if sin(x)+cos(x)=1/3 then Sin(x)^5+Cos(x)^5=?
What do you think it is? We do not do your homework for you here. Show some work, please.
One more thing: Please post homework problems in one of the homework forums.
x5+ y5= (x+ y)(x4- x3y+ x2y2+ xy3+ y4)
You have a sign error, Halls.
5+ y5= (x+ y)(x4- x3y+ x2y2- xy3+ y4)
Let u=sin(x) v=cos(x). We know u+v, and since (u+v)^2=1+2uv, we know uv, hence we know a basis of the symmetric functions in u,v, and we can, if we are bothered work out, u^5+v^5. That I feel is far less interesting than the knowledge about symmetric function theory...
But the OP seems to be posting all these questions as challenges, rather than questions for which he seeks the answer.
sin(x)+cos(x)=1/3 then 1+2sin(x)cos(x)=1/9= 1+sin(2x)
so sin(2x)=-8/9 then 2x=arcsin(-8/9) once you have got 'x' the rest is easy
Separate names with a comma.