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Math Help - Pre-calc Idenities & math Help me understand this please

  1. #1
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    Smile Pre-calc Idenities & math Help me understand this please

    Ok I don't understand how to do these 3 types of questions.
    Please help me understand how to do them. Thanks for your time.


    1. Verify the identity (Can only change one side, and end result must equal each other)
    cos^3Xsin^2X=(sin^2X-sin^4X)cosX

    2. Sole the equation.
    4cosX=1+2cosX

    3. Find the exact values of the sine, cosine, and tangent of the angle by using the sum or difference formula.
    345 degrees = 300 degrees + 45 degrees

    Thanks for any help.
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  2. #2
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    Quote Originally Posted by mrjameslat View Post
    Ok I don't understand how to do these 3 types of questions.
    Please help me understand how to do them. Thanks for your time.


    1. Verify the identity (Can only change one side, and end result must equal each other)
    cos^3Xsin^2X=(sin^2X-sin^4X)cosX

    left side ... cosx(cos^2x sin^2x) = cosx[(1 - sin^2x)sin^2x] = cosx(sin^2x - sin^4x)

    2. Sole the equation.
    4cosX=1+2cosX

    subtract 2cosx from both sides ...

    2cosx = 1

    cosx = 1/2 ... x = pi/3, x = 5pi/3


    3. Find the exact values of the sine, cosine, and tangent of the angle by using the sum or difference formula.
    345 degrees = 300 degrees + 45 degrees

    use the sum/difference identities ...

    cos(a+b) = cosa cosb - sina sinb

    sin(a+b) = sina cosb + cosa sinb

    tan(a+b) = (tana + tanb)/(1 - tana tanb)

    in addition, you should be very familiar with the values on the unit circle ...

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    Thanks for the input.
    I guess i can make a paradigm of what you said. Thanks a lot.


    however with the last one would this be right

    345 degrees = 300 degrees + 45 degrees

    cos(300+45) = cos(300) cos(45) - sin(300) sin(45)

    sin(300+45) = sin(300) cos(45) + cos(300) sin(45)

    tan(300+45) = (tan(300) + tan(45))/(1 - tan(300) tan(45))

    and would they equal this.
    cos=1/2(√2/2)--√3/2(√2/2)=( √2+√6 )/4

    sin=-√3/2(1/2)+1/2(√2/2)=(√2-√6)/4

    and how would i do the tangent one? thanks again.
    Last edited by mrjameslat; February 9th 2011 at 08:12 AM.
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    Did i do it right?
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  5. #5
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    Quote Originally Posted by mrjameslat View Post
    however with the last one would this be right

    345 degrees = 300 degrees + 45 degrees

    cos(300+45) = cos(300) cos(45) - sin(300) sin(45)

    sin(300+45) = sin(300) cos(45) + cos(300) sin(45)

    tan(300+45) = (tan(300) + tan(45))/(1 - tan(300) tan(45))

    and would they equal this.
    cos=1/2(√2/2)--√3/2(√2/2)=( √2+√6 )/4 ... ok

    sin=-√3/2(1/2)+1/2(√2/2)=(√2-√6)/4 ... ok

    and how would i do the tangent one? tanx = sinx/cosx
    ... you can also check your results w/ a calculator.
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    Quote Originally Posted by skeeter View Post
    in addition, you should be very familiar with the values on the unit circle ...

    I can get all that information by knowing the 45-45-90 triangle, the 30-60-90 triangle, and the graphs of sin and cosine.
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    Quote Originally Posted by Ackbeet View Post
    I can get all that information by knowing the 45-45-90 triangle, the 30-60-90 triangle, and the graphs of sin and cosine.
    ... and all I need is the 45-45-90 triangle, the 30-60-90 triangle, and the beauty of symmetry.
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    Out of curiosity, how would you get the values of sin and cosine on the x and y axes via symmetry?
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    Quote Originally Posted by Ackbeet View Post
    Out of curiosity, how would you get the values of sin and cosine on the x and y axes via symmetry?
    the picture of the unit circle is color-coded to show that symmetry from quadrant to quadrant.
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    Yeah, but I'm talking about values that are on the x axis, or on the y axis. By symmetry, how do you know that \cos(\pi/2)=0?
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    I guess I would need to remember the basic definition ... \cos{\theta} = \dfrac{x}{r}.
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    Ah, so you'd have to remember something in addition to the 45-45-90, the 30-60-90, and symmetry.
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    Ok when i try the tan one i get stuck please help?

    tan(300+45) = (sina/cosa + sinb/cosb)/(1 – sina/cosa*sinb/cosb) =

    ( -√3/2/1/2 + √2/2/1/2 )/(1- -√3/2/1/2 * √2/2/1/2)=

    (-2√3+2√2) / (1+2√3*2√2)

    What do I need to do, or what have I done wrong?
    thanks
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    Quote Originally Posted by mrjameslat View Post
    Ok when i try the tan one i get stuck please help?

    tan(300+45) = (sina/cosa + sinb/cosb)/(1 sina/cosa*sinb/cosb) =

    ( -√3/2/1/2 + √2/2/1/2 )/(1- -√3/2/1/2 * √2/2/1/2)=

    (-2√3+2√2) / (1+2√3*2√2)

    What do I need to do, or what have I done wrong?
    thanks
    recheck your values and simplify them before using the sum identity for tangent ...

    \tan(300) = -\sqrt{3}

    \tan(45) = 1

    \tan(300+45) = \dfrac{1 - \sqrt{3}}{1 + \sqrt{3}}

    if you decide to rationalize the denominator, the value will simplify to \sqrt{3} - 2
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    "work smart, not hard" thanks a lot!
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