5
$\begingroup$

How can I simplify 1/2 (Sqrt[3] Cos[x] - Sin[x]) in order to get Cos[x + Pi/6] on Mathematica? I saw that Wolfram|Alpha lists this simplification on the AlternateForms.

Thanks.

$\endgroup$

2 Answers 2

6
$\begingroup$

Use the TrigFactor[] function, as in

TrigFactor[1/2*(Sqrt[3]*Cos[x] - Sin[x])]

Output

Cos[Pi/6 + x].
$\endgroup$
6
  • $\begingroup$ That's great! But TrigFactor was not able to simplify the following expression: 1/2*((Sqrt[3]*Cos[x]-Sin[x])*m-n). I expect m*Cos[x+Pi/6]-n/2 as the result. $\endgroup$
    – Hermano
    Commented May 22, 2015 at 16:21
  • $\begingroup$ Did you mean to have a parentheses around that $m-n$? $\endgroup$ Commented May 22, 2015 at 16:24
  • $\begingroup$ no. The expression is correct. $\endgroup$
    – Hermano
    Commented May 22, 2015 at 16:28
  • $\begingroup$ Well, that expression you have in comment is $\frac{1}{2} \left(m \left(\sqrt{3} \cos (x)-\sin (x)\right)-n\right)$. See that you can pull out the constant. $-\frac{n}{2}$. The remaining part will factor. $\endgroup$ Commented May 22, 2015 at 16:30
  • $\begingroup$ In other words, that's just $\frac{1}{2} m \left(\sqrt{3} \cos (x)-\sin (x)\right)-\frac{n}{2}$. The trig part will factor, the constant will not. $\endgroup$ Commented May 22, 2015 at 16:32
1
$\begingroup$

You could call a function that fetches all WolframAlpha alternate expression forms:

AlternateExpressionForms[expression_]:=Module[{alternateFormData},
    alternateFormData={};
    alternateFormData=Quiet[Check[TimeConstrained[ReleaseHold[WolframAlpha[ToString[expression,InputForm],{"AlternateForm","Input"}]],60],{}]];
    Flatten[Table[alternateFormData[[i,2]],{i,Length[alternateFormData]}]]
]
SetAttributes[AlternateExpressionForms,Listable];

Example:

AlternateExpressionForms[1/2 (Sqrt[3] Cos[x] - Sin[x])]
(*==> {Cos[\[Pi]/6 + x], 1/2 Sqrt[3] Cos[x] - Sin[x]/2, -(1/4) I E^(-I x) + 1/4 Sqrt[3] E^(-I x) + 1/4 I E^(I x) + 1/4 Sqrt[3] E^(I x)}*)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.