Choice of Rcos(A+-B) or Rsin(A+-B)

[Skelly4444]

I've solved this equation using 2 different methods but one doesn't work and I cannot see what I've done wrong?

I have attached my workings.

 

[topsquark]

If you can't even post the attachment right side up I'm not going to bother to read it.

-Dan

 

[Skelly4444]

The document should be the correct way round now.

 

[Dr.Peterson]

I've solved this equation using 2 different methods but one doesn't work and I cannot see what I've done wrong?

I have attached my workings.

The problem is here:

There are two different (non-coterminal) angles with this tangent; the one you chose has a positive cosine and negative sine.

To get the other angle, add (or subtract) 180 degrees. Then be sure to check that it has the right sine and cosine.

 

[Skelly4444]

I am still confused here to be honest. How can we solve this equation using the alternative form of Rcos(A+B) instead?
I know how to solve the equation using the form Rsin(A-B) but I just want to know why it's not working using the alternative form?

When I try and use this form, I end up with RcosB = -12 and RsinB = -5 giving A as 22.6 degrees.

Now solving gives: 13cos(A + 22.6) = 6
Producing angles of 39.9 and 274.9 degrees.

These are obviously wrong answers but why? If I wanted to use the form Rcos(A+B) what signs would need to be on sinB and cosB to match it up ?

 

[Deleted member 4993]

I am still confused here to be honest. How can we solve this equation using the alternative form of Rcos(A+B) instead?
I know how to solve the equation using the form Rsin(A-B) but I just want to know why it's not working using the alternative form?

When I try and use this form, I end up with RcosB = -12 and RsinB = -5 giving A as 22.6 degrees.

Now solving gives: 13cos(A + 22.6) = 6
Producing angles of 39.9 and 274.9 degrees.

These are obviously wrong answers but why? If I wanted to use the form Rcos(A+B) what signs would need to be on sinB and cosB to match it up ?

Your process is "good" - but you have made mistakes about "sin or cos" and "+ or -".

cos(Θ + α) = cos(Θ) * cos(α) - sin(Θ) * sin(α)

sin(Θ + α) = sin(Θ) * cos(α) + cos(Θ) * sin(α)

cos(Θ - α) = cos(Θ) * cos(α) + sin(Θ) * sin(α)

sin(Θ - α) = sin(Θ) * cos(α) - cos(Θ) * sin(α)

Start with a "fresh sheet" of paper and do not look back at your previous work. Pay super extra attention to "signs" and "sin or cos". Please show us what you get.

 

[BigBeachBanana]

Thought this would be an interesting solution to share. Here' another methodology suggested by an engine: transform the equation into a rational function via Weierstrass substitution.

Spoiler

The Weierstrass substitution: Let [imath]y=\tan\left(\frac{x}{2}\right)[/imath]. Then, [imath]\sin x=\frac{2y}{y^2+1}[/imath] and [imath]\cos x=\frac{-y^2+1}{y^2+1}[/imath]
It follows
[math]5\left(\frac{2y}{y^2+1}\right)-12\left(\frac{-y^2+1}{y^2+1}\right)-6=0\\ \frac{6y^2+10y-18}{y^2+1}=0\\ 3y^2+5y-9=0\\ y^2+\frac{5}{3} y -3=0\\ \text{Complete the square}\\ \left(y+\frac{5}{6}\right)^2=\frac{133}{36}\\ \implies y=\frac{\sqrt{133}-5}{6}\quad \land\quad y=\frac{-\sqrt{133}-5}{6}\\ \text{Substitute back}\quad y=\tan\left(\frac{x}{2}\right)\quad\text{ and solve for }x.[/math]

 

[Skelly4444]

Fresh sheet of paper as suggested but same answer unfortunately.
Obviously I'm misunderstanding the matching up process somewhere but not sure how?

Please see attached.

 

[Dr.Peterson]

Fresh sheet of paper as suggested but same answer unfortunately.
Obviously I'm misunderstanding the matching up process somewhere but not sure how?

Please see attached.

Do you not see that when you write

​

your angle -22.6 degrees does not have the sine and cosine you want?

cos(-22.6) = +0.923, rather than what you need, -12/13 = -0.923​

sin(-22.6) = -0.384, rather than 5/13 = +0.384​

Don't use the inverse tangent of -5/12! Use the OTHER angle with the same tangent, namely -22.6 + 180 = 157.4 degrees. That angle has the correct sine and cosine.

 

[Skelly4444]

Thank you for this. So just to clarify that we use the other angle of 157.4 because it is negative for cos and positive for sin and therefore matches the original equation?

In other words, 157.4 sits in the 2nd quadrant as opposed to the angle of -22.6 that I was using, which sits in the 4th quadrant?

 

[Dr.Peterson]

Right.

You are not looking merely for an angle with the given tangent, but for the angle in the proper quadrant with that tangent, in order to match both the sine and the cosine.