Question 1099317
<pre>
To prove:

{{{(tan(5x/2))/(tan(x/2))}}}{{{""=""}}}{{{(sin(3x)+sin(2x))/(sin(3x)-sin(2x))}}}

Work with left side:

{{{tan(5x/2)}}}{{{"÷"}}}{{{tan(x/2)}}}

Change to sines and cosines:

{{{(sin(5x/2))/(cos(5x/2))}}}{{{"÷"}}}{{{(sin(x/2))/(cos(x/2))}}}

Invert and multiply

{{{(sin(5x/2))/(cos(5x/2))}}}{{{""*""}}}{{{(cos(x/2))/(sin(x/2))}}}

{{{(sin(5x/2)cos(x/2))/(cos(5x/2)sin(x/2))}}}

{{{(sin(5x/2)cos(x/2)^"")}}}{{{"÷"}}}{{{(cos(5x/2)sin(x/2)^"")}}}

Use a 'product to sum' formula, which is:

         {{{sin(A)cos(B)}}}{{{""=""}}}{{{expr(1/2)(sin(A+B)^""+sin(A-B))}}}
         [That's easy to prove using the formulas for sin(AħB)]

{{{expr(1/2)(sin(5x/2+x/2)^""+sin(5x/2-x/2))}}}{{{"÷"}}}{{{expr(1/2)(sin(x/2+5x/2)^""+sin(x/2-5x/2))}}}

{{{expr(1/2)(sin(6x/2)^""+sin(4x/2))}}}{{{"÷"}}}{{{expr(1/2)(sin(6x/2)^""+sin(-4x/2))}}}

{{{expr(1/2)(sin(3x)^""+sin(2x))}}}{{{"÷"}}}{{{expr(1/2)(sin(3x)^""+sin(-2x))}}}

Use the "odd-even" identity sin(-A) = -sin(A)

{{{expr(1/2)(sin(3x)^""+sin(2x))}}}{{{"÷"}}}{{{expr(1/2)(sin(3x)^""-sin(2x))}}}

{{{(expr(1/2)(sin(3x)^""+sin(2x)))/(expr(1/2)(sin(3x)^""-sin(2x)))}}}

{{{(expr(cross(1/2))(sin(3x)^""+sin(2x)))/(expr(cross(1/2))(sin(3x)^""-sin(2x)))}}}

{{{(sin(3x)^""+sin(2x))/(sin(3x)^""-sin(2x))}}}

Edwin</pre>AKA AnlytcPhil