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Math Help - r formula visualization

  1. #1
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    r formula visualization

    I am having problems with understanding the R formula of which I was required to self study for. I know how to convert basic equations using the formula but I have problems understanding the max and min. points.

    could anyone explain to me when do I use Rcos or Rsin and (\theta + or - \alpha) ?
    It would be good if you can provide me with the graphs so I can visualise the question better.

    example questions : find the maximum and minimum values of  4cos\theta - 3sin\theta , and 1 - 2sin\theta -cos\theta and their corresponding values of theta between 0 and 360 degrees.

    sorry about the paragraphing,I'm typing from my phone.
    thanks in advance
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  2. #2
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    Quote Originally Posted by arccos View Post
    I am having problems with understanding the R formula of which I was required to self study for. I know how to convert basic equations using the formula but I have problems understanding the max and min. points.

    could anyone explain to me when do I use Rcos or Rsin and (\theta + or - \alpha) ?
    It would be good if you can provide me with the graphs so I can visualise the question better.

    example questions : find the maximum and minimum values of  4cos\theta - 3sin\theta , and 1 - 2sin\theta -cos\theta and their corresponding values of theta between 0 and 360 degrees.

    sorry about the paragraphing,I'm typing from my phone.
    thanks in advance
    When the trigonometric equation is in the form of a cos x + b sin x or
    a sin x + b cos x where a, b are constants and x the angle in degrees or radians. Graphs are not of any help here.

    For your example questions, put them in the r-formula form. Then note maximums and minimums when the cosine(or sine) is 1 and -1 respectively.
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    When the trigonometric equation is in the form of a cos x + b sin x or
    a sin x + b cos x where a, b are constants and x the angle in degrees or radians. Graphs are not of any help here.

    For your example questions, put them in the r-formula form. Then note maximums and minimums when the cosine(or sine) is 1 and -1 respectively.
    This is not always the case(as it seems so to me)
    For example,
    4cos\theta - 3sin\theta = Rsin(\theta - \alpha)
    ....
    Maximum value = 5 when, sin(\theta - 53.1) = -1
    \theta = 270(as \, sin-90 = sin(360-90)) + 53.1 = 323.1(323.1 was the answer given in the textbook)

    I know this may sound weird but the way I look at it is that if the brackets containing theta and alpha has a negative sign in it, i treat the sine graph as inverted(opens downwards before curving upwards),in this way, the minimum point coordinate on the x axis is smaller than the maximum point and hence I can deduce if my answer is correct or not. Seems to be the only pattern that I can find from the questions I have done involving Rsin.
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  4. #4
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    Quote Originally Posted by arccos View Post
    This is not always the case(as it seems so to me)
    For example,
    4cos\theta - 3sin\theta = Rsin(\theta - \alpha)
    ....
    Maximum value = 5 when, sin(\theta - 53.1) = -1
    \theta = 270(as \, sin-90 = sin(360-90)) + 53.1 = 323.1(323.1 was the answer given in the textbook)

    I know this may sound weird but the way I look at it is that if the brackets containing theta and alpha has a negative sign in it, i treat the sine graph as inverted(opens downwards before curving upwards),in this way, the minimum point coordinate on the x axis is smaller than the maximum point and hence I can deduce if my answer is correct or not. Seems to be the only pattern that I can find from the questions I have done involving Rsin.
    For convenience, i will use 'a' for alpha, and 'x' for theta. It's more appropriate to use R cos (x+a) in this case.

    In your example, 4 cos x - 3 sin x = R cos (x+a) = 5 cos (x+36.87)

    so R=5 and tan a = 3/4 , a= 36.87 degree

    The maximum of it is 5, when cos (x+36.87)=1

    If the domain is given as 0<x<360(which is very likely to be in this case)

    Then, x+36.87=360 and x=323.1
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  5. #5
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    Quote Originally Posted by mathaddict View Post
    For convenience, i will use 'a' for alpha, and 'x' for theta. It's more appropriate to use R cos (x+a) in this case.

    In your example, 4 cos x - 3 sin x = R cos (x+a) = 5 cos (x+36.87)

    so R=5 and tan a = 3/4 , a= 36.87 degree

    The maximum of it is 5, when cos (x+36.87)=1

    If the domain is given as 0<x<360(which is very likely to be in this case)

    Then, x+36.87=360 and x=323.1
    It seems that sometimes when you use the wrong R(trig function)(x(+/-)a) you can't substitute =1 for maximum and =-1 for minimum points.
    About that, how would you know which to use in any given question?
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  6. #6
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    Quote Originally Posted by arccos View Post
    It seems that sometimes when you use the wrong R(trig function)(x(+/-)a) you can't substitute =1 for maximum and =-1 for minimum points.
    About that, how would you know which to use in any given question?
    I would suggest by looking at the leading term of the expression. For example, if the leading term in the expression is a sine, then the formula involving sine is used.

    In your example, the leading term is a cosine, so the formula with cosine is used. Also, note that you will have to flip signs when dealing with cosines.

    a cos x + b sin x = R cos (x - a)
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  7. #7
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    Quote Originally Posted by mathaddict View Post
    I would suggest by looking at the leading term of the expression. For example, if the leading term in the expression is a sine, then the formula involving sine is used.

    In your example, the leading term is a cosine, so the formula with cosine is used. Also, note that you will have to flip signs when dealing with cosines.

    a cos x + b sin x = R cos (x - a)
    What about questions like these?
    1-2sinx - cosx (where x = theta)
    It can either be 2sinx + cosx = 1 or cosx + 2sinx = 1
    For this question i rearranged it into 2sinx + cosx = 1 and used Rsin(x+a).

    2 = Rcosa , 1 = Rsina
    a = 26.6
    \displaystyle \sqrt5sin(\theta+26.6) =1
    How do i continue when a number is present on the RHS even before I equated it to 1 or -1 to find the max and min. points?

    Edit:

    I read somewhere before that we ignore the number for these type of questions. In that case I tried,
    \displaystyle 1 - 2 sin\theta - cos\theta

    \displaystyle - 2 sin\theta - cos\theta = Rsin(\theta-\alpha)

    \displaystyle -2 =Rcos\alpha

    \displaystyle 1 =Rsin\alpha

    \displaystyle \alpha = arctan(-0.5) = -26.6
    Stuck here.

    \displaystyle sin(\theta-(-26.6)) = 1

    will give 90-26.6 = 63.4 which is not the maximum point(243.4)
    Last edited by arccos; November 14th 2010 at 08:38 PM.
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  8. #8
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    Quote Originally Posted by arccos View Post
    What about questions like these?
    1-2sinx - cosx (where x = theta)
    It can either be 2sinx + cosx = 1 or cosx + 2sinx = 1
    For this question i rearranged it into 2sinx + cosx = 1 and used Rsin(x+a).

    2 = Rcosa , 1 = Rsina
    a = 26.6
    \displaystyle \sqrt5sin(\theta+26.6) =1
    How do i continue when a number is present on the RHS even before I equated it to 1 or -1 to find the max and min. points?

    Edit:

    I read somewhere before that we ignore the number for these type of questions. In that case I tried,
    \displaystyle 1 - 2 sin\theta - cos\theta

    \displaystyle - 2 sin\theta - cos\theta = Rsin(\theta-\alpha)

    \displaystyle -2 =Rcos\alpha

    \displaystyle 1 =Rsin\alpha

    \displaystyle \alpha = arctan(-0.5) = -26.6
    Stuck here.

    \displaystyle sin(\theta-(-26.6)) = 1

    will give 90-26.6 = 63.4 which is not the maximum point(243.4)
    Actually, it doesn't matter.

    1-2 sin x- cos x = 1- (2 sin x+ cos x)= 1-(cos x+2sin x)

    In order for the expression to be maximum, 2 sin x+ cos x or cos x+2sin x has to be minimum.

    2 sin x + cos x = sqrt(5)sin (x+26.57)

    cos x+ 2 sin x = sqrt(5) cos (x-63.43)

    When sin (x+26.57) = -1 , x=243.4

    When cos (x-63.43) = -1 , x=243.4

    So it doesn't matter which one you use, they will arrive at the same result.

    Therefore, the maximum is 1+sqrt(5)
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  9. #9
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    Quote Originally Posted by mathaddict View Post
    Actually, it doesn't matter.

    1-2 sin x- cos x = 1- (2 sin x+ cos x)= 1-(cos x+2sin x)

    In order for the expression to be maximum, 2 sin x+ cos x or cos x+2sin x has to be minimum.

    2 sin x + cos x = sqrt(5)sin (x+26.57)

    cos x+ 2 sin x = sqrt(5) cos (x-63.43)

    When sin (x+26.57) = -1 , x=243.4

    When cos (x-63.43) = -1 , x=243.4

    So it doesn't matter which one you use, they will arrive at the same result.

    Therefore, the maximum is 1+sqrt(5)

    So for questions involving a number, i just factor out all negatives amongst the sinx and cosx terms and treat the maximum as = -1 and vice versa?
    Oh and just to check, when finding alpha,lets say tan(alpha) = -3 , when we take the arctan we always take the absolute value(positive) of the number(sorry,bad explanation)?

    Eg tan(alpha) = -3 , alpha = arctan 3 ?

    Thanks for your help.
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  10. #10
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    Quote Originally Posted by arccos View Post
    So for questions involving a number, i just factor out all negatives amongst the sinx and cosx terms and treat the maximum as = -1 and vice versa?
    Oh and just to check, when finding alpha,lets say tan(alpha) = -3 , when we take the arctan we always take the absolute value(positive) of the number(sorry,bad explanation)?

    Eg tan(alpha) = -3 , alpha = arctan 3 ?

    Thanks for your help.
    No, not in all case. For example,

    1/(3+ cos x) , the maximum for this expression is when 3+cos x is as small as possible and this happens when cos x=-1

    Yes, you take the absolute values.

    Just bear in mind that the max and min for both cos and sin are 1 and -1 respectively .
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  11. #11
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    Quote Originally Posted by mathaddict View Post
    No, not in all case. For example,

    1/(3+ cos x) , the maximum for this expression is when 3+cos x is as small as possible and this happens when cos x=-1

    Yes, you take the absolute values.

    Just bear in mind that the max and min for both cos and sin are 1 and -1 respectively .
    I guess I wouldn't encounter a fraction in my trigonometry class for the R formula(yet).
    This minimum and maximum value thing is still a little confusing for me cause its like sometimes it's max = 1 and sometimes it's = -1. I tried playing with the graphs using Wolfram Alpha and I noticed that for any cosx and sinx expression,the graph seems to always follow the shape of the sine graph and if sinx is negative,the graph will be inverted(eg -cosx-sinx) , regardless of the sign that is infront of the cosx.
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  12. #12
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    Quote Originally Posted by arccos View Post
    I guess I wouldn't encounter a fraction in my trigonometry class for the R formula(yet).
    This minimum and maximum value thing is still a little confusing for me cause its like sometimes it's max = 1 and sometimes it's = -1. I tried playing with the graphs using Wolfram Alpha and I noticed that for any cosx and sinx expression,the graph seems to always follow the shape of the sine graph and if sinx is negative,the graph will be inverted(eg -cosx-sinx) , regardless of the sign that is infront of the cosx.
    The maximum and minimum for cosine is ALWAYS 1 and -1 respectively. It's the same for sine.
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  13. #13
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    Quote Originally Posted by mathaddict View Post
    The maximum and minimum for cosine is ALWAYS 1 and -1 respectively. It's the same for sine.
    I get that. This is the part that im confused with : "In order for the expression to be maximum, 2 sin x+ cos x or cos x+2sin x has to be minimum. "
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  14. #14
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    Quote Originally Posted by arccos View Post
    I get that. This is the part that im confused with : "In order for the expression to be maximum, 2 sin x+ cos x or cos x+2sin x has to be minimum. "
    1-2 sin x- cos x = 1- (2 sin x+ cos x) = 1 - sqrt(5)sin (x+26.57)

    Ask yourself, when is the expression maximum?

    1- sqrt(5)(1) or 1-sqrt(5)(-1)
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  15. #15
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    Quote Originally Posted by mathaddict View Post
    1-2 sin x- cos x = 1- (2 sin x+ cos x) = 1 - sqrt(5)sin (x+26.57)

    Ask yourself, when is the expression maximum?

    1- sqrt(5)(1) or 1-sqrt(5)(-1)
    Well I already looked at the answer so i would pick 1-sqrt(5)(-1).So the point here is to notice that only by equating it to -1 can we obtain all positive values in the maximum value?
    and if so, for the expression to be maximum, there must only be positive values?
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