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Thread: Getting exact answer from cos(u+-v) & sin(u+-v). Simple problem.

  1. March 1st 2009, 07:14 PM #1
    vane505
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    Getting exact answer from cos(u+-v) & sin(u+-v). Simple problem.

    Having a simple problem again.

    First: I'm really sorry, I've looked and I can't find and post on how to insert the correct "Math-Code" (well, I don't know how to insert degrees, square root and other things).

    * = degrees

    Example:
    Determine the EXACT value of cos(15*) by rewriting cos(15*) = cos(45*-30*):

    I can get exact values like:
    cos15* = cos(45*-30*) = cos45*cos30* + sin45*sin30* = 0.965925......
    But, ofcourse, this won't do. The correct answer is:
    (squareroot(6) + squareroot(2))/4

    How can I get the correct (well, I guess both are correct if I write the whole answer, but I mean the "better looking" one) later answer instead of the first? Appreciate help!!
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  2. March 1st 2009, 07:41 PM #2
    sbcd90
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    as u could see,that sqrt.(6) is a non-recurring & non-terminating decimal ,so this problem arises.
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  3. March 1st 2009, 08:03 PM #3
    vane505
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    Hi, and thank you for your response.

    I'm not sure if I understand you correct. You mean that sqrt.(6) is a never-ending decimal right?

    I probably wasn't clear with my problem, or . I can get the correct answer, but I don't know how to calculate/convert the result so that it shows (sqrt(6) + sqrt(2))/4 instead of 0.965925......

    I know how to calculate cos(45*-30*) = cos45*cos30*+sin45*sin30* = =0.965925......

    This is how the problem is solved in the book:
    cos(45* - 30*)= (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2) =
    = (sqrt(3) + 1)/(2sqrt(2)) = sqrt(2)(sqrt(3) + 1)/ = (sqrt(6)+sqrt(2))/4

    My issue is that I don't know how to rewrite the cos45*cos30*+sin45*sin30* into (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2)
    and so on.

    I'm sorry if my explanation is bad. I haven't studied much math before.

    Again, appreciate help!
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  4. March 1st 2009, 08:31 PM #4
    Reckoner
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    Quote Originally Posted by vane505 View Post
    Example:
    Determine the EXACT value of cos(15*) by rewriting cos(15*) = cos(45*-30*):

    I can get exact values like:
    cos15* = cos(45*-30*) = cos45*cos30* + sin45*sin30* = 0.965925......
    But, ofcourse, this won't do. The correct answer is:
    (squareroot(6) + squareroot(2))/4
    You should know the sine and cosine of 30^\circ and 45^\circ. If you do not, then you must review the unit circle.

    \cos15^\circ=\cos(45^\circ-30^\circ)=\cos45^\circ\cos30^\circ+\sin45^\circ\si n30^\circ

    =\left(\frac{\sqrt2}2\right)\left(\frac{\sqrt3}2\r ight)+\left(\frac{\sqrt2}2\right)\left(\frac12\rig ht)

    =\frac{\sqrt2\sqrt3}4+\frac{\sqrt2}4=\frac{\sqrt6} 4+\frac{\sqrt2}4=\frac{\sqrt6+\sqrt2}4.

    Edit: From your latest post it seems that you do not know your unit circle, so I suggest you review that section in the book. Wikipedia has an overview. These values are found using special properties of 45^\circ\text{-}45^\circ\text{-}90^\circ and 30^\circ\text{-}60^\circ\text{-}90^\circ triangles, discussions of which can be found in topics 6 and 7 on this page.
    Last edited by Reckoner; March 1st 2009 at 08:45 PM.
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  5. March 1st 2009, 09:00 PM #5
    vane505
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    Thank you for your help! I get it now
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  6. March 2nd 2009, 04:42 AM #6
    HallsofIvy
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    Quote Originally Posted by vane505 View Post
    Hi, and thank you for your response.

    I'm not sure if I understand you correct. You mean that sqrt.(6) is a never-ending decimal right?

    I probably wasn't clear with my problem, or . I can get the correct answer, but I don't know how to calculate/convert the result so that it shows (sqrt(6) + sqrt(2))/4 instead of 0.965925......

    I know how to calculate cos(45*-30*) = cos45*cos30*+sin45*sin30* = =0.965925......

    This is how the problem is solved in the book:
    cos(45* - 30*)= (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2) =
    = (sqrt(3) + 1)/(2sqrt(2)) = sqrt(2)(sqrt(3) + 1)/ = (sqrt(6)+sqrt(2))/4

    My issue is that I don't know how to rewrite the cos45*cos30*+sin45*sin30* into (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2)
    and so on.

    I'm sorry if my explanation is bad. I haven't studied much math before.

    Again, appreciate help!
    Can we at least assume that you know that cos(45)= \frac{\sqrt{2}}{2}, sin(45)= \frac{\sqrt{2}}{2}, cos(30)= \frac{\sqrt{3}}{2}, and sin(30)= \frac{1}{2}? This problem assumes that you already know that!

    Then cos(45)cos(30)= \frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}= \frac{\sqrt{6}}{4} and sin(45)sin(30)= \frac{\sqrt{2}}{2}\frac{1}{2}= \frac{\sqrt{2}}{4}
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  7. March 2nd 2009, 05:06 AM #7
    vane505
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    The strange thing is that the book never mentioned (yet) that it's important to know these specific values, so I thought there was some almost-impossible magic way to get these numbers by doing some kind of calculation.

    As said before, I'm a math-beginner
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