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Math Help - Problems From Trig Test

  1. #1
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    Problems From Trig Test

    Could someone please post a solution to these 4 questions:
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  2. #2
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Could someone please post a solution to these 4 questions

    Hello,

    to #3.a)

    \cos^3(2x)=\cos(2x)
    \cos^3(2x)-\cos(2x)=0
    \cos(2x)(\cos^2(2x)-1)=0
    \cos(2x)=0 \vee \cos^2(2x)-1=0
    \cos(2x) = 0  \vee \cos(2x) = -1 \vee \cos(2x) = 1

    (For some reason the Latex doesn't work properly anymore. I'll repeat the equations in plain text)

    cos(2x) = 0 or cos(2x) = -1 or cos(2x) = 1

    This character π means pi
    From the 1rst equ.:

    2x = -π/2 --> x = -π/4 or
    2x = -3π/2 ---> x = -3π/4

    from the 2nd equ.:

    2x = -π ---> x = -π/2

    from the 3rd equ.:

    2x = 0 ---> x = 0 or
    2x = -2π ---> x = -π

    EB
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  3. #3
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    Hello, SportfreundeKeaneKent!

    Here's the very first one . . .


    1. If cos(x) = -12/13, where π < x < 3π/2

    then find the exact value of sin(2x)

    We'll use the identity: .sin(2x) .= .2Ěsin(x)Ěcos(x)

    We know that: cos(x) = -12/13 . . . we need sin(x).


    Angle x is in Quadrant 3.
    Since cos(x) = -12/13 = adj/hyp,
    . . we use Pythagorus and find that: opp = -5
    Hence: .sin(x) = -5/13


    Therefore: .sin(2x) .= .2(-5/13)(-12/13) .= .120/169

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  4. #4
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    Hello again, SportfreundeKeaneKent!

    Here's #4(b) . . .

    Prove:
    . . . . . . . . . . . . . .tan(x) - tan(y)
    . . tan(x - y) . = . --------------------
    . . . . . . . . . . . . .1 + tan(x)Ětan(y)
    We will use these identities:

    . . sin(x - y) .= .sin(x)Ěcos(y) - sin(y)Ěcos(x)
    . . cos(x - y) .= .cos(x)Ěcos(y) + sin(x)Ěsin(y)


    . . . . . . . . . . . sin(x - y) . . . . . sin(x)Ěcos(y) - sin(y)Ěcos(x)
    tan(x - y) . = . ------------ . = . -----------------------------------
    . . . . . . . . . . . cos(x - y) . . . . .cos(x)Ěcos(y) + sin(x)Ěsin(y)


    Divide top and bottom by cos(x)Ěcos(y):

    . . .sin(x)Ěcos(y) . .sin(y)Ěcos(x) . . . . . . sin(x) . sin(y)
    . . .--------------- - --------------- . . . . . . ------- - -------
    . . .cos(x)Ěcos(y) . cos(x)Ěcos(y) . . . . . . cos(x) . cos(y)
    . . ------------------------------------ . = . ---------------------
    . . .cos(x)Ěcos(y) . .sin(x)Ěsin(y) . . . . . . . . sin(x) sin(y)
    . . .--------------- + --------------- . . . . .1 + -------Ě--------
    . . .cos(x)Ěcos(y) . .cos(x)Ěcos(y) . . . . . . . .cos(x) cos(y)


    . . . . . . . tan(x) - tan(y)
    . . . = . ---------------------
    . . . . . . 1 + tan(x)Ětan(y)

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  5. #5
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Could someone please post a solution to these 4 questions[/IMG]
    Hello, Sportsfreund!

    #3.b)

    \sec^4(x)-4\sec^2(x)+4=0,\ 0 < x < 2\pi

    Use substitution: y=\sec^2(x). Then you'll get the equation:

    y^2-4y+4=0 \Longleftrightarrow (y-2)^2=0 \Longleftrightarrow y=2

    Now re-substitute:

    \sec^2(x)=2\Longleftrightarrow \sec(x)=\sqrt{2}\  \vee \ \sec(x)=-\sqrt{2}

    From the 1rst equation you get:

    \sec(x)=\sqrt{2} \Longleftrightarrow x=\frac{1}{4} \cdot \pi\ \vee \ x=\frac{7}{4} \cdot \pi

    From the 2nd equation you get:

    \sec(x)=-\sqrt{2} \Longleftrightarrow x=\frac{3}{4} \cdot \pi\ \vee \ x=\frac{5}{4} \cdot \pi

    EB
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  6. #6
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Could someone please post a solution to these 4 questions
    Hello, Sportsfreund,

    #2

    Use the property:

    \cos(w-x)=\cos(w) \cos(x)+\sin(w) \sin(x)

    As Soroban has demonstrated use the Pythagoran theorem:

    \sin^2(x)+\cos^2(x)=1 \Longrightarrow \sin(x)=\sqrt{1-\left(\frac{4}{5} \right)^2}=-\frac{3}{4}. Because x is in the 4. quadrant the sine-value must be negative.

    \sin^2(w)+\cos^2(w)=1 \Longrightarrow \cos(w)=\sqrt{1-\left(\frac{7}{25} \right)^2}=-\frac{24}{25}. Because w is in the 2. quadrant the cosine-value must be negative.

    Now you know all values necessary:

    \cos(w-x)=\left(-\frac{24}{25}\right)\cdot \left(\frac{4}{5}\right)+\left(\frac{7}{25}\right)  \cdot \left(-\frac{3}{4}\right)=\left(-\frac{489}{500}\right)

    EB
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  7. #7
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    Hello, SportfreundeKeaneKent!

    There's something wrong with 4(a).
    . . Is there a typo?


    4a)\;\sec2x(1 - \sin2x) \:=\:\sec^2x - \tan^2x

    The right side is always equal to 1.
    . . So it's a silly way to write a problem.


    Besides, the statement is not an identity.

    Try x = π/3
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
    The left side is: .sec(π/3)[1 - sin(π/3)] .= .2(1 - √3/2) . . 1

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