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  1. #1
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    1 Last question Guys

    Guys I really appreciate all of your help today, as I am trying very, very hard to learn all of this. Thanks to Soroban I am actually starting to grasp this concept, but I have one last question.

    Suppose I had

    Cos(2x)=3/5 and I know that 2x is in quadrant I.

    My Unknown is: tan(x)=?

    Since this is equal to 3/5 how would I go about appyling the double angle formula so than I can say cos(x)=adj/hyp, and so fourth. I like putting it into this form because it makes it way easier for me to solve the problem. I know that Cos(2x)=Cos(2+x) and that Cos(2+x)= Cos(2)*Cos(x)-Sin(2)*Sin(x) but I am just not really sure how to implement the 3/5 what would I substitue it for?

    Again Thanks for all of your help!

    I love this site!
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Guys I really appreciate all of your help today, as I am trying very, very hard to learn all of this. Thanks to Soroban I am actually starting to grasp this concept, but I have one last question.

    Suppose I had

    Cos(2x)=3/5 and I know that 2x is in quadrant I.

    My Unknown is: tan(x)=?

    Since this is equal to 3/5 how would I go about appyling the double angle formula so than I can say cos(x)=adj/hyp, and so fourth. I like putting it into this form because it makes it way easier for me to solve the problem. I know that Cos(2x)=Cos(2+x) and that Cos(2+x)= Cos(2)*Cos(x)-Sin(2)*Sin(x) but I am just not really sure how to implement the 3/5 what would I substitue it for?

    Again Thanks for all of your help!

    I love this site!
    you had made just one mistake
    it is
    Quote Originally Posted by qbkr21 View Post
    I know that Cos(2x)=Cos(2+x) and that Cos(2+x)= Cos(2)*Cos(x)-Sin(2)*Sin(x)
    2x =x+x
    2x is not equal to 2+x
    Correct it and you will easily get the answer.
    While solving, remember that 2x is in first quadrant.
    If you are not able to solve, I willpost it for you. But just try, you will enjoy doing it.

    Keep Smiling
    Malay
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Guys I really appreciate all of your help today, as I am trying very, very hard to learn all of this. Thanks to Soroban I am actually starting to grasp this concept, but I have one last question.

    Suppose I had

    Cos(2x)=3/5 and I know that 2x is in quadrant I.

    My Unknown is: tan(x)=?

    Since this is equal to 3/5 how would I go about appyling the double angle formula so than I can say cos(x)=adj/hyp, and so fourth. I like putting it into this form because it makes it way easier for me to solve the problem. I know that Cos(2x)=Cos(2+x) and that Cos(2+x)= Cos(2)*Cos(x)-Sin(2)*Sin(x) but I am just not really sure how to implement the 3/5 what would I substitue it for?

    Again Thanks for all of your help!

    I love this site!
    There may be a fancier and more direct way to do this, but since we know what quadrant the angle 2x is in, this will work.

    cos(2x) = 2cos^2(x) - 1 will let you find cos(x):

    \frac{3}{5} = 2cos^2(x) - 1

    2cos^2(x) = \frac{8}{5}

    cos^2(x) = \frac{4}{5}

    cos(x) = \frac{2}{\sqrt{5}} (which we know is positive, since if 2x is in QI, then x must also be.)

    Again:
    cos(2x) = 1 - 2sin^2(x) will let you find sin(x):

    \frac{3}{5} = 1 - 2sin^2(x)

    2sin^2(x) = \frac{2}{5}

    sin^2(x) = \frac{1}{5}

    sin(x) = \frac{1}{\sqrt{5}} (which we know is positive since x is in QI)

    So
    tan(x) = \frac{sin(x)}{cos(x)} = \frac{ \frac{1}{\sqrt{5}} }{ \frac{2}{\sqrt{5}} }

    tan(x) = \frac{1}{2}

    -Dan
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  4. #4
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    Re:

    Ok so when Cos(2x)=3/5

    I break the problem about using the Double Angle Identity:


    so...

    Cos(2x)= Cos(x+x) so:

    Cos(x+x)= Cos(x)*Cos(x)-Sin(x)*Sin(x)

    but I am still trying to figure out out you implement the 3/5 where would I put this? Where each of the "x's" are?

    Thanks
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Ok so when Cos(2x)=3/5

    I break the problem about using the Double Angle Identity:


    so...

    Cos(2x)= Cos(x+x) so:

    Cos(x+x)= Cos(x)*Cos(x)-Sin(x)*Sin(x)

    but I am still trying to figure out out you implement the 3/5 where would I put this? Where each of the "x's" are?

    Thanks
    There are 3 forms of the cosine double angle formula.
    cos(2x) = cos^2(x) - sin^2(x)
    cos(2x) = 2cos^2(x) - 1
    cos(2x) = 1 - 2sin^2(x)

    If you know that cos(2x) = 3/5 you don't substitute x = 3/5, what you have is an equation that you need to solve for x. (Like x^2 + 5 = 10, you don't put x = 10.)

    What I did in the previous post gets around that. Rather than solving for x (which in this case is an inexact proceedure since x is (very likely) irrational and thus would need to be approximated) I solved for cos(x) and sin(x) and used them to find tan(x).

    However if you like you can use your calculator to do the work for you:
    cos(2x) = 3/5

    Use the cos^{-1} button to find
    2x = cos^{-1}(3/5) = 0.927295218

    Thus
    x = \frac{0.927295218}{2} = 0.463647609

    Thus
    tan(0.463647609) = 0.5

    -Dan
    Last edited by topsquark; November 21st 2006 at 12:53 PM.
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  6. #6
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    Re:

    But tan(.927) doesn't = .5 it equals 1.3



    in you spare time could you please show me how to use it in this formula

    cos(2x)=cos(x)^2-sin2(x) or is this not possible.

    She doesn't want decimals she wants straight up fractions

    Thanks D
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by qbkr21 View Post
    But tan(.927) doesn't = .5 it equals 1.3



    in you spare time could you please show me how to use it in this formula

    cos(2x)=cos(x)^2-sin2(x) or is this not possible.

    She doesn't want decimals she wants straight up fractions

    Thanks D
    First I copied and pasted the wrong number. I fixed it in the post. Thanks for catching that.

    I couldn't figure out a way to use the formula for cos(2x) in that form, since it involves both sine and cosine functions at the same time. Is there something about the solution I gave you earlier that you are having trouble with?

    -Dan
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  8. #8
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    Talking Re:

    First off I have a crummy textbook that doesn't do a good job of expaining Trig., if I did have a good one I would read it before I posted so many questions, but needless to say I don't.

    For tests and future references I want to know how to solve problems when you are given a Sin(2x), Cos(2x), Tan(2x), CSC(2x), Sec(2x), Cot(2x). The problems are really easy for me when there is just Sin(x), Cos(x), Tan(x), CSC(x), Sec(x), Cot(x). If you take a look at this post Soroban created he did a real nice job of of putting each in terms of (opp)^2+ (adj)^2= (Hyp)^2. I really had never though of using Path Theorm this way. Here is the post is there any you can corollate his reasoning with this problem?

    http://www.mathhelpforum.com/math-he...-problems.html
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by qbkr21 View Post
    First off I have a crummy textbook that doesn't do a good job of expaining Trig., if I did have a good one I would read it before I posted so many questions, but needless to say I don't.

    For tests and future references I want to know how to solve problems when you are given a Sin(2x), Cos(2x), Tan(2x), CSC(2x), Sec(2x), Cot(2x). The problems are really easy for me when there is just Sin(x), Cos(x), Tan(x), CSC(x), Sec(x), Cot(x). If you take a look at this post Soroban created he did a real nice job of of putting each in terms of (opp)^2+ (adj)^2= (Hyp)^2. I really had never though of using Path Theorm this way. Here is the post is there any you can corollate his reasoning with this problem?

    http://www.mathhelpforum.com/math-he...-problems.html
    The problem with the double angle formulas is that the triangle where you have an angle of 2x doesn't have a simple relation to the triangle where you have an angle x. There is possibly a way to do it this way, but it is beyond my ken. Sometimes you just have to go algebraic instead of visual, I'm afraid.

    -Dan
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  10. #10
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    Re:

    Ok thanks Dan... so if you were me would you just suck it up and memorize those three formula's, and use the calculator. I mean to me this is the easiest way I have seen. Also I take it that you divided by 2 because you are solving for x? So if it was a 3x,4x, or 5x of cos or what not you would divide by 3,4, or 5 as well? Right now I am sitting a nearly a page of trig functions deviations all of which I memorized today. My head hurts. I use the TI voyage 200 Calculator and it has all of the trig fuctions so I guess I can just use it that way. Also if she wants all of this in degrees instead of radians, I probably could just set my calc to degrees instead of constantly mulitplying by (180/pi)? Thanks Dan and I look forward to hearing back from you...
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  11. #11
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    Don't memorize all the formulas.

    If you know \sin^2(x)+\cos^2(x)=1 and \cos(2x)=\cos(x+x)=\cos^2(x)-\sin^2(x) you can derive them.
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  12. #12
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    Re:

    When using decimals how many significant figures would you use?

    Also if I have sec(alpha/2)=-4 and I am looking for tan(alpha) is there a way on the calculator to find the inverse like we did earlier?
    Last edited by qbkr21; November 21st 2006 at 04:36 PM. Reason: Cold hands
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  13. #13
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    Sure. \frac{1}{\cos(\frac{\alpha}{2})}=-4

    \cos(\frac{\alpha}{2})=-\frac{1}{4}

    See it now?
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  14. #14
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    Re:

    So from here if I was look for the tan(x) what formula would I use?
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  15. #15
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    Re:

    For that problem in particular I got tan(x)= 3/5


    I set the problem up like this:

    alpha/2=arcsec(-4)

    got the arcsec(-4) then multiplied it by 2

    then took that answer and put it inside the tan() button on my calculator which then gave me= .6 which is about 3/5?


    Is this right...Can I it this way?
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