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Solving trigonometry problem in Wolfram|Alpha

Posted 1 month ago
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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

POSTED BY: i d
Answer
13 Replies

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)

Posted 1 month ago

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

POSTED BY: i d
Answer
Posted 1 month ago

enter image description here

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

Posted 1 month ago

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

POSTED BY: i d
Answer

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

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
Posted 1 month ago

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.

POSTED BY: i d
Answer

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
Posted 1 month ago

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

POSTED BY: i d
Answer

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

Posted 1 month ago

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?

POSTED BY: i d
Answer

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

https://reference.wolfram.com/language/ref/ArcTan.html https://reference.wolfram.com/language/ref/Manipulate.html https://reference.wolfram.com/language/workflow/BuildAManipulate.html

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.

Posted 1 month ago

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

POSTED BY: i d
Answer
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