Solving trigonometry problem in Wolfram|Alpha

Questions
1

Hello,

I have the following wave..

4 cos(3 theta)-6 sin(3 theta)

…and express it in the following form…

Rsin(3 theta + beta)

I’d be grateful if anyone can explain how to do this in Wolfram|Alpha?

(BTW - I have typed theta and beta as I don’t know how to type the symbols.)

Regards,
Ian

2

This exactly shows how to solve this problem. I was wondering about it for a while myself and a few years ago found that solution:

Trigonometry/Simplifying a sin(x) + b cos(x)

3

Thanks for the link Alexander, but my question was about how to do this in Wolfram|Alpha.

4

7721Polarform.png
7721Polarform.png347×417 17 KB

EDIT :

R=-2 \sqrt{13}

\beta =-\tan ^{-1}\left(\frac{2}{3}\right)

A negative R (radius) looks strange, but the expression is equal to:

2 \sqrt{13} \sin \left(\tan ^{-1}\left(\frac{2}{3}\right)-3 \theta \right)
5

The problem asks for the answer in the form Rsin(3 theta + beta), is this possible and can the steps be shown?

6

What about this?

Manipulate[
 Plot[
  {4 Cos[3 th] - 6 Sin[3 th], f Sin[3 th + b]},
  {th, 0, 3 Pi}, PlotStyle -> {Blue, {Thick, Red}}],
 {f, 0, 10}, {b, 0, 2 Pi}]

Look at f = 7.16 and b = 2.613

7

Or do this

data = Table[{th, 4 Cos[3 th] - 6 Sin[3 th]}, {th, 0, 2 Pi, .1}];
fit = NonlinearModelFit[
   data, f Sin[3 th + b], {{f, 7.16}, {b, 2.16}}, th] // Normal
Table[
  {4 Cos[3 th] - 6 Sin[3 th], fit}, {th, 0, 2 Pi, .05}];
% // TableForm
8

Hi Hans,

I’ve tried both suggestions in Wolfram|Alpha. The first returned a definition of the word ‘manipulate’ and the second described a table and displayed a photo of a coffee table.
I’m probably doing something wrong but what you’ve suggested is quite complicated and requires special knowledge that I neither have nor know where to find.
I thought Wolfram|Alpha would be more straightforward than this given that no documentation is provided. I don’t think it will ever be of use to me, which is disappointing as I’ve spent quite a lot of money on the subscription.

9

Ok, I know next to nothing about Wolfram|Alpha.

But there is an analytical solution as well, obtained with Mathematica, and I think, you could get it with Wolfram|Alpha as well.

Try

aa = -1/Cos[ArcTan[-2/3]]
6 aa Sin[x + ArcTan[-2/3]] // TrigExpand
10

Too much for me Hans.

I think WA can only be used by people who don’t really need it.

I’ve tried my best but can’t see how it can help me

11

Ok. Pity. Perhaps you should consider to switch to Mathematica.

I typed

- 6/Cos[ArcTan[-2/3]] Sin[x + ArcTan[-2/3]]

into Wolfram Alpha and got everything you wanted.

x is 3 theta and ArcTan[-2/3] is your beta

12

Do you think Mathmatica is a better choice, is there documentation available?

I’ve heard good things about Symbolab, do you think this would be a good option?

13

I don’t know anything about Symbolab. I think Mathematica is a good choice. Documentation is availabel. sometimes not satisfactory, but in general good.

Look for example at

I am not quite sure but I think you can download a testversion of Mathematica without costs for one month. But I do not know that.

14

Thanks Hans, I’ll take a look at Mathematica when time permits. I’m going to take a break from Wolfram|Alpha for a while as it’s slowing me down and I need to move on with my course .

Regards,
Ian