Feb 6, 2006 #2
    Start with the pythagorean identity for cos and sin.
  4. Feb 6, 2006 #3
    hmm.. im not sure what the pythagrean identies are... and i haven't learned about them either.. is there a simpler way to do this?
  5. Feb 6, 2006 #4
    Think about what happens to the sin of an angle when it is greater than 90. Also an obtuse angle is greater than 90 soooo. What happens to the sign of sin in the second quadrant? Do you see what I am getting at here?
  6. Feb 6, 2006 #5
    sin^2x + cos^2x = 1

    I believe this is what Plastic Photon is speaking of.
    However, another way of going at this is to use the sin-1 or arcsin function.

    About the obtuse angle thingy, where are the values of sin positive? Also, which quadrant would it have to lay in in order for the angle to be obtuse?

    EDIT: Valhalla said it.... I'm too slow
  7. Feb 6, 2006 #6
    at risk of sounding stupid, what's second quadrant?

    edit: ok, i think i might need to show the full question, as this is a multiple choice question..

    What is the value of cos A if sin A = 0.9063 and angle A is an obtuse angle?
    A. -0.4226
    B. -0.0158
    C. 0.0158
    D. 0.4226
  8. Feb 7, 2006 #7


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    Homework Helper

    No, I don't. :frown:
    Maybe I'm missing something... But the problem has given you the sin value of that angle.
    Look at the first image, it shows you the four quadrants, labeled I, II, III, IV.
    The Pythagorean identity is:
    sin2x + cos2x = 1, where x is an angle. Do you know this identity?
    Now you are given sin A = 0.9063, can you find cos2A using the above identity?
    From there, you know that A is an obtuse angle, is cos A positive or negative?
    From there, can you find cos A? :)
    Last edited: Feb 7, 2006
  9. Feb 7, 2006 #8
    Vietdao place the angle at the origin. That means the angle for the sin is 90 + theta. Take the sin inverse of .9063 which gives you 65. 90-65 is equal to 25. therefore 90+25=115. sin 115=.9063=sin65

    It helps with a picture.

    Therefore just take the cos 115
  10. Feb 8, 2006 #9


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    Homework Helper

    Nope, in your post, you say about the sign of the sin value for some A, which does not make much sense. Because, as given in the problem, it's positive (the problem states that: sin A = 0.9063). You may want to have a look at your previous post. Here it is:
    Let's look at the bolded part.
    By the way, you may want to try The Pythagorean identity, it does not involve the inverse of sin function, though. :)
    @ soggybread, have you worked out the problem?
  11. Feb 8, 2006 #10
    Well, a simpler (cheat yourself/cop out) method would be to know that .9063 is equal to 9063/1000. So you have r - y. 1000²-9063²=x (adj). But I would not recommend you attempt this method.
  12. Feb 9, 2006 #11
    To the OP: you know what the laws of sines and cosines are but you don't know the Pythagorean Identities or what the quadrants are? what kind of school are you going to?

    Ok, it's been a couple days so I'm going to solve this in its entirety. The best way to solve this has been mentioned earlier, the Pythagorean Identities:
    then we rearrange (subtracting sin2 x from each side of the equation) to give us
    [tex]cos^2 x=1-sin^2 x[/tex]
    Now, we know that [tex]sin x=0.9063[/tex], so we square it to get [tex]sin^2 x=0.8214[/tex]
    Plugging that back into our equation we have [tex]cos^2 x=1-0.8214[/tex] or [tex]cos^2 x=0.1786[/tex]
    Taking the square root of each side,
    [tex]\sqrt {cos^2x} = \sqrt{0.1786}[/tex]
    gives us
    [tex]cos x = 0.4226[/tex] or [tex]-0.4226[/tex]

    The angle terminates in the second quadrant, where the cosine is negative.
    So [tex]cos x = -0.4226[/tex]
    Last edited: Feb 9, 2006
  13. Feb 9, 2006 #12
    A school like mine. My high school changed from standard math to "integrated" mathematics about 10 years ago. Everyone knows that this math is subpar, but some of our teachers pushed for it anyways. You take four years of math - Math 1, Math 2, Math 3, and Math 4 - and learn a little bit of everything every year. (AP Statistics and AP Calculus are offered though, but Math 4 is a prerequisite for those classes). It just doesn't work well and it's ridiculous. We learned the Law of Sine and Law of Cosine in Math 3, but we have yet to learn the Pythagorean Identities. We are finally offering both the standard math and integrated math next year so that freshmen next year have a choice. It's a bummer that I'm a junior right now...the math I'm taking now hasn't allowed me to reach my potential, and I can see how other students will definitely be unprepared for college.[/RANT]
    Last edited: Feb 9, 2006
  14. Feb 9, 2006 #13
    I'm not exactly an eloquent person. What I meant was to ask whether or not the OP was in high school or college, but I guess it wouldn't matter much with the point I was trying to communicate. I certainly didn't mean to come off as being condescending if it seemed like it.

    And if your average high school is anything like my old school, I'd say the vast majority of the students entering university aren't prepared mathematically. Two of the three math teachers I had in high school were football coaches who didn't have the faintest idea of what they were talking about.

    But don't let your current situation discourage you. It's up to you how far you go, so keep at it!
  15. Feb 9, 2006 #14
    I guess that's the problem with communicating via internet :wink:. Sometimes the meaning of what someone types is different than what you think it is simply because you can't hear the tone of their voice.

    Thanks for the advice!
    Last edited: Feb 9, 2006
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