Question 1039994
<pre>
The good lady Ikleyn didn't finish. She didn't take it all the 
way to a product.  When you start with a sum or difference of 
2 trig functions as in the case of this one, cos(A-B)-sin(A+B), 
it's normally assumed that you are to change it to a product
of 2 trig functions, not as a product of 2 sums or differences of
trig functions. 

I'll start from scratch, and take it all the way to a product
involving only two trig functions.

{{{cos(A-B)-sin(A+B)}}}

{{{cos(A)cos(B)-sin(A)sin(B)-(sin(A)cos(B)+cos(A)sin(B))}}}

{{{sin(A)sin(B)-sin(A)cos(B)-cos(A)sin(B)-cos(A)cos(B)}}}

{{{(sin(A)^""-cos(A))(sin(B)^""-cos(B))}}}

That's essentially what she got.  That's a product, but a product of 
two differences.  We started with an expression that contained 
only two trig functions.  We aren't done when we get it to 
and expression containing four trig functions. 

So we can keep going, using the trick that both the sine and 
cosine of {{{pi/4}}} are the same since both equal {{{sqrt(2)/2}}}.

We can multiply both parentheses through by {{{sin(pi/4)}}},
which is legitimate as long as we divide by {{{sin(pi/4)}}}
twice also. 

{{{expr(1/sin^2(pi/4))(sin(A)red(sin(pi/4))^""-cos(A)sin(pi/4))(sin(B)red(sin(pi/4))^""-cos(B)sin(pi/4))}}}

Since the cosine and sine of {{{pi/4}}} are the same we can
change the red sines to cosines. 

{{{expr(1/sin^2(pi/4))(sin(A)cos(pi/4)^""-cos(A)sin(pi/4))(sin(B)cos(pi/4))^""-cos(B)sin(pi/4))}}}

Now we use the identity {{{sin(alpha+beta)=sin(alpha)cos(beta)-cos(alpha)sin(beta)}}} to replace the parentheses:

{{{expr(1/sin^2(pi/4))(sin(A-pi/4)^"")(sin(B-pi/4)^"")}}}

And since {{{sin(pi/4)=sqrt(2)/2}}}, the constant out front,

{{{1/sin^2(pi/4) = 1/(sqrt(2)/2)^2 = 1/(2/4) = 1/(1/2)=1*(2/1) = 2}}}

So here's the final answer as a product of two trig functions:

{{{2sin(A-pi/4)sin(B-pi/4)}}}

Edwin</pre>