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Math Help - Trigo identities--Pls Help!!

  1. #1
    Newbie xeno_vert's Avatar
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    Exclamation Trigo identities--Pls Help!!

    I beg all of you to pls help me to solve this trigonometric identities equation..the problems are 10, i already solve 5 and obviously, there's a remaining 5 problems..I can't solve them i mean i've been solving those remaining 5 questions for 5 hours and still no success..pls help me...these are the questions...

    * those enclosed in parentheses indicated that the number inside is an exponent)

    * / = over e.g. cosx/sinx (cosx over sinx)


    1.) cos(4)x - sin(4)x = 2cos(2)x - 1

    2.) 1 - (sinx / cscx) = cosx / secx

    3.) tan x / (1 - tan(2)x) = cosx / (cscx - 2sinx)

    4.) secx sinx / (tanx +cotx) = sin(2)x

    5.) cos(2)x - sin(2)x = 1 - 2sin(2)x


    plss help me...thank you in advance for those who'll helped...

    Godbless you all!!!
    Last edited by mr fantastic; November 29th 2008 at 04:02 AM. Reason: Added some brackets to try and make this pea soup easier to understand
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  2. #2
    Member great_math's Avatar
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    Quote Originally Posted by xeno_vert View Post
    1.) cos(4)x - sin(4)x = 2cos(2)x - 1

    2.) 1 - sinx / cscx = cosx / secx

    3.) tan x / 1 - tan(2)x = cosx / cscx - 2sinx

    4.) secx sinx / tanx = cotx = sin(2)x

    5.) cos(2)x - sin(2)x = 1 - 2sin(2)x
    1) \cos^4x-\sin^4x=2\cos^2x-1

    LHS =\cos^4x-\sin^4x

    =(\cos^2x+\sin^2 x)(\cos^2x-\sin^2 x)

    =\cos^2x-\sin^2x

    =\cos^2x-(1-\cos^2x)

    =\cos^2x-1+\cos^2x

    =2\cos^2x-1=RHS


    2) 1-\frac {\sin x}{\csc x}=\frac{\cos x}{\sec x}

    LHS=1-\frac {\sin x}{\csc x}

    =1-\frac{\sin x}{\frac 1{\sin x}}

    =1-\sin^2 x=\cos^2 x

    RHS=\frac{\cos x}{\sec x}

    =\frac{\cos x}{\frac 1{\cos x}}

    =\cos^2 x=LHS
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  3. #3
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    Quote Originally Posted by xeno_vert View Post

    3.) tan x / 1 - tan(2)x = cosx / cscx - 2sinx

    4.) secx sinx / tanx = cotx = sin(2)x

    5.) cos(2)x - sin(2)x = 1 - 2sin(2)x
    3) \frac{\tan x}{1-\tan^2x}=\frac{\cos x}{\csc x-2\sin x}

    LHS=\frac{\tan x}{1-\tan^2x}

    =\frac{\frac{\sin x}{\cos x}}{1-\frac{\sin^2x}{\cos^2x}}

    =\frac{\sin x}{\cos x}\times \frac{\cos^2x}{\cos^2-\sin^2x}

    =\frac{\sin x\cdot \cos x}{\cos^2x-\sin^2x}

    RHS=\frac{\cos x}{\csc x-2\sin x}

    =\frac{\cos x}{\frac{1}{\sin x}-2\sin x}

    =\frac{\cos x\cdot \sin x}{1-2\sin ^2 x}

    =\frac{\cos x\cdot \sin x}{\cos^2 - \sin^2 x}=LHS

    4th question seems to be incorrect

    5) \cos^2x  - \sin^2x  = 1 - 2\sin^2x

    <br />
RHS=1-2\sin^2x

    =1-\sin^2x-\sin^2x

    =\cos^2x-\sin^2x

    PROVED
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  4. #4
    Newbie xeno_vert's Avatar
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    3)





    I'm just confused how did u get
    cos(2)x / cos(2)x - sin(2) x ..
    sorry...


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  5. #5
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    OK so now the edited question
    4) \frac{\sec x\cdot\sin x}{\tan x+\cot x}=\sin^2x<br />

    LHS=\frac{\sec x\cdot\sin x}{\tan x+\cot x}

    =\frac{\frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}

    =\frac{\frac{\sin x}{\cos x}}{\frac{\sin^2x+\cos^2x}{\sin x\cdot\cos x}}

    =\frac{\sin x}{\cos x}\cdot \sin x\cdot\cos x

    =\sin^2x=RHS
    Last edited by great_math; November 29th 2008 at 02:20 AM.
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  6. #6
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    Quote Originally Posted by xeno_vert View Post
    3)



    I'm just confused how did u get
    cos(2)x / cos(2)x - sin(2) x ..
    sorry...
    1-\frac {\sin^2 x}{\cos^2x}

    =\frac{\cos^2x-\sin^2x}{\cos^2x}

    So the fraction becomes

    \frac{\frac{\sin x}{\cos x}}{\frac{\cos^2x-\sin^2x}{\cos^2x}}



    Got it?
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  7. #7
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    sorry..I'm confused again..

    Regarding Question 3:









    where did u get the cos(2)x??? (the denominator of cosx . sinx / cos(2) - sin(2)x = LHS
    Last edited by mr fantastic; November 29th 2008 at 04:07 AM. Reason: Added the question number to this mess masquerading as a thread.
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  8. #8
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    Quote Originally Posted by xeno_vert View Post
    sorry..I'm confused again..








    where did u get the cos(2)x??? (the denominator of cosx . sinx / cos(2) - sin(2)x = LHS

    Check the 5th question keeping in mind the denominator...
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    i'm terribly sorry for i ask a lot of questions..i'm really sorry..i'm really analyzing it but i need to be guided sometimes..i'm really sorry.. im confused again with this one...

    Regarding Question 5:








    how did u express secx sinx to sinx / cosx?
    sinx /cosx in terms of quotient relation is equivalent to tanx

    another things is that secx sinx in terms of product relation is equivalent to tanx (the original form is sinx secx, i just interchange them)

    tnx for the reminder...huhuhuh i'm really sorry...
    Last edited by mr fantastic; November 29th 2008 at 04:05 AM. Reason: Aded a question number
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  10. #10
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    Quote Originally Posted by xeno_vert View Post
    i'm terribly sorry for i ask a lot of questions..i'm really sorry..i'm really analyzing it but i need to be guided sometimes..i'm really sorry.. im confused again with this one...








    how did u express secx sinx to sinx / cosx?
    sinx /cosx in terms of quotient relation is equivalent to tanx

    another things is that secx sinx in terms of product relation is equivalent to tanx (the original form is sinx secx, i just interchange them)

    tnx for the reminder...huhuhuh i'm really sorry...
    Sorry I messed up Latex.. You can check it now..

    <br />
\sec x=\frac 1{\cos x}

    \sec x\cdot\sin x=\frac{\sin x}{\cos x}=\tan x<br />
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  11. #11
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    Quote Originally Posted by great_math View Post
    Sorry I messed up Latex.. You can check it now..

    <br />
\sec x=\frac 1{\cos x}

    \sec x\cdot\sin x=\frac{\sin x}{\cos x}=\tan x<br />

    will it changed the equation?
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  12. #12
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    So the fraction becomes






    im sorry...the question supposed to be is that where does the cos(2)x / cos(2) - sin(2)x was derived?
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    Quote Originally Posted by xeno_vert View Post
    will it changed the equation?
    No..We will land again where were are...

    Quote Originally Posted by xeno_vert View Post




    So the fraction becomes






    im sorry...the question supposed to be is that where does the cos(2)x / cos(2) - sin(2)x was derived?
    well what is \frac{\frac{1}{15}}{\frac{1}{5}} ?

    that is same as \frac{1}{15}\div\frac{1}{5}

    =\frac{1}{15}\times 5

    =\frac 1{3}

    Do you now get from where cos(2)x / cos(2) - sin(2)x was derived?
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  14. #14
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    Regarding the last part of the solution to Question 5:






    im sorry again..i'll just asked what happened to cosx?

    i min if that's the form then i guess it wud be sin(2)x / cos(2)x
    Last edited by mr fantastic; November 29th 2008 at 04:09 AM. Reason: Added a question number to this morass of confusion
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  15. #15
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    Quote Originally Posted by xeno_vert View Post





    im sorry again..i'll just asked what happened to cosx?

    i min if that's the form then i guess it wud be sin(2)x / cos(2)x
    Read it more carefully. The cos x's cancel.


    This whole thread is a headache because of the number of questions originally posted and the number of follow-up questions that have been asked.

    Next time: Do not ask more than one question in your original post.
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