Can someone explain to me the R-Formula?(trigonometry)

[equilibrum]

Can someone explain to me the R formula?(expressing \displaystyle a cos \theta + b sin \theta in terms of R sin(\theta + \alpha))

I don't really understand my textbook at all which jumped straight to the questions. Could someone please enlighten me on the logic behind this concept and how it can be used to solve problems? I couldn't find any reference videos on youtube and/or on Wikipedia..


Any help at all would be appreciated.

 

[Mentallic]

Do you understand how to convert asin\theta+bcos\theta into Rsin(\theta+\alpha) ?

All you need to do is expand the right side, and notice that sin\alpha and cos\alpha are just some constants, the variable is \theta so what you can do is equate the coefficients of each side to find the value of R and \alpha.


It helps to solve problems because if you have asin\theta+bcos\theta=k then it is difficult to know how to solve for \theta but if we convert the left side to Rsin(\theta+\alpha) then \theta=sin^{-1}(k/R)-\alpha

 

[equilibrum]

Do you understand how to convert asin\theta+bcos\theta into Rsin(\theta+\alpha) ?

All you need to do is expand the right side, and notice that sin\alpha and cos\alpha are just some constants, the variable is \theta so what you can do is equate the coefficients of each side to find the value of R and \alpha.


It helps to solve problems because if you have asin\theta+bcos\theta=k then it is difficult to know how to solve for \theta but if we convert the left side to Rsin(\theta+\alpha) then \theta=sin^{-1}(k/R)-\alpha

Yes i do know how to convert.
Can you tell me all of the possible variants of the R formula and the required arctan function for α?
I found a sin θ + b cos θ = R sin(θ + α) online and in my textbook,but the website states that α = arctan(b/a) whereas my textbook states that α = arctan(a/b)

Which one of them is wrong? I don't really understand how they derived the arctan either.

 

[Mentallic]

Well if you know how to convert then you should understand where they're getting that equality from, and which one is the correct one. By the way, don't memorize those formulae, it is most likely going to be expected of you that you can reproduce this in an exam:

Rsin(\theta+\alpha)=Rsin\theta cos\alpha +Rcos\theta sin\alpha

Equating coefficients, we get

Rcos\alpha=a
and
Rsin\alpha=b

Now solve for \alpha. The fastest way would be to divide one equation by the other.

Can you tell me all of the possible variants of the R formula and the required arctan function for α?

Sorry I don't know what you mean by this.

 

[equilibrum]

Well if you know how to convert then you should understand where they're getting that equality from, and which one is the correct one. By the way, don't memorize those formulae, it is most likely going to be expected of you that you can reproduce this in an exam:

Rsin(\theta+\alpha)=Rsin\theta cos\alpha +Rcos\theta sin\alpha

Equating coefficients, we get

Rcos\alpha=a
and
Rsin\alpha=b

Now solve for \alpha. The fastest way would be to divide one equation by the other.



Sorry I don't know what you mean by this.

I saw a few types of R formula. There are the ones with asin +- bcos and those with acos +- bsin = Rcos/sin(theta +- alpha)

 

[Mentallic]

Well then what's the expansion of Rsin(\theta-\alpha)? And acos\theta-bsin\theta=-(bsin\theta-acos\theta) so the process is really the most important part, and you can adjust for your specific question accordingly.

Oh and the constants a,b can be negative too. It wasn't specified that they were just positive or anything like that so really there is no difference between any of the expressions you've shown.

 

[equilibrum]

Well then what's the expansion of Rsin(\theta-\alpha)? And acos\theta-bsin\theta=-(bsin\theta-acos\theta) so the process is really the most important part, and you can adjust for your specific question accordingly.

Oh and the constants a,b can be negative too. It wasn't specified that they were just positive or anything like that so really there is no difference between any of the expressions you've shown.

So to decide which R formula to use, we must first determine which terms come first(eg, asin + bcos or acos + bsin) and the required output(Rsin or Rcos) and expand accordingly to get the required R formula to solve the problem , am I right?

EDIT: Oh i think I saw somewhere that alpha must be a positive acute angle,but that doesn't seem to be the case for all questions. I have some answers in my book that is like -341.5 degrees etc. Is the range for alpha determined by the question or is it the same throughout this whole concept?

Sorry if I am asking too much :/

 

[Mentallic]

Do you realize that asin+bcos and bsin+acos are no different at all? a,b are just some constants, for all I care we could make it ksin+pcos. And as for whether it is asin+bcos or asin-bcos, that doesn't matter either. Again, b is just some constant which can be positive or negative, so -b can just be any constant as well.

For example, let's make it

asinx-bcosx=Rsin(x+k)

Expanding the right side gives

Rsinxcosk+Rcosxsink

Equating coefficients

Rcosk=a
Rsink=-b

Dividing the second by the first

tank=-b/a
k=atan(-b/a)

Substituting back into find R

Rcos(atan(-b/a))=a

Now do you know how to find the value of cos(tan(x)) in terms of x?

Oh and by the way, the value of alpha is generally between -180 degree and 180 degrees. If you find the value of alpha to be, say, 200 degrees then compute 200-360=-160 degrees to get in the required range.

 

[equilibrum]

I have started to attempt the questions in my book and for now I have no problems with the basic questions. However, I am still a little unsure of questions that require you to find the maximum and minimum value. For one,I do not know when to use Rcos,Rsin etc. and I don't really get the intuition behind some of the questions.

For example,
Find the maximum and minimum value of 20cos\theta + 21sin\theta and the corresponding values of theta where theta is between 0 and 360 degrees.

I understand R is the amplitude of the sine graph etc.
For the maximum value I had no problems but for the minimum value it was a little counter-intuitive for me.

Minimum Value = -R = -29 (no problems with this)

sin(\theta + 43.6) = -1

\theta + 43.6 = -90

sin(\theta + 43.6) = sin(360-90)

remove the sin from both sides,

\theta + 43.6 = 270

\theta = 226.4

I had to do this to fit theta into the required range. I thought that we have to introduce a trigonometric function if theta doesn't fit into the range given until I did the next question,Find the maximum and minimum value of 4cos\theta - 3sin\theta
For the maximum value,I got to
\theta - 53.1 = 90
Why can't I just bring the 53.1 over? It will still fit into the required range as stated in the question but that isn't the answer given in the answer sheet(max value = 5, theta = 323.1) (all digits are in degrees,i don't know the latex code for it)