# Thread: Application Problem (Trig Identities)

1. ## Application Problem (Trig Identities)

When the two voltages V1 = 30sin(120Πt) and V2 = 40cos(120Πt)are applied to the same circuit, the resulting voltage V will be equal to their sum.

a. Graph the sum in the window [0 .05] by [-60,60] and use the graph to estimate values for a and Φ so that V = a sin(120Πt + Φ)

b. Use identities to verify that your expression for V is valid.

Okay........I got the graph on my calculator and can easily find a, it's just the amplitude of the sinusoid, which is 50.
So a=50, but how can I estimate a value for Φ from the graph? The answer in the back of the book is Φ = -5.353, but I can't see how to get there.

And for part b, how can I use identities to verify that 30sin(120Πt)+ 40cos(120Πt) = 50 sin(120Πt + (-5.353))????

2. ## Re: Application Problem (Trig Identities)

$\displaystyle 30\sin(120\pi t)+40\cos(120\pi t) = 50\left(\dfrac{3}{5}\sin(120\pi t)+\dfrac{4}{5}\cos(120\pi t)\right)$

The sum of angles formula for sine is $\displaystyle \sin(\alpha+\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$, so you want $\displaystyle \cos \phi = \dfrac{3}{5}$ and $\displaystyle \sin \phi = \dfrac{4}{5}$. Hence, $\displaystyle \phi = \arcsin\left(\dfrac{3}{5}\right) = \arccos \left(\dfrac{4}{5}\right)$.

3. ## Re: Application Problem (Trig Identities)

I think you Φ = arcsin (4/5) = arcos (3/5). But still I get Φ= .927295, and the answer should be -5.353.

4. ## Re: Application Problem (Trig Identities)

Are you using degrees or radians? Angles are periodic (either 360 degrees or $\displaystyle 2\pi$ radians). Hence, there are an infinite number of answers. Either subtract 360 degrees or $\displaystyle 2\pi$ radians to see if your answer matches the book's.

5. ## Re: Application Problem (Trig Identities)

Yes that was it. Using radians, arcsin (3/5) - 2Π = -5.353. But how would you know to do this within the context of the problem?

6. ## Re: Application Problem (Trig Identities)

Originally Posted by cdbowman42
Yes that was it. Using radians, arcsin (3/5) - 2Π = -5.353. But how would you know to do this within the context of the problem?
To find phi, you would solve for it. You have $\displaystyle 30\sin(120\pi t)+40\cos(120\pi t) = 50\sin(120\pi t+\phi)$

So, solving for $\displaystyle \phi$ gives

$\displaystyle \phi = \arcsin\left(\dfrac{3}{5}\sin(120\pi t) + \dfrac{4}{5}\cos(120\pi t)\right) - 120\pi t$

That should give a step function (it will have an infinite number of answers). So, pick any answer that satisfies the equations. I don't understand what you mean "how do you know to do that". Sine and cosine are always periodic. The book's answer is correct, but there are infinite correct answers. Your answer (before subtracting 2 pi) is correct. There is no need to subtract anything.

Edit: in the given range of $\displaystyle t \in [0,0.05]$, the graph gives the book's answer as the solution.

7. ## Re: Application Problem (Trig Identities)

Right, looking a the graph must somehow make -5.353 stand out, but I cannot see how to interpret this value from the graph. I think it must be a phase shift of the graph of
50 sin(120Πt), but I can't see how to interpret this. Make sense? I'm just trying to understand how looking at the graph will give that value rather than using identities.

8. ## Re: Application Problem (Trig Identities)

Graph $\displaystyle \phi$ as a function of t on the interval you are given.

9. ## Re: Application Problem (Trig Identities)

Here is a table of possible values for $\displaystyle \phi$ in the given interval: Table[Arcsin[(3/5)*Sin[120*Pi*t]+(4/5)*Cos[120*Pi*t]]-120*Pi*t,{t,0,0.05,0.001}] - Wolfram|Alpha
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