Results 1 to 4 of 4

Thread: cos32 + cos58 + cos20 + cos70

  1. February 19th 2006, 08:13 PM #1
    unit
    unit is offline
    Newbie
    Joined
    Feb 2006
    Posts
    2

    cos32 + cos58 + cos20 + cos70

    I ran into this problem on a worksheet:

    Simplify and Evaluate: (without using a calculator)
    cos32 + cos58 + cos20 + cos70


    I have tried using difference of sine and difference of cosine by
    cos (30 +2) = cos 30cos2 - sin30sin2
    etc.. with the others

    still nothing comes to mind


    anyone want to give it a shot?
    Follow Math Help Forum on Facebook and Google+
  2. February 19th 2006, 08:43 PM #2
    earboth
    earboth is offline
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,741
    Thanks
    90
    Quote Originally Posted by unit
    I ran into this problem on a worksheet:

    Simplify and Evaluate: (without using a calculator)
    cos32 + cos58 + cos20 + cos70

    I have tried using difference of sine and difference of cosine by
    cos (30 +2) = cos 30cos2 - sin30sin2
    etc.. with the others
    still nothing comes to mind

    anyone want to give it a shot?
    Hello,

    1. use \cos(\alpha)+\cos(\beta)=2\cos \frac{\alpha+\beta}{2}\ \cdot \ 2\cos \frac{\alpha-\beta}{2}

    2. The first 2 angles add up to 90°, the 2nd two too. So the first factor on the RHS of the equation will be \sqrt{2}

    I hope this will help you a little bit further on. And now I've to hurry: Duty is calling!

    Greetings

    EB
    Follow Math Help Forum on Facebook and Google+
  3. February 20th 2006, 04:06 PM #3
    unit
    unit is offline
    Newbie
    Joined
    Feb 2006
    Posts
    2
    Are there any other ways of doing this problem?

    I am not familiar with the identity cosA + cos B = 2cos((a+b)/2)*2cos((a-b)/2)

    Maybe you could show me how to derive this? All help greatly appreciated! Thanks!
    Follow Math Help Forum on Facebook and Google+
  4. February 21st 2006, 02:42 PM #4
    ThePerfectHacker
    ThePerfectHacker is offline
    Global Moderator ThePerfectHacker's Avatar
    Joined
    Nov 2005
    From
    New York City
    Posts
    10,603
    Quote Originally Posted by unit
    Are there any other ways of doing this problem?

    I am not familiar with the identity cosA + cos B = 2cos((a+b)/2)*2cos((a-b)/2)

    Maybe you could show me how to derive this? All help greatly appreciated! Thanks!
    Notice that,
    2\cos x\cos y=\cos x\cos y-\sin x\sin y+\cos x\cos y+\sin x\sin y
    Thus, using the formula:
    \cos(x\pm y)=\cos x\cos y\mp\sin x\sin y
    Thus, we have that:
    2\cos x\cos y=\cos(x+y)+\cos(x-y)
    Now let,
    A=x+y
    B=x-y
    Thus,
    x=\frac{A+B}{2}
    y=\frac{A-B}{2}
    Substituting this we get,
    \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
    Q.E.D.
    Follow Math Help Forum on Facebook and Google+
« tri | Stumped by this one »

Search Tags


/mathhelpforum @mathhelpforum