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Adding sin plus cos with fractions containing square roots

  1. Sep 5, 2008 #1
    How do I solve:

    [sin (pi/3)] + [cos (pi/6)]? <--- "pi" is 3.14......

    I think that [sin (pi/3)]= (square root 3) divided by 2

    AND that [cos (pi/6)]= (square root 3) divided by 2.

    Now I can't remember how to add fractions containing square roots.

    My textbook says the answer is: square root 3.

    I just have NO IDEA how to get there.

    Please help! Thanks in advance.
     
  2. jcsd
  3. Sep 6, 2008 #2
    Sorry... I just read that this forum isn't for getting homework help. But I don't know how to move this post, so if somebody in here could help me, I'd appreciate it.
     
  4. Sep 6, 2008 #3

    berkeman

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    Staff: Mentor

    No worries. You may have noticed that it took only nanoseconds for your post to get moved to Homework Help :-)

    Are you allowed to use a calculator, or do you need to use tables, or are you supposed to use some other methods?
     
  5. Sep 6, 2008 #4
    Wow... A+ for getting my post the heck out of dodge.

    Calculators are permitted; but I can't get mine to cooperate. There is a "Trigonometric Functions of Special Angles" table in our textbook and I see that the sin pi/3 = (square root 3)/2, and that cos pi/6 = (square root 3)/2.

    Sine my original post, I converted radians to degrees to see if that would work, and it did... kind of.

    For the sin part of the problem, the radians = 60 degrees. The sin of 60 degrees is .8660.

    For the cos part of the problem, the radians = 30 degrees. The cos of 30 degrees is .8660.

    I realize .8660 + .8660 = 1.732 which is THE SQUARE ROOT OF 3.

    But, how am I supposed to know 1.732 = the square root of 3 on my own? (The book says the answer is the square root of 3.)

    I'm assuming I went about the problem differently, so I would now like to know how to work it using square roots.
     
  6. Sep 6, 2008 #5

    HallsofIvy

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    Staff Emeritus
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    You are going to (or should!) feel very silly about this. You add fractions with square roots the same way you add any fractions: by getting a common denominator. Since you already have a common denominator, 2, you just add the numerators. Or, since the fractions happen to be identical here, use "a+ a= 2a"! What is 2 times sqrt(3)/2?
     
  7. Sep 6, 2008 #6
    Could the equation look like this?

    sqrt(3)/2 + sqrt(3)/2 = sqrt(6)/2 = sqrt(3)/1 = sqrt(3)
     
  8. Sep 6, 2008 #7

    Borek

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    Staff: Mentor

    [tex]\frac {\sqrt 3} 2 + \frac { \sqrt 3} 2 = \frac { 2 \sqrt 3} 2 = \sqrt 3[/tex]

    It doesn't get any simpler :smile:

    Note: what you did was awfully wrong :frown: You can't add square roots this way. Think about it: if sqrt(4) + sqrt(4) = sqrt(8) then 2+2=sqrt(8), or 42=8. Obviously that's wrong.
     
  9. Sep 6, 2008 #8
    Oh!! I finally get it!! Thank you SO MUCH! I have seen the light of adding square roots.
     
  10. Sep 7, 2008 #9

    HallsofIvy

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    Yes, it is exactly like adding anything else: x/2+ x/2= 2x/2= x no matter what x is.
     
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