Adding trigonometric waveforms with different phase angles

freckles101 said:
Express f(t) in the form: Acos(ωt-ϕ)
f(t)=-12 sin⁡(5t+2)-9 cos⁡(5t-8)
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Expand using the Compound Angle formulae and gather like terms,
then convert to Wave Function.
Continue....
NB: Just state your work is to 2 (or 3) decimal places to permit the use of \(\displaystyle =\text{instead of} \approx \)
 
The Highlander said:
Expand using the Compound Angle formulae and gather like terms,
then convert to Wave Function.
Continue....
Click to expand...
That's great advice

The Highlander said:
NB:[/B] Just state your work is to 2 (or 3) decimal places to permit the use of \(\displaystyle =\text{instead of} \approx \)
Click to expand...
I'd keep the answer exact by declaring two constants [imath]c_1=-(?\sin(?) + ?\cos(?))\text{ and } c_2=[/imath] something similar
 
Cubist said:
That's great advice


I'd keep the answer exact by declaring two constants [imath]c_1=-(?\sin(?) + ?\cos(?))\text{ and } c_2=[/imath] something similar
Click to expand...
Whilst mathematically 'sound', I find introducing more 'letters' can cause confusion (they can start to look like more variables/unknowns to be dealt with) and pupils find it easier to both understand & manipulate when they are dealing with actual numbers.
Working to 2/3 d.p. throughout produces a final answer for the phase angle that is (usually) in keeping with what you get either way; since it needs to be declared to that kind of level of accuracy anyway in the end. ?
 
(I meant to say both Amplitude and Phase Angle in that last post, ofc.)
Though I suspect Freckles has given up and gone elsewhere (perhaps to seek a 'ready-made' answer)! ?
 
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thank you so much!! please excuse my messy writing! but does this look okay? :)
 

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Hi again,

I looked at your working and you started off correctly, by Expanding using the Compound Angle formulae (as I suggested) but then you went completely awry I’m afraid.

Rather than pick through everything you have written, it is easier for me to show you the method I used. I haven’t bothered to check your actual arithmetic (or sign handling) because you diverged from the correct approach right after expanding the initial expression, so here’s one way to get the right answer:-

I would begin by stating that all angles are in radians and I am working to 2 d.p.

Starting with:
\(\displaystyle -12sin (5t+2)-9cos(5t-8)\)​
You (correctly) expanded this to:

\(\displaystyle -12(sin5tcos2+cos5tsin2)-9(cos5tcos8+sin5tsin8)\)
​
However, after that, you appear to have attempted to “simplify” your new expression into two separate ones that you wished to convert simultaneously (into the form: Acos(ωt-ϕ) and subsequently re-combine these!
When I said “gather like terms” what I meant was to, first, replace all calculable terms with their numerical values, eg: sin2=0.91 (to 2d.p.) and then, further expand & simplify your expression into a form (eg: Xsinθ+Ycosθ) that may be converted to Acos(ωt-ϕ).

Cubist's suggestion of using constants (like c1, c2) is intended to preserve the “accuracy” throughout but, since you need to find numeric values for A & ϕ at the end anyway, I find it easier just to insert the numbers right away. (You could always work to 3 (or 4) decimal places if you were fanatical about it.! ?)

Thus we can now replace cos2, sin2, cos8 & sin8 with numbers in our expression to get:

\(\displaystyle -12(-0.42sin5t+0.91cos5t)-9(-0.15cos5t+0.99sin5t)\)​

And we can now expand the brackets by multiplying by the -12 & the -9 to get:

\(\displaystyle 4.99sin5t-10.91cos5t+1.31cos5t-8.90sin5t\)​

Now we are ready to gather like terms (ie: cos5t’s & sin5t’s) to simplify our expression to:

\(\displaystyle -9.6cos5t-3.91sin5t\)​

ie: a form you know how to convert to Acos(ωt-ϕ), (I believe?)

So the complete working should look like this:

All angles are in radians and I am working to 2 d.p.

\(\displaystyle -12sin (5t+2)-9cos(5t-8)\)
\(\displaystyle =-12(sin5tcos2+cos5tsin2)-9(cos5tcos8+sin5tsin8)\)
\(\displaystyle =-12(-0.42sin5t+0.91cos5t)-9(-0.15cos5t+0.99sin5t)\)
\(\displaystyle =4.99sin5t-10.91cos5t+1.31cos5t-8.90sin5t\)
\(\displaystyle =-9.60cos5t-3.91sin5t\)

I would strongly recommend that you should do the conversion to Acos(ωt-ϕ), before you look at my working, just to make sure you can get the same answer. (Don't cheat, now. ??)
Converting to Acos(ωt-ϕ)

\(\displaystyle A=\sqrt{9.60^2+3.91^2}=10.37\) (Ignoring the minus signs because you are squaring.)

and
​
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\(\displaystyle θ=tan^{-1}\left(\frac{-3.91}{-9.60}\right)=0.39\)

\(\displaystyle \Rightarrow \text{ϕ}=\pi+0.39=3.53\)

\(\displaystyle \Rightarrow -12sin (5t+2)-9cos(5t-8)=10.37cos(5t-3.53)\)    Q.E.D.
Hope that helps. ?
 
PS: If you don't get the same answers for A & ϕ, share your work so we can explain the correct way to do it. ?
The "working" you provided also gave me some concerns about how you were calculating the Phase Angle (which is why I said you should do it yourself before looking at my calculation).
 
The Highlander said:
Hi freckles,
I'm afraid that's not what I get. ?
Let me check through your working and I'll try to explain where you've gone wrong.
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Haha yeah, I rushed through that one, and have noticed a few algebraic mistakes... I've since typed it out with the corrections i saw that were required... does this look closer to the answer you got? Thank you so much for your support and assistance!!
1652737128190.png
 
freckles101 said:
Haha yeah, I rushed through that one, and have noticed a few algebraic mistakes... I've since typed it out with the corrections i saw that were required... does this look closer to the answer you got? Thank you so much for your support and assistance!!
View attachment 32753
Click to expand...
No, I’m sorry but that’s still not right.

Now that you have typed it out nice and clearly, with explanations, I can see what you were trying to do and your method (if applied correctly) would, indeed, work.

However, your “terminology” is all wrong and could be construed as showing a lack of understanding of what you are about; have you copied this ‘method’ from somewhere?

For example, you cannot say that your ‘Factorised' form is “analogous to:
Acos(ωt-ϕ) = Acos(ωt)cos(ϕ) + Asin(ωt)sin(ϕ)”​
because that is simply not true! ?

You could (should) have said that your factorized expression was analogous to:

Acos(ωt-ϕ) = Xcos(α)cos(ωt) + Ysin(β)sin(ωt) ?​

Ignoring your persistent (terminological) mistakes that (all) arise from this confusion, your subsequent calculations do arrive at the correct coefficients for cos(ωt) & sin(ωt) at the finish (ie: -9.6021 & -3.9105) which, in turn, allowed you calculate the correct Amplitude of the waveform.

Using those numbers you could then have written:

\(\displaystyle V= -9.6021cos5t-3.9105sin5t\) ?​
(by the way, you should also have posted the complete question in its entirety in your OP.)
(Doing so can save a lot of time (& confusion) when we are considering your problem(s)!)

which is analogous to: Xcos(α)cos(ωt) + Ysin(β)sin(ωt)        [or just Xcos(ωt) + Ysin(ωt)] ?
which is the form that converts (by the 'usual' method) to Acos(ωt-ϕ) ?

Basically, what you are doing is what Cubist suggested and delaying the evaluation of these coefficients right to the end but you are naming everything wrongly (even though you get the right results, this would be very bad “Maths”!)

Unfortunately, however, just as I suspected, (see my comments above about your calculation of the Phase Angle; at the end of Post #10 and, again, in Post #11), you don’t appear to have grasped how to do that correctly.

Did you read my posts before you made your latest post (#12)??? ?

If you are likely to face this kind of problem again, I would suggest that you have a good read through my post (#10) and compare what I have laid out there with the method you have pursued. IMNSHO, my method is a lot simpler to both follow and execute and I have provided you with the correct way to get the Phase Angle (sketching that diagram is the surefire way to get it right!)

Please read, carefully, what I posted previously (#10) and let us know that you understand what is there, how it compares with your “method” and also confirm that you now know how to get the Phase Angle right.

Cheers, B. ?
 
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Correction (before someone else points it out ?).

You could (should?) have said that your factorized expression was analogous to:

Acos(ωt-ϕ) = X[fsin(α) + gcos(β)]cos(ωt) + Y[hcos(γ) + isin(δ)]sin(ωt)​

Sheesh! Even more complicated! ?
(I'm not 'fixing' the bit further down! ?)
 
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