r formula visualization

**arccos** · November 14th 2010, 01:43 AM

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

**mathaddict** · November 14th 2010, 02:20 AM

Originally Posted by arccos

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.

**arccos** · November 14th 2010, 03:34 AM

Originally Posted by mathaddict

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.

**mathaddict** · November 14th 2010, 05:41 PM

Originally Posted by arccos

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

**arccos** · November 14th 2010, 05:47 PM

Originally Posted by mathaddict

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?

**mathaddict** · November 14th 2010, 06:33 PM

Originally Posted by arccos

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)

**arccos** · November 14th 2010, 06:51 PM

Originally Posted by mathaddict

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)

**mathaddict** · November 15th 2010, 03:34 AM

Originally Posted by arccos

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)

**arccos** · November 15th 2010, 04:32 AM

Originally Posted by mathaddict

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.

**mathaddict** · November 15th 2010, 04:46 AM

Originally Posted by arccos

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 .

**arccos** · November 15th 2010, 05:07 AM

Originally Posted by mathaddict

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.

**mathaddict** · November 15th 2010, 06:41 AM

Originally Posted by arccos

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.

**arccos** · November 15th 2010, 06:46 AM

Originally Posted by mathaddict

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. "

**mathaddict** · November 15th 2010, 06:54 AM

Originally Posted by arccos

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)

**arccos** · November 15th 2010, 07:15 AM

Originally Posted by mathaddict

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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