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