Question 869068
Yes, those are the only ones you can use.
{{{sin(a-b)=sin(a)cos(b)-cos(a)sin(b)}}}
{{{sin(a+b)=sin(a)cos(b)+cos(a)sin(b)}}}
So,
{{{sin(a-b)-sin(a+b)=sin(a)cos(b)-cos(a)sin(b)-sin(a)cos(b)-cos(a)sin(b)}}}
{{{sin(a-b)-sin(a+b)=-2cos(a)sin(b)}}}
Then,
{{{sin(a-b)-sin(a+b)-sin^2(b)-cos^2(a)=-2cos(a)sin(b)- sin^2(b)-cos^2(a)}}}
{{{sin(a-b)-sin(a+b)-sin^2(b)-cos^2(a)=-(cos(a)^2+2cos(a)sin(b)+ sin^2(b))}}}
{{{sin(a-b)-sin(a+b)-sin^2(b)-cos^2(a)=-(cos(a)+sin(b))^2}}}
So now this can't equal the right hand side {{{(sin(b)-cos(a))^2}}}
Check the problem setup and make sure it's correct.