Cos(A-B) vs sin(A+B) [acos(x) + bsin(x) question]

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Homework Help Overview

The discussion revolves around solving the equation 3 cosx + 4 sinx = 2 for values of X from 0 to 360 degrees. Participants are exploring the implications of using the cosine of a difference formula, cos(A-B), versus the sine of a sum formula, sin(A+B), in the context of this problem.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the advantages of using cos(A-B) over sin(A+B) in solving the equation, with some expressing uncertainty about the benefits. There are attempts to compare the results obtained from both methods and questions about the steps taken in the solution process.

Discussion Status

The conversation is ongoing, with participants sharing their findings and questioning the effectiveness of each approach. Some have noted that the cos(A-B) method yields correct answers, while others are still considering the implications of their results from the sin(A+B) method.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the depth of exploration into the methods discussed. There is a focus on understanding the reasoning behind the choice of formulas rather than arriving at a definitive conclusion.

lionely
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Homework Statement


I already solved the question but there was a question at the end just for thought I guess.

Solve the equation 3 cosx + 4 sinx = 2, for values of X from 0 to 360, inclusive.

Again I already solved it ,the thing that I am curious about is the question below in bold.

" What advantage is there in using the formula for cos(A-B) , rather than that for sin(A+B) in the above question?"


Homework Equations

The Attempt at a Solution


I don't see any advantages. But I am not entirely sure?
 
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lionely said:

Homework Statement


I already solved the question but there was a question at the end just for thought I guess.

Solve the equation 3 cosx + 4 sinx = 2, for values of X from 0 to 360, inclusive.

Again I already solved it ,the thing that I am curious about is the question below in bold.

" What advantage is there in using the formula for cos(A-B) , rather than that for sin(A+B) in the above question?"

Homework Equations



The Attempt at a Solution


I don't see any advantages. But I am not entirely sure?
What steps did you follow when you solved it?
 
cosycosx + sinysinx = constant

Comparing this with
3cosx + 4sinx = 2

so
(cosy)/3 = (siny)/4 that is tany = 4/3

so y = 53.13
and siny = 4/5 and cos y = 3/5 therefore

3/5cosx + 4/5sinx = 2/5

therefore cosxcosy +sinxsiny = 0.4

cos(x-y) = 0.4
x- 53.13 = 66.42 or 293.58
so x = 119.6 and 346.7
 
lionely said:
cosycosx + sinysinx = constant

Comparing this with
3cosx + 4sinx = 2

so
(cosy)/3 = (siny)/4 that is tany = 4/3

so y = 53.13
and siny = 4/5 and cos y = 3/5 therefore

3/5cosx + 4/5sinx = 2/5

therefore, cosxcosy +sinxsiny = 0.4

cos(x-y) = 0.4
x- 53.13 = 66.42 or 293.58
so x = 119.6 and 346.7
So, that's the solution using cos(A - B).

The solution using sin(A + B) must have sin(y)cos(x) + cos(y)sin(x) = constant .

Giving tan(y) = 3/4, so that y ≈ 36.87° and sin(x + y) = sin(x + 36.87°) =0.4

Thus x + 36.87° ≈ 23.58° or 156.42° , plus multiples of 360° for each .

So, why is the cos(x-y) method easier to work with in this case?
 
Hmm because the cos(x-y) gives the write answers? When I put the answers gotten from the sin(A+B) it doesn't satisfy the equation.
 
lionely said:
Hmm because the cos(x-y) gives the write answers? When I put the answers gotten from the sin(A+B) it doesn't satisfy the equation.
You will get the right numbers eventuality.
x + 36.87° ≈ 23.58° gives -13.27° . Add 360° to that to get346.71° .

So, what's the advantage/disadvantage?
 
Oh , well all I can say is that the cos(A-B) has less steps.
 

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