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    Hi need help with this c3 question:

    Given that f(x)=5sinx+12cosx

    find f(x) in the form of Rsin(x+alpha) where R and alpha are positive constants to be found. (0<alpha<90degrees)

    I found R which is equal to 13, but how do i find alpha?

    and how do i find the minimum value of f(x)+4??? thank you
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    I'm actually not sure why but this works for me (this is a bit silly to not understand, but we've got exams to pass...!). If you're converting to a sin graph I do tan^{-1}\frac{coeff. cos}{coeff. sin} and if converting to a cos graph tan^{-1}\frac{coeff. sin}{coeff. cos}.

    Consider the minimum value of a sin graph, for the last part...
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    (Original post by QED)
    I'm actually not sure why but this works for me (this is a bit silly to not understand, but we've got exams to pass...!). If you're converting to a sin graph I do tan^{-1}\frac{coeff. cos}{coeff. sin} and if converting to a cos graph tan^{-1}\frac{coeff. sin}{coeff. cos}.

    Consider the minimum value of a sin graph, for the last part...
    haha thats true, anwyways many thanks it still doesn't make sense to my because how can cos/sin=tan , but thanks anyways REP+ remind me to give u one tomorow ok I already gave one

    edit: in the markin scheme for this question it gives tan(alpha)=12/5 >> alpha=67.38 degrees

    i didn't get this i got tan(alpha)=5/12 it's so confusin isn't tan=Sin/cos??
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    Well Rsin(x+alpha)= R(sinx. cosalpha+cosx. sinalpha), so equate coeffiecients, sinx: 5=Rcosalpha and cosx: 12=Rsinalpha. divide so that Rsinalpha/Rcosalpha= 5/12, therefore tanalpha= 5/12 then just get rid of the tan to find alpha. sorry but not sure how to find the minimum value
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    (Original post by Remarqable M)
    Hi need help with this c3 question:

    Given that f(x)=5sinx+12cosx

    find f(x) in the form of Rsin(x+alpha) where R and alpha are positive constants to be found. (0<alpha<90degrees)

    I found R which is equal to 13, but how do i find alpha?

    and how do i find the minimum value of f(x)+4??? thank you
    You find alpha by drawing a right angled triangle. If you know what O and A are, and you know tan = O/A sub in the values of O and A.

    Then, tan a (alpha) = O/A

    Do inverse tan to find what alpha is.
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    5sinx+12cosx = 13sin(x+\alpha)
    You use the special angle formula:
    5sinx+12cosx = 13sinxcos\alpha + 13cosxsin\alpha
    nd you take either the cosx or sinx part of the equation:
    12cosx = 13cosxsin\alpha
    Divide by cosx or sinx (which ever you've chosen beforehand)
    12 = 13sin\alpha
    and solve for alpha
    \frac{12}{13} = sin\alpha

    Hope it helped.
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    (Original post by dinkyness_blob)
    Well Rsin(x+alpha)= R(sinx. cosalpha+cosx. sinalpha), so equate coeffiecients, sinx: 5=Rcosalpha and cosx: 12=Rsinalpha. divide so that Rsinalpha/Rcosalpha= 5/12, therefore tanalpha= 5/12 then just get rid of the tan to find alpha. sorry but not sure how to find the minimum value
    I did that, but in the mark scheme it says tan(alpha)=12/5 not 5/12 and this is confusing because how can tan=cos/sin??
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    QED is right (I think :hmmm:) about finding the value of alpha using tan, if you want to understand how it works, using the sine addition formula:

    \sin(x+\alpha)={\sin}x \ \cos \alpha + {\cos}x \sin \alpha

    and compare coefficients of both:

     \sin x and  \cos x

    You should then see how you can eliminate R, and find  \tan \alpha in one easy step

    EDIT: Too slow, damn me and my slow latex hands! :mad2:
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    (Original post by maxfire)
    5sinx+12cosx = 13sin(x+\alpha)
    You use the special angle formula:
    5sinx+12cosx = 13sinxcos\alpha + 13cosxsin\alpha
    nd you take either the cosx or sinx part of the equation:
    12cosx = 13cosxsin\alpha
    Divide by cosx or sinx (which ever you've chosen beforehand)
    12 = 13sin\alpha
    and solve for alpha
    \frac{12}{13} = sin\alpha

    Hope it helped.
    aah i see thanks for the help remind me after the exams to rep+

    edit: actually you know what remind me tomorrow to rep+++ you lol your explanation is wicked
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    Thanks to everyone for your input, but can someone answer the last question?? find the minimum?
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    The minimum is always the R value.
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    Max or min is found by setting differential of equation to zero, i.e. solve

     \frac{d}{dx}[ 5 sin x + 12 cos x + 4] = 0

    or solve in the other form

     \frac{d}{dx}[ 13sin(x + \alpha) + 4] = 0

    Either should work... Then sub in value of x you get from this into the original equation - I get 17 as answer!
 
 
 
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