1. ## 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.

2. 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.
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.

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.

4. 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

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?

6. 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)

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)
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)

8. Originally Posted by arccos
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)

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 ?

10. 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 ?

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 .

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.

12. 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.

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. "

14. 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)

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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