# Homework Help: 3cosx+4sinx=5, confused about R

1. Feb 15, 2012

### solve

1. The problem statement, all variables and given/known data

3cos(x)+4sin(x)=5

Solve for X.

2. Relevant equations

acos(x)+bsin(x)=c

f(x)=Rsin(x+θ)

3. The attempt at a solution

Rsin(x+θ)=5

Rsin(x)cos(θ)+Rcos(x)sin(θ)=5

Rcos(θ)=4

Rsin(θ)= 3

R^2*cos^2(θ)+R^2*sin^2(θ)=R^2=3^2+4^2=25

R=+/-2

5sin(x+θ)=5

sin(x+θ)=1

x+θ=pi/2+/-2npi

x=pi/2-θ+/-2npi

Rsin(θ)/Rcos(θ)=tan(θ)=3/4

so X=pi/2-0.64+/-2npi

What confuses me is that I get a different result if I use negative R:

x=-pi/2-θ+/-2npi

Are -pi/2 and pi/2 equal?

Thank You.

2. Feb 15, 2012

### ehild

When you wrote Rsin(x+θ)=5 it was assumed that R>0. A negative R would correspond to a different θ, θ'=θ±pi.

ehild

3. Feb 15, 2012

### solve

Thank You for the response, Ehild. I am just given f(x)=Rsin(x+θ) to solve certain type of trig equation without any explanation as to where it came from. Honestly, when I wrote Rsin(x+θ)=5 I didn't assume that R>0. Just going by formula. So why is is it assumed R>0? And What's that R? Better yet, where did f(x)=Rsin(x+θ) pop out from? Sorry for too many questions.

I actually asked this question on another forum and someone said R<0 and R>0 work out to the same answer. I'll go get it.

Last edited: Feb 15, 2012
4. Feb 15, 2012

### solve

""pi/2 + M * pi is the same as -pi/2 + N * pi.

The first one is a set of numbers of the form pi/2 + M * pi, where M runs ofver all integers (i.e. ...-2, -1, 0, 1, 2, ...)...so, these numbers are:

{...., pi/2 - 3pi, pi/2 - 2pi, pi/2 - pi, pi/2, pi/2 + pi, pi/2 + 2pi, pi/2 + 3pi,...} = {..., -5/2pi, -3/2pi, -pi/2, pi/2, 3/2pi, 5/2pi, 7/2pi,...}

The second set is comprised of the numbers of the form -pi/2 + N * pi, where N runs over all integers....so, these numbers are:

{...., -pi/2 - 3pi, -pi/2 - 2pi, -pi/2 - pi, -pi/2, -pi/2 + pi, -pi/2 + 2pi, -pi/2 + 3pi,...} = {..., -7/2pi, -5/2pi, -3/2pi, -pi/2, pi/2, 3/2pi, 5/2pi,...}

So, you can see that two sets are the same.

Another way (easier) is to rewrite both sets as

pi/2 + M * pi =pi (1/2 + M)

-pi/2 + N * pi = pi (-1/2 + N)

and, again, notice that 1/2+M defines the same set of numbers as -1/2 + N. ""

It sure looks intimidating, so I didn't look at it more closely...Will look now.

5. Feb 15, 2012

### ehild

Allowing both R=5 and R=-5, you would get the same set of solutions.

Your mistake is that the tangent function is periodic with pi, instead of 2pi.

So X=pi/2-(0.64±Npi): X=pi/2-0.64, -pi/2-0.64, 3pi/2-0.64, -3pi/2-0.64,...

If R=-5, X=-pi/2-(0.64±Npi): X=-pi/2-0.64, pi/2-0.64, 3pi/2-0.64, -3pi/2-0.64....

ehild

6. Feb 15, 2012

### solve

Ok, so, is it true that pi/2-(0.64±Npi)=-pi/2-(0.64±Npi) ?

7. Feb 15, 2012

### ehild

No with the same N. But it is true that for each N there is an M so that pi/2-(0.64+Npi)=-pi/2-(0.64+Mpi) (M and N are both positive and negative integers).

For example, if N=-2, x= pi/2-0.64+2pi=5/2pi-0.64. 5/2pi-0.64=-pi/2-0.64-Mpi=>M=-3

ehild

8. Feb 15, 2012

### solve

Thanks. Do I need some kind of background to understand this? Something about sets? My textbook doesn't explain R<0 situation, nor does it require me to go over that.

9. Feb 15, 2012

### solve

3cos(x)+4sin(x)=5

3cos(0.930796327)+4sin(0.930796327)= 4.99996936...when R>0

3cos(-2.21079633)+4sin(-2.21079633)= -4.99996936...when R<0

Does the solution for the equation when R<0 still hold true because "pi/2 + M * pi is the same as -pi/2 + N * pi" ?

Also, just because "pi/2 + M * pi is the same as -pi/2 + N * pi" it doesn't mean pi/2=-pi/2 ?

So does it mean that "pi/2 + M * pi is the same as -pi/2 + N * pi" is something redundant that has nothing to do with solutions of this equation?

Thanks.

10. Feb 15, 2012

### ehild

I am sorry, I mislead you a bit. There is only a single θ between 0 an 2pi.

When you have equations of form a cosx + b sinx =c the easiest way to solve them is to write the left hand side in the form Rsin(x+θ). You can do it, so why not.
The idea comes from the sum law of the sine function: sin(x+θ)=sinx cosθ +cosx sin θ.

From Rsinθ=3 and Rcosθ=4 it follows that R=5 and sinθ=0.6, cosθ=0.8. There is a single angle between 0 and 2pi for which both are true: θ=0.6435.
At the same time, Rsin(x+θ)=5 =>sin(x+θ)=1. x+θ=pi/2±2piN, x=-0.6435+pi/2±2piN.

If you take R=-5, sinθ=-0.6 and cosθ=-0.8, θ=0.6435-pi. -5sin(x+θ)=5, so sin(x+θ)=-1,
(x+θ)=-pi/2±2piN. x=pi-0.6435-pi/2±2piN=pi/2-0.6435±2piN. The same as for positive R.
You do not need to consider negative R-s.

You can apply other methods, for example expressing sinx as ±sqrt(1-cos2x) and solving the equation for cos(x). There will be invalid solutions, you have to replace back the results into the original equation to exclude them.

ehild

Last edited: Feb 15, 2012
11. Feb 15, 2012

### solve

Aside from the fact that I have no idea how the left side of acosx + bsinx =c gets to be written as Rsin(x+θ), with R>0 (My textbook doesn't show that- just gives the formula Rsin(x+θ) without the provision R>0 to solve the equation of the form acosx + bsinx =c), I am curious as to why R<0 can't be used.

12. Feb 15, 2012

### ehild

I edited my previous post meanwhile, read it. You can use negative R-s, and you get the same solution. No reason to use both positive and negative R-s.

ehild

13. Feb 16, 2012

### solve

Thank You. I have couple questions:

1. θ=0.6435-pi. Where did -pi come from?
2. So, I see how Rsin(x+θ) came to be, except, how did R get there? What is it? What does it do in practice? What does this R really mean? :shy:

Thank You.

14. Feb 16, 2012

### ehild

1. Sorry, it is a typo. When sinθ=-0.6 and cosθ=-0.8, θ=0.6435+pi. It comes from the unit circle.

2.The linear combination of a sine and a cosine of the same angle always equals to a single sine or a cosine of an angle multiplied with a number. In Physics, it appears when two vibrations add up, or two waves interfere. The result is a new vibration or wave, with a new amplitude (it is R) and phase constant (it is theta).
In Maths, you have two functions f1(x)=acosx, f2(x)=bsinx . If you add them, you get the function g=f1+f2, which has the same shape - a sine or cosine function, but shifted and having a certain amplitude, different from a and b. See attachment.

ehild

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15. Feb 16, 2012

### solve

sinθ=0.6, θ=0.6435,

sinθ=-0.6, θ=-0.6435

How/why does arcsin(-0.6) = -0.6435 become θ=0.6435+pi and arcsin(0.6) is just θ=0.6435, without any pi's attached and/or signs changed?

Thank You.

16. Feb 16, 2012

### solve

Awesome explanation. Thanks.

Is there also an algebraic way to see how R jumps in front of sin(x+θ)=f(x) ?

Thank You.

17. Feb 16, 2012

### ehild

R does not jumps in front of sin(x+θ): you write it there. As the sine function can not exceed 1, you must put a factor R which multiplies sin(x+θ) so as it equals to f(x)=a cosx +b sinx, with arbitrary a and b.

ehild

18. Feb 16, 2012

### ehild

The domain of the arcsin function is [-pi/2, pi/2], that of the cosine function is [0, pi].
The sin and cosine do not determine the angle completely. sin(θ)=sin(pi-θ), so the equation sinθ=-0.6 has two groups of solutions, θ=arcsin(-0.6)+2Npi =-0.6435+2Npi and pi-arcsin(-0.6)+2Npi=pi+0.6435+2Npi. Between 0 and 2pi you have 2pi-0.6435=5.6397 and 3.7851.

cos(θ)=cos(-θ). Therefore the equation cos(θ)=-0.8 has two solutions: θ=arccos(-0.8)+2Npi and θ=-arccos(-0.8)+2Npi. arccos(-0.8)=2.4981. Between 0 and 2pi the solutions are 2.4981 and 2pi-2.4981=3.7851.

So θ=3.7851 is the angle which cosine is -0.8 and sine is -0.6.

In case sinθ=0.6 and cosθ=0.8, arcsin(0.6)=0.6435, so θ=0.6435 + 2Kpi or θ=pi-0.6435 + 2Kpi=2.4981+2Kpi.
arccos(0.8)=0.6435, so θ=0.6435 + 2Kpi or θ=-0.6435 + 2Kpi, that is 5.6397 between 0 and 2pi. The common solution is θ=0.6435 + 2Kpi.

Draw the unit circle and see the relation between angles which have the same cosine or sine.

ehild

19. Feb 16, 2012

### solve

Thank You, Ehild.