Question

Hi.
Can someone help me with problem 6a on the first assignment. I understand how they are finding A, but I don't understand how they are finding c. Any help would be appreciated.

Answer 1

The problem is:
Express in the form A sin (x+ c)
sinx + √3cosx

Answer 2

multiply and divide by 2..so u get 2(six/2 +sqrt3 cosx/2)... now sin 30 = 1/2 and cos 30 =sqrt 3 /2 so it becomes 2sin(x+30)

Answer 3

I'm not sure your explanation is correct. If I double check the result using x = pi/12 (and rad mode) then I get #1 = 1.932 and #2 = 1.414. But most likely I'm just missing something

Answer 4

I looked at the solution, and it is clear how A is found, but unclear how c = pi/3 is found.

Answer 5

Since r=√a squared+b squared, √1 squared+√3 squared gives r=2.
Since tan=sinx/cosx, arctan=cosx/sinx, which gives π/3.
Plugging these values into the given formula, we get the answer: 2sin(x+π/3).
I only found this after looking it up. I appreciate everyone who helped!

Answer 6

isn't cosx/sinx = cot, not arctan ?
I see tan = sin / cos = 1 / sqrt(3), but I still dont see how it can be sqrt(3)/1

Answer 7

I think I finally figured it out.
what is y if x = 0?, y = sqrt(3), since sin(x) = 0 and sqrt(3)cos(x) = sqrt(3).
Since A = 2, the value for sin(x + c) = sqrt(3) / 2 = pi/3. So c = pi/3.

Answer 8

It's inverse tangent. Sorry. Another way is jkristia's way. I think both get you the same answer.

Answer 9

ireneb i'm having some trouble understanding this and was wondering where you looked it up?

Answer 10

I honestly cannot remember where I looked it up. I remember from my high school PreCalc that inverse tangent helps you find angle measures, and is different than just regular tangent to
find the side of a triangle. It's tangent to the -1, not tangent as in o/a or a/o.

Answer 11

It is correct that \[\arctan (\sqrt{3}) = \frac{\pi}{3}\]
Since \[\sin(\pi/3) = \frac{\sqrt{3}}{2}, \cos(\pi/3) = \frac{1}{2}\]
so\[\tan(\pi/3) = \frac{\sin(\pi/3)}{\cos(\pi/3)}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\]

So by dividing sin(x) + sqrt(3)cos(x) by cos(x), you get\[\tan(x) + \sqrt{3}\] but I still cannot make the last connection. Why is tan c = sqrt(3) ?

Answer 12

tan c should be π/3, which is the given solution. If you have
\[\tan^{-1} =(\sqrt{3/1})\]

you should get the answer of c=π/3, or 60 degrees.

Answer 13

I know it is pi/3, and I know how to solve it by setting x = 0, and I know that arctan(sqrt(3)) is pi/3, - but I still dont see why it is tan c.

Answer 14

You are expressing in the form A sin(x +c). We found A to be 2, and x is not what we are needing to solve for. C is just the variable that we solved for. It just means that we are going to put in π/3 for c in the required form.

Answer 15

thanks ireneb ! i will try it again and c if i can't figure it out.