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Math Help - difficult trig identity help plz!

  1. #1
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    difficult trig identity help plz!

    I am having a lot of trouble proving these trig identities! If anyone can help me out, that would be fantastic!


    2 csc 2x = sec x csc x
    2 cot 2x = cot x - tan x

    and ...

    cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x ( question editted!!! )


    THANKS
    Last edited by finalfantasy; November 3rd 2007 at 08:30 AM.
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  2. #2
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    2 csc 2x = sec x csc x
    LHS = 2 csc 2x
    = 2 / sin 2x
    = 2 / 2 sin x cos x
    = 1 / sin x cos x
    = csc x sec x
    = RHS

    Thefore, proven.

    2 cot 2x = cot x - tan x
    LHS = 2 cot 2x
    = 2 / tan 2x
    = 2 / [2 tan x / 1 - tan^2 x]
    = 2[1 -tan^2 x] / 2 tan x
    = 1 - tan^2 x / tan x
    = cot x -tan x
    = RHS

    Therefore, proven.

    and ...

    cos6x + sin6x = 1 - 3 sin2x + 3 sin4x


    Let us check if that is true when x = 30 degrees,
    cos(6*30deg) +sin(6*30deg) =? 1 -3sin(2*30deg) +3sin(4*30deg)
    -1 +0 =? 1 -3(0.866) +3(0.866)
    -1 =? 1
    No.
    So, cos6x + sin6x = 1 - 3 sin2x + 3 sin4x is not an identity.
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  3. #3
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    Hello, finalfantasy!

    Another approach . . .


    1)\;\;2\csc 2x \:= \:\sec x \csc x

    We have: . \sec x\csc x \;=\;\frac{1}{\cos x}\cdot\frac{1}{\sin x} \;=\;\frac{2}{2\sin x\cos x} \;=\;\frac{2}{\sin2x} \;=\;2\csc2x



    2)\;\;2\cot 2x \:= \:\cot x - \tan x

    We have: . \cot x - \tan x \;=\;\frac{1}{\tan x} - \tan x \;=\;\frac{1-\tan^2\!x}{\tan x} \;=\;\frac{2(1-\tan^2\!x)}{2\tan x}

    . . = \;\frac{2}{\frac{2\tan x}{1-\tan^2\!x}} \;=\;\frac{2}{\tan2x} \;=\;2\cot2x<br /> <br />

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  4. #4
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    Hey guys! I editted the last question, can you solve it now? THANKS SO MUCH!

    @ Soroban: how does 1 / cos x * 1 / sin x give you 2 / 2sinxcosx

    1 times 1 is 1 right? So where does the 2 come in?


    @ticbol:

    how do you get this step highlighted in bold?

    LHS = 2 csc 2x
    = 2 / sin 2x
    = 2 / 2 sin x cos x
    = 1 / sin x cos x
    = csc x sec x
    = RHS

    what happens to the first 2 in 2 csc 2x?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by finalfantasy View Post
    cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x
    Recall that a^3 + b^3 = (a + b)(a^2 - ab + b^2)

    Here I am setting a = cos^2(x) and b = sin^2(x)

    Thus
    (cos^2(x) + sin^2(x))(cos^4(x) - cos^2(x)sin^2(x) + sin^4(x)) =? ~ 1 - 3sin^2(x) + 3sin^4(x)

    cos^4(x) - cos^2(x)sin^2(x) + sin^4(x) =? ~ 1 - 3sin^2(x) + 3sin^4(x)

    (1 - sin^2(x))^2 - (1 - sin^2(x))sin^2(x) + sin^4(x) =? ~ 1 - 3sin^2(x) + 3sin^4(x)

    Now expand out the LHS and simplify.

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by finalfantasy View Post
    Hey guys! I editted the last question, can you solve it now? THANKS SO MUCH!

    @ Soroban: how does 1 / cos x * 1 / sin x give you 2 / 2sinxcosx

    1 times 1 is 1 right? So where does the 2 come in?


    @ticbol:

    how do you get this step highlighted in bold?

    LHS = 2 csc 2x
    = 2 / sin 2x
    = 2 / 2 sin x cos x
    = 1 / sin x cos x
    = csc x sec x
    = RHS

    what happens to the first 2 in 2 csc 2x?
    He noted that 2sin(x)cos(x) = sin(2x). He only had a sin(x)cos(x), so he multiplied the numerator and denominator by 2 (essentially multiplying the fraction by 1.)

    You can get the same result by noting that sin(x)cos(x) = \frac{1}{2}sin(2x), putting this into the denominator and then simplifying the fraction. Since many people really don't like complex fractions, I think Soroban chose the best course.

    -Dan
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  7. #7
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    Hello, finalfantasy!

    \cos^6\!x  + \sin^6\!x \;= \;1 - 3\sin^2\!x + 3\sin^4\!x
    On the left side, use: . a^3 + b^3 \:=\:(a+b)(a^2-ab +b^2)

    \text{We have: }\;\cos^6\!x +\sin^6\!x \:=\:\underbrace{(\cos^2\!x + \sin^2\!x)}_{\text{This is 1}}(\cos^4\!x - \cos^2\!x\sin^2\!x + \sin^4\!x)

    . . . = \;\cos^4\!x - \cos^2\!x\sin^2\!x\,{\color{blue} + \,3\cos^2\!x\sin^2\!x} + \sin^4\!x\,{\color{blue} -\, 3\cos^2\!x\sin^2\!x}

    . . . = \;\underbrace{\cos^4\!x + 2\cos^2\!x\sin^2\!x + \sin^4\!x}  -  3\cos^2\!x\sin^2\!x

    . . . = \qquad\underbrace{(\cos^2\!x + \sin^2\!x)^2}_{\text{This is 1}} - 3\cos^2\!\sin^2\!x


    We have: . 1 - 3\cos^2\!x\sin^2\!x \;=\;1 - 3(1-\sin^2\!x)\sin^2\!x\;=\; 1 - 3\sin^2\!x + 3\sin^4\!x

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  8. #8
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    Quote Originally Posted by finalfantasy View Post
    Hey guys! I editted the last question, can you solve it now? THANKS SO MUCH!

    @ Soroban: how does 1 / cos x * 1 / sin x give you 2 / 2sinxcosx

    1 times 1 is 1 right? So where does the 2 come in?


    @ticbol:

    how do you get this step highlighted in bold?

    LHS = 2 csc 2x
    = 2 / sin 2x
    = 2 / 2 sin x cos x
    = 1 / sin x cos x
    = csc x sec x
    = RHS

    what happens to the first 2 in 2 csc 2x?
    Huh? How would I answer that?

    2csc(2x)
    = 2[1/sin(2x)
    = 2 / sin(2x)

    sin(2x) = 2sin(x)cos(x)

    So,
    = 2 / 2sin(x)cos(x)
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