Question 926756
<pre>
To prove the identity:

{{{(sin(10y)-sin(4y))/(cos(4y)+cos(10y))}}}{{{""=""}}}{{{tan(3y)}}}

The average of 10y and 4y is 7y, so write 10y as 7y+3y and 4y as 7y-3y


{{{(sin(7y+3y)-sin(7y-3y))/(cos(7y-3y)+cos(7y+3y))}}}

Use the double angle formulas for sin(AħB) and cos(AħB):

{{{( ( sin(7y)cos(3y)+cos(7y)sin(3y)^"" )-( sin(7y)cos(3y)-cos(7y)sin(3y)^"" ) )
/
( (cos(7y)cos(3y)+sin(7y)sin(3y)^"" )+(cos(7y)cos(3y)-sin(7y)sin(3y)^"" ) )

}}}

{{{(  sin(7y)cos(3y)+cos(7y)sin(3y) - sin(7y)cos(3y)+cos(7y)sin(3y)  )
/
( cos(7y)cos(3y)+sin(7y)sin(3y) +cos(7y)cos(3y)-sin(7y)sin(3y)  )


}}}

{{{( cross(sin(7y)cos(3y))+cos(7y)sin(3y) - cross(sin(7y)cos(3y))+cos(7y)sin(3y)  )
/
( cos(7y)cos(3y)+cross(sin(7y)sin(3y)) +cos(7y)cos(3y)-cross(sin(7y)sin(3y))  )


}}}

{{{( cos(7y)sin(3y)+cos(7y)sin(3y)  )
/
( cos(7y)cos(3y)+cos(7y)cos(3y))  )


}}}

{{{(2cos(7y)sin(3y)  )
/
(2cos(7y)cos(3y))  )


}}}

{{{(cross(2)cross(cos(7y))sin(3y)  )
/
(cross(2)cross(cos(7y))cos(3y))  )


}}}

{{{sin(3y)/cos(3y)}}}

{{{tan(3y)}}}

Edwin</pre>