# 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 Answer
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Posted 1 month ago
 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) Answer
Posted 1 month ago
 Thanks for the link Alexander, but my question was about how to do this in Wolfram|Alpha. Answer
Posted 1 month ago 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: $$2 \sqrt{13} \sin \left(\tan ^{-1}\left(\frac{2}{3}\right)-3 \theta \right)$$ Answer
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? Answer
Posted 1 month ago
 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 Answer
Posted 1 month ago
 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 Answer
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. Answer
Posted 1 month ago
 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 Answer
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 Answer
Posted 1 month ago
 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 Answer
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? Answer
Posted 1 month ago
 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 atI 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. Answer
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 Answer
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