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Math Help - Addition and Subtraction Formulas

  1. #1
    Member purplec16's Avatar
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    Addition and Subtraction Formulas

    tan 60 + tan 225

    \frac{tan60+tan225}{1-tan60 tan225}

    \frac{\sqrt{3}+1}{1-\sqrt{3}\times1}

    \frac{(\sqrt{3}+1)(1+\sqrt{3})}{1-\sqrt{3}}

    \frac{2\sqrt{3}+4}{-2}

    The answer is suppose to be -2-\sqrt{3}

    and im not gettin it can anyone help me?

    Another question, What is the difference between tan60+tan 225 and tan(60+225)
    because they give difference answers?
    Last edited by purplec16; March 15th 2010 at 06:37 PM.
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  2. #2
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    (2*sqrt3 + 4)/-2 = 2*sqrt3/-2 + 4/-2 =.........?
    And
    tanA + tanB is not equal to tan(A+B)
    tan(A+B) = (tanA+tanB)/[1-tanA*tanB]
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  3. #3
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    1) "The answer is supposed to be..."

    That is what you have. Try some algebra. There is a reason why you studied "simplification" over and over and over.


    2) What is the difference.

    Simply, you do not understand the concept fo funciton notation.

    f(a) is a function operating on a value 'a'.
    f(b) is a function operating on a value 'b'.
    f(a+b) is a function operating on the value 'a + b'.
    f(a) + f(b) is ths sum of a function operating on a value 'a' and that same function operating on a value 'b'.

    I am not encouraged that 1) You started the problem with the wrong expression and 2) you only got so far as to wonder "they give different answers". You really should start understanding what you are doing. One day you will discover the mathematics is not just a class to get through.
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  4. #4
    Member purplec16's Avatar
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    I know how to do the question, Thanks alot I understand now I was just kind of confused
    Last edited by purplec16; March 15th 2010 at 07:30 PM.
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  5. #5
    Member purplec16's Avatar
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    I have another question that I really need help with

    If \alpha  and  \beta are acute angles such that cos\alpha=\frac{4}{5} and tan\beta=\frac{8}{15}
    Find:
    sin(\alpha+\beta) and cos(\alpha+\beta)

    Which quadrant contains \alpha+\beta?
    Last edited by purplec16; March 15th 2010 at 07:58 PM.
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  6. #6
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    hi purplec16,

    x(60)+x(225)=x(60+225)

    That's algebraic multiplication.

    Tan(angle) is a "trigonometric function".
    So you must study the difference.
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  7. #7
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    Quote Originally Posted by purplec16 View Post
    I have another question that I really need help with

    If \alpha  and  \beta are acute angles such that csc\alpha=\frac{4}{5} and tan\beta=\frac{8}{15}
    Find:
    sin(\alpha+\beta) and cos(\alpha+\beta)

    Which quadrant contains \alpha+\beta?
    You will need to work these out using Pythagoras' Theorem
    and also the following trigonometric identities

    cos(A+B)=cosAcosB-sinAsinB

    sin(A+B)=sinAcosB+cosAsinB

    To start, draw 2 right-angled triangles.

    For the first csc\alpha=\frac{4}{5}=\frac{1}{sin\alpha}

    Therefore sin\alpha=\frac{5}{4}=\frac{opposite}{hypotenuse}

    So there is something wrong here as the opposite cannot be longer than the hypotenuse
    since the hypotenuse is the longest side of a right-angled triangle.

    Then, when you use Pythagoras' theorem, you find the 3rd side
    and you can write cos\alpha

    Do the same with the 2nd triangle, using tan\beta=\frac{8}{15}=\frac{opposite}{adjacent}

    Again you get the 3rd side of that to help you write cos\beta and sin\beta
    Last edited by Archie Meade; March 16th 2010 at 05:07 AM. Reason: incomplete
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  8. #8
    Member purplec16's Avatar
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    sorry it is [tex]cos\alpha[\math], I see why cos has to do with it, but I dont see how tan comes into play?
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  9. #9
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    ok,

    just use cos\alpha=\frac{4}{5}=\frac{adjacent}{hypotenuse}

    Adjacent=4, hypotenuse=5.

    Draw the triangle.

    Pythagoras' theorem gives (opposite)^2+(adjacent)^2=(hypotenuse)^2

    Use this to find the length of the 3rd side.

    Then write sin\alpha

    Finally use the trigonometric identity.

    You need sin\beta and cos\beta also.

    You get these when you have the 3rd side of the 2nd triangle.

    tan\beta gives you 2 of the 3 sides of that one.

    Pythagoras theorem gets you the 3rd.

    Then you can write cos\beta and sin\beta

    before using the trigonometric identity.
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  10. #10
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    Hello purplec16
    Quote Originally Posted by purplec16 View Post
    I have another question that I really need help with

    If \alpha  and  \beta are acute angles such that cos\alpha=\frac{4}{5} and tan\beta=\frac{8}{15}
    Find:
    sin(\alpha+\beta) and cos(\alpha+\beta)

    Which quadrant contains \alpha+\beta?
    For future reference, it's more helpful if you post additional questions as a new thread. (See Forum Rules, #14)

    Grandad
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  11. #11
    Member purplec16's Avatar
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    Quote Originally Posted by Archie Meade View Post
    ok,

    just use cos\alpha=\frac{4}{5}=\frac{adjacent}{hypotenuse}

    Adjacent=4, hypotenuse=5.

    Draw the triangle.

    Pythagoras' theorem gives (opposite)^2+(adjacent)^2=(hypotenuse)^2

    Use this to find the length of the 3rd side.

    Then write sin\alpha

    Finally use the trigonometric identity.

    You need sin\beta and cos\beta also.

    You get these when you have the 3rd side of the 2nd triangle.

    tan\beta gives you 2 of the 3 sides of that one.

    Pythagoras theorem gets you the 3rd.

    Then you can write cos\beta and sin\beta

    before using the trigonometric identity.
    Why does quandrant I contain \alpha+\beta
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  12. #12
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    It isn't a matter of figuring out which quadrant the angle (\alpha+\beta) is in, purplec.

    The question does not ask you to find the angle.

    It's asking for sin(\alpha+\beta) and cos(\alpha+\beta)

    Are you able to use Pythagoras theorem to write

    Sin\alpha

    cos\alpha

    sin\beta

    cos\beta ?
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  13. #13
    Member purplec16's Avatar
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    Yes I am i got the question but it ask the quadrant of alpha + beta?
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  14. #14
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    Ok,

    well, once you have found out what cos(\alpha+\beta)
    and sin(\alpha+\beta) are

    you find out the quadrant of the angle using the following...

    sin(angle) gives the vertical co-ordinate of a point on the unit circle circumference, centred at (0,0).

    cos(angle) gives the horizontal co-ordinate.

    Hence,

    sin + and cos + is the 1st quadrant 0-90 degrees

    sin + and cos - is the 2nd quadrant 90-180 degrees

    sin - and cos - is the 3rd quadrant 180-270 degrees

    sin - and cos + is the 4th quadrant 270-360 degrees
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