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Math Help - Need sum and difference on trigonomtery

  1. #1
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    Need sum and difference on trigonomtery

    Hi,

    I'm getting close to the right answer, but one little thing is off. I think I'm missing a negative somewhere...

    "Find the exact value for each of the following under the given conditions:
    (a) sin(alpha + beta) ... etc...


    Sin alpha = 5/13 , -3pi/2 < alpha < -pi ; Tan beta = -sqrt of 3 , pi/2 < beta < pi
    Here's my work:

    I make a couple of pictures to find the missing side. For alpha I get a triangle in quadrant II. I get the missing, base, side is equal to 12 after using the Pythagorean theorem. The 12 is also negative because of the quadrant.

    For beta I get another triangle in quadrant II. I get the missing, hypotenuse, side is equal to 2.

    I then plug the numbers into the sum and difference formula for Sin(alpha + beta) which is sin alpha cos beta + cos alpha sin beta

    I get (5/13) (-1/2) + (-12/13) (sqrt3/2)

    = (-5/26) + (-12sqrt3/26)

    = -5 - 12sqrt3 all over 26.

    The answer should be 5 + 12sqrt3 all over 26.

    Where are my negatives coming from? Did I use the wrong quadrant or did I miss something in the work?

    Thanks,


    -Tom
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  2. #2
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    Double angle formula help as well...

    This one is giving me issues too...

    Use the information given about the angle theta, 0 < theta < 2pi, to find the exact value of
    (a) sin(2theta) ... etc...


    cos theta = - sqrt6/3, pi/2 < theta < pi
    I find the missing side... -sqrt6 + y^2 = -3^2

    -6 + y^2 = -9
    +6 +6

    y^2 = -3

    y = -sqrt3

    Then I plug them into the double angle formula... 2 sin theta cos theta.

    2 (sqrt3/3)(- sqrt6/3)

    Then I get 2( - sqrt18/3)

    The answer should be -2 sqrt2/3 ... Did I just make a simple mistake or something?

    Thanks again,


    -Tom
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TomCat View Post
    Hi,

    I'm getting close to the right answer, but one little thing is off. I think I'm missing a negative somewhere...



    Here's my work:

    I make a couple of pictures to find the missing side. For alpha I get a triangle in quadrant II. I get the missing, base, side is equal to 12 after using the Pythagorean theorem. The 12 is also negative because of the quadrant.

    For beta I get another triangle in quadrant II. I get the missing, hypotenuse, side is equal to 2.

    I then plug the numbers into the sum and difference formula for Sin(alpha + beta) which is sin alpha cos beta + cos alpha sin beta

    I get (5/13) (-1/2) + (-12/13) (sqrt3/2)

    = (-5/26) + (-12sqrt3/26)

    = -5 - 12sqrt3 all over 26.

    The answer should be 5 + 12sqrt3 all over 26.

    Where are my negatives coming from? Did I use the wrong quadrant or did I miss something in the work?

    Thanks,


    -Tom
    Your work looks good to me.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TomCat View Post
    2 (sqrt3/3)(- sqrt6/3)

    Then I get 2( - sqrt18/3)
    Careful.
    2 (sqrt3/3)(- sqrt6/3) = -2*sqrt(18)/9

    Now, sqrt(18) = sqrt(2*9) = 3*sqrt(2)

    So
    2 (sqrt3/3)(- sqrt6/3) = -2*sqrt(18)/9 = -2*3*sqrt(2)/9 = -2*sqrt(2)/3

    -Dan
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