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Trigonometric Transformations

To rewrite a product of sinuses and cosinuses as a finite Fourier sum, we can use the function trigreduce:

trigreduce(sin(x)^3*cos(x));

We obtain:

   2 sin(2 x) - sin(4 x)
   ---------------------
            8

To obtain a representation that does not contain trigonometric functions of multiple angles, we write:

trigexpand(%);
               3           3
   4 cos(x) sin (x) - 4 cos (x) sin(x) + 4 cos(x) sin(x)
   -----------------------------------------------------
                             8

This looks still a bit unsatisfactory, but we can now use trigsimp to perform simplifications that use the Pythagorean identity
sin(x)^2 + cos(x)^2 = 1:

trigsimp(%);
       3
 cos(x) sin (x)

Here we obtained exactly the exression that we started with. This is not always possible. Let us look at this example:






A lot of questions on various mailing lists are about problems
that require sophisticated simplification of trigonometric
expressions. Some time ago, a maxima user ran into trouble when
he solved this integral:


result:integrate(9/cos(7*x)^2, x);
 

Maxima answers


							   18 sin(14 x)
(%o20) 					  ----------------------------------------------
					       2	      2
					  7 sin (14 x) + 7 cos (14 x) + 14 cos(14 x) + 7


which is not a very beautiful solution. The user knew that


9*tan(7*x)
----------
    7


is an integral for


   9
---------
        2
cos(7*x)



but found it difficult to simplify the answer from Maxima into this form.
He found it also difficult to prove that the derivative of the answer is in
fact the integrand. Let us look at both problems:


We can try to simplify the answer with trigsimp:


simp1: trigsimp(result);


and obtain


			    9 sin(14 x)
			  ---------------
			  7 cos(14 x) + 7


The use of trigexpand is not helpful: This function would rewrite sin(14*x)
and cos(14*x) in terms of sin(x) and cos(x). We would rather wish to rewrite sin(14*x)
and cos(14*x) in terms of sin(7*x) and cos(7*x). This is possible, but only with
a trick: We have to replace  14*x with 2*y:



simp1, 14*x=2*y;


			    9 sin(2 y)
			  --------------
			  7 cos(2 y) + 7


Now we can apply trigexpand:


trigexpand(%);


			 18 cos(y) sin(y)
		     -------------------------
			   2	     2
		     7 (cos (y) - sin (y)) + 7


trigsimp(%);


			     9 sin(y)
			     --------
			     7 cos(y)


A final trigreduce gives the desired result:


trigreduce(%);


		    9 tan(y)
		    --------
			7


To obtain the original integrand, we have to undo the variable substitution:


%, y = 7*x;


		    9 tan(7*x)
		    ----------
			7



A variable substitution is a powerful instrument to influence simplification with 
trigexpand.

It should be mentioned that the given problem can be simplified much faster:



trigrat(result);


		    9 sin(7 x)
		    ----------
		    7 cos(7 x)



trigreduce(%);


		    9 tan(7*x)
		    ----------
			7



Now let us look at the derivative of the result:



(%i7)deriv: diff(result, x);


		      252 cos(14 x)
      ----------------------------------------------
	   2		  2
      7 sin (14 x) + 7 cos (14 x) + 14 cos(14 x) + 7

						       2
					       3528 sin (14 x)
			    + -------------------------------------------------
				    2		   2			      2
			      (7 sin (14 x) + 7 cos (14 x) + 14 cos(14 x) + 7)



First let us try trigrat, the function that turned out to be so
helpful in the previous example:


trigrat(deriv);


				      18
(%o8) 				 -------------
				 cos(14 x) + 1


Well, this is simplified, but it is still not what we hoped for.
Let us try a variable substitution and a trigexpand:


 %, 14*x = 2*y;
				      18
(%o9) 				 ------------
				 cos(2 y) + 1
trigexpand(%);
				      18
(%o10) 			    -----------------------
				 2	   2
			    - sin (y) + cos (y) + 1
trigsimp(%);
				       9
(%o11) 				    -------
				       2
				    cos (y)
%, y = 7*x;
				       9
(%o15) 				   ---------
				      2
				   cos (7 x)




What we have learned:









  
  

    
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