Expressing in R sin(3θ + β)

[Chris_inkubate]

Hi there,
I am working on a submission which asks the following question:

The third harmonic of a sound wave is given by;
4cos(3?) − 6sin(3?)

Express this sound wave in the form;
? sin(3?+?)

I managed to work out R without any huge problems. However working out ? has been an issue. I managed to find a value that worked but only by getting the result of (3?-?) and using some intuition of radians that pi minus this value would be the difference and work. It appears to have worked on this occasion but I feel I have fudged it a little to get this and expect there is a better way. Any assistance would be much appreciated to improve my workings. Many thanks.

 

[Dr.Peterson]

I don't understand. Your work started, appropriately, with an expression for ? sin(3?+?), so why do you think the result was for ? sin(3?-?)?

Ah, I see; without explanation you ignored the sign on -6 at line 5. There's no need to have done that. If you keep the sign, then you'll end up with [MATH]\beta = \tan^{-1}(-4/6) = -0.588[/MATH].

Adding [MATH]\pi[/MATH] gives another angle with the same tangent; all you need to do is to decide which is appropriate. To do that, go back to the signs of [MATH]\sin(\beta)[/MATH] and [MATH]\cos(\beta)[/MATH] and see what quadrant you need. No fudge required.

 

[Chris_inkubate]

Many thanks for your explanation. I wasn't sure if the usage of pi to manipulate the quadrant as you say was considered an appropriate action. It was something I worked out in my head when I could see the phase was out once graphed with 0.588. It just made sense as the angle is subtracted.

Good point re. the sign on line 5 - I'll keep an eye for this and commit this all to my notes.

 

[LCKurtz]

I think you are mostly right but not quite. I see Dr. Peterson replied while I was typing this so I will throw in my 2 cents worth anyway. You started with [MATH]\sin(3\theta + \beta) = \sin(3\theta)\cos\beta + \cos(3\theta)\sin\beta[/MATH]. The expression you are given is
[MATH]4\cos(3\theta) - 6\sin(3\theta)[/MATH]. To make this look like an addition formula you need to multiply and divide by [MATH]\sqrt{4^2+6^2}=\sqrt{52}[/MATH] giving
[MATH]\sqrt{52}\left[ \frac 4 {\sqrt{52}} \cos(3\theta) - \frac 6 {\sqrt{52}}\sin(3\theta) \right ][/MATH]. To match this up with the given you need [MATH]\sin\beta = \frac 4 {\sqrt{42}}[/MATH] and [MATH]\cos\beta = -\frac 6 {\sqrt{52}}[/MATH]. That makes it clear that [MATH]\beta[/MATH] is in the second quadrant and [MATH]\beta = \arccos(- \frac 6 {\sqrt{52}})[/MATH].

 

[Chris_inkubate]

Thank you for this explanation. I did wonder if there was something needed to deal with one side being addition and the other subtraction. However decided to work through anyway and see what happens. I will have a play using the method you show and get a feel for it.

 

[Steven G]

Rcos(B) = -6 NOT 6 and sqrt(52) is NOT 7.2111

 

[Chris_inkubate]

Rcos(B) = -6 NOT 6 and sqrt(52) is NOT 7.2111

Many thanks for the message. Expressing the sqrt(52) as would be more accurate and this works well when the waveform is graphed.