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Consider the following function:

$a Sin(x)+b Cos(x)$

I tried to obtain the maximum value of this function using MaxValue[]:

MaxValue[a Sin[x] + b Cos[x], x, Reals]

I expect Mathematica to return the following answer:


But Mathematica cannot find the answer. Why does this happen?

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note it cant even do this: MaxValue[a Sin[x], x, Reals] – george2079 Aug 26 '13 at 15:53
up vote 4 down vote accepted

This is essentially what b.g did, except by using Max instead of the second derivative constraint we get the result without rewriting the original expression.

f[x_] = a Sin[x] + b Cos[x];
    Max[f[x] /. Solve[ {f'[x] == 0 }, x] ],
       Assumptions -> {Element[a, Reals], Element[b, Reals], 
         Element[C[1], Integers]}]
Sqrt[a^2 + b^2]

I'm a bit puzzled why the conditional expression on C[1] doesn't simplify out on its own when the C[1] is gone from the expression..?

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You can use a more down to earth method :

func = Sqrt[a^2 + b^2] Sin[x + f];
sol = First@Solve[{D[func, {x, 1}] == 0, D[func, {x, 2}] < 0, 
 f \[Element] Reals, a \[Element] Reals, b \[Element] Reals}, x, Reals];

Simplify[func /. sol, Assumptions -> C[1] \[Element] Integers]
(* Sqrt[a^2 + b^2] *)
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Would you please explain why a Sin[x] + b Cos[x] is defined as Sqrt[a^2 + b^2] Sin[x + f]. – M6299 Aug 24 '13 at 10:42
You can do TrigExpand[ Sin[x + f]] and match Cos[f] to a/Sqrt[a^2 + b^2] and Sin[f] to b/Sqrt[a^2 + b^2]. – b.gatessucks Aug 24 '13 at 11:36

Try this:

MaxValue[a Sin[x] + b Cos[x] /. x -> ArcTan[t], t, Reals]


enter image description here


Reduce[{y == a Sin[x] + b Cos[x], a != 0, b != 0}, y, {x}, Reals]
Reduce[%, y, Reals] // LogicalExpand
% // BooleanConvert[#, "CNF"] &

enter image description here

share|improve this answer
not quite.. if a==0 the max is abs[b]. I suppose because ArcTan doesn't admit the full range of angles. – george2079 Aug 24 '13 at 20:11
Thanks for the answer. The ranges of x and ArcTan[t] are different. How come x can be substituted with ArcTan[t]? – M6299 Aug 25 '13 at 5:52
@M6299, Update complete. – chyaong Aug 25 '13 at 6:32

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