I see no trig function for this

[StrawHat]

Homework Statement

244.10746 = 845.9064sin(θ) - 274.87cos(θ)

Homework Equations

Is there? I have no clue.

The Attempt at a Solution

I plugged it into WolframAlpha and I got an incomprehensible answer.

 

[Villyer]

Have you attempted to use the property 1 = sin^{2}(\theta) + cos^{2}(\theta) to get an equation in terms of only sin or cos?

 

[StrawHat]

Have you attempted to use the property 1 = sin^{2}(\theta) + cos^{2}(\theta) to get an equation in terms of only sin or cos?

But I can't use that trig function for this equation.

 

[Chestermiller]

Homework Statement

244.10746 = 845.9064sin(θ) - 274.87cos(θ)

Homework Equations

Is there? I have no clue.

The Attempt at a Solution

I plugged it into WolframAlpha and I got an incomprehensible answer.

Try using sin(A-B) = sinA cosB - cos A sinB. Start out by dividing both sides of the equation by sqrt (845.9064² + 274.87²)

Chet

 

[Muphrid]

Of course you can. Let x = \cos \theta. Then \sin \theta = \sqrt{1-x^2}. All it takes is a little more algebra to massage this into a polynomial equation in x that you can then convert back into \theta.

 

[StrawHat]

Of course you can. Let x = \cos \theta. Then \sin \theta = \sqrt{1-x^2}. All it takes is a little more algebra to massage this into a polynomial equation in x that you can then convert back into \theta.

I got x = 809.919 after plugging everything in. But I can't use that x-value for the sin(θ) = √(1-x^2) because that would become a negative number inside of that root.

 

[Muphrid]

I suspect you did some algebra wrongly; I get a value for x that is between -1 and 1 as required. Why don't you type out a couple steps of algebra?

 

[StrawHat]

I suspect you did some algebra wrongly; I get a value for x that is between -1 and 1 as required. Why don't you type out a couple steps of algebra?

I must be squaring it wrong. This time I got x = 0.91059, but I know it's supposed to be 0.82917.

244.10746 = 845.9064sin(θ) - 274.87cos(θ)

244.10746 = 845.9064\sqrt{1-x^2} - 274.87x

*multiply everything by the power of 2*

59588.4520 = 715557.6376 - 715557.6376x² - 75553.5169x²

I'm guessing I squared the above equation incorrectly...

 

[Muphrid]

Yes, you did. On the right hand side, you've squared each term, but that's not how squaring works.

(a+b)^2 = a^2 + 2ab + b^2 \neq a^2 + b^2

 

[StrawHat]

But then if I square it, I would still be left with \sqrt{1-x^2}, wouldn't I?

 

[Muphrid]

Yes. How might you isolate that square root so you don't get any cross terms when you square both sides?

 

[StrawHat]

I put the square root by itself and got this:

\sqrt{1-x^2} = \frac{244.10746 + 274.87x}{845.9064}

I did the calculations wrong again, and got this:

0 = 70.4433 + 158.6412x + 90.3166x²

It's probably because I didn't square the 845.9064 as well.

 

[Muphrid]

Well, once you're able to square these numbers correctly with a calculator, you should be getting close to something that you an arrange and solve by the quadratic formula.

 

[Chestermiller]

I'm surprised that you guys didn't even try to solve this problem using the much simpler method that I alluded to in Response #4 (which does not even require solving a quadratic equation). Here is an additional hint:

A sinθ - B cosθ = sqrt(A² + B²) (sinθ cos\phi -cos\theta sin\phi) = sqrt(A² + B²) sin(\theta-\phi)

where \phi = arctan (B/A)

Chet

 

[vela]

Here's where Chestermiller's suggestion comes from. The idea is to find R and φ such that
\begin{align*}
A &= R \cos \varphi \\
B &= R \sin \varphi
\end{align*} so that
\begin{align*}
A \sin\theta - B \cos \theta &= (R \cos\varphi)\sin\theta - (R\sin\varphi)\cos\theta \\
&= R(\cos\varphi\sin\theta - \sin\varphi\cos\theta) \\
&= R \sin(\theta-\varphi).
\end{align*} If you divide the second equation by the first, you get
$$\frac{B}{A} = \frac{R\sin\varphi}{R\cos\varphi} = \tan\varphi,$$ and using the Pythagorean identity, you get
$$1 = \cos^2 \varphi + \sin^2\varphi = \left(\frac{A}{R}\right)^2 + \left(\frac{B}{R}\right)^2,$$ or equivalently, $R^2 = A^2+B^2$.

This trick is a good one to know. As he noted, it's a lot less tedious to solve the problem using this method than dealing with the quadratic equation, especially when you have non-integer numbers like the ones in this problem.