Mathematica 9 is now available
Student Support Forum
-----
Student Support Forum: 'a*sin(x)+b*cos(x)' topicStudent Support Forum > General > "a*sin(x)+b*cos(x)"

< Previous CommentHelp | Reply To Comment | Reply To Topic
Author Comment/Response
Bill Simpson
10/11/12 02:15am

Mathematica usually uses LeafCount to decide which of two alternatives is "simpler", smaller count means simpler.

In[1]:= LeafCount[a*Sin[x]+b*Cos[x]]
Out[1]= 9

In[2]:= LeafCount[Sqrt[a*b]*Sin[x+c]]
Out[2]= 12

So Mathematica thinks your preferred form is more complicated. That means something more than Simplify will be needed.

Here is one method you might use

In[3]:= a*Sin[x]+b*Cos[x]/.a_*Sin[x_]+b_*Cos[x_]->Sqrt[a*b]*Sin[x+c]

Out[3]= Sqrt[a*b]*Sin[c + x]

This even works when you do not literally have a,b,x, but the same variable or constant must appear on both your x positions.

In[4]:= 7*Sin[q]+3*Cos[q]/.a_*Sin[x_]+b_*Cos[x_]->Sqrt[a*b]*Sin[x+c]

Out[4]= Sqrt[21]*Sin[c + q]

You might want to enhance this by using the appropriate expression for c as a function of a and b.

URL: ,

Subject (listing for 'a*sin(x)+b*cos(x)')
Author Date Posted
a*sin(x)+b*cos(x) Qi Tang 10/10/12 11:52am
Re: a*sin(x)+b*cos(x) Bill Simpson 10/11/12 02:15am
< Previous CommentHelp | Reply To Comment | Reply To Topic