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Math Help - [SOLVED] Value of sin2x and cos2x when tanx=5/3

  1. #1
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    [SOLVED] Value of sin2x and cos2x when tanx=5/3

    if tan x=5/3

    a) What is the value of sin (2x)
    b) What is the value of cos (2x)

    my awnser have to be in fraction

    Im not allowed to use a calculater to get an awnser and im unsure what im suppose to do...

    I think that sin x=5
    and cos x=3
    but i really don't know...

    Thanks in advance
    Janne the Swede
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  2. #2
    Super Member malaygoel's Avatar
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    tanx=\frac{sinx}{cosx}

    hence, \frac{sinx}{cosx}=\frac{5}{3}

    also, sin^2x+cos^2x=1

    (\frac{5cosx}{3})^2+cos^2x=1

    34cos^2x=9

    17cos2x+17=9

    hence,
    cos2x=\frac{-8}{17}
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  3. #3
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    Quote Originally Posted by JannetheSwede View Post
    if tan x=5/3

    a) What is the value of sin (2x)
    b) What is the value of cos (2x)

    my awnser have to be in fraction

    Im not allowed to use a calculater to get an awnser and im unsure what im suppose to do...

    I think that sin x=5
    and cos x=3
    but i really don't know...

    Thanks in advance
    Janne the Swede
    Hi

    Be careful : sin x and cos x are always between -1 and 1

    \sin 2x = 2 \sin x \cos x = 2 \tan x \cos^2 x

    A very useful relationship is \cos^2 x = \frac{1}{1+\tan^2 x}

    \sin 2x = 2  \frac{\tan x}{1+\tan^2 x} = \frac{15}{17}
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  4. #4
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    Are you allowed to use these (not so well known) multiple angle formulas?

    \sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta}

    \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}

    If so, just plug in 5/3 for \tan \theta and simplify.


    01
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  5. #5
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    Quote Originally Posted by malaygoel View Post
    tanx=\frac{sinx}{cosx}

    hence, \frac{sinx}{cosx}=\frac{5}{3}

    also, sin^2x+cos^2x=1

    (\frac{5cosx}{3})^2+cos^2x=1

    34cos^2x=9


    17cos2x+17=9

    hence,
    cos2x=\frac{-8}{17}
    (\frac{5cosx}{3})^2+cos^2x=1

    34cos^2x=9

    what do you do in between these steps?? i don't get it
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  6. #6
    Super Member malaygoel's Avatar
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    Quote Originally Posted by JannetheSwede View Post
    (\frac{5cosx}{3})^2+cos^2x=1

    34cos^2x=9

    what do you do in between these steps?? i don't get it
    Multiply both sides of the equation by 9...then recollect like terms
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  7. #7
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    Quote Originally Posted by malaygoel View Post
    Multiply both sides of the equation by 9...then recollect like terms
    Ahhh, i see ^^ but why 9?

    Never mind i saw it now
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  8. #8
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    I was unable to correctly use your way of getting cos2x

    I tested it on another question.

    tan(x)=1/7
    find a. a*cos(2x)
    =>
    ((1*cosx)/7)²+cos²x=1
    2cos²x=49
    and i just saw what I did wrong... i forgot to multiply the ...+cos²x with 49.. nm xD

    after i got 50cos²x=49
    is the awnser then cos2x+1=7
    =cos2x=-6/25

    or did I do it wrong somewhere?
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  9. #9
    Super Member malaygoel's Avatar
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    Quote Originally Posted by JannetheSwede View Post
    I was unable to correctly use your way of getting cos2x

    I tested it on another question.

    tan(x)=1/7
    find a. a*cos(2x)
    =>
    ((1*cosx)/7)²+cos²x=1
    2cos²x=49
    and i just saw what I did wrong... i forgot to multiply the ...+cos²x with 49.. nm xD

    after i got 50cos²x=49
    is the awnser then cos2x+1=7
    =cos2x=-6/25

    or did I do it wrong somewhere?
    50cos^2x=49

    50\frac{cos2x+1}{2}=49
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  10. #10
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    so cos2x=((49-1)*2)/50?

    =48/25
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  11. #11
    Super Member malaygoel's Avatar
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    Quote Originally Posted by JannetheSwede View Post
    so cos2x=((49-1)*2)/50?

    =48/25
    you made calculation mistake

    Spoiler:


    <br />
50\frac{cos2x+1}{2}=49<br />

    <br />
25cos2x+25=49<br />

    cos2x=\frac{24}{25}

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  12. #12
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    ok, thanks a lot
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