Sin^5(x)+Cos^5(x) Equation Solution for Mathematicians

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The discussion revolves around solving the equation Sin^5(x) + Cos^5(x) given that Sin(x) + Cos(x) = 1/3. Participants emphasize the importance of showing work rather than simply providing answers, highlighting that this is not a homework help forum. There is a correction regarding a sign error in a mathematical identity related to the sum of fifth powers. The conversation also touches on using symmetric function theory to derive the solution, with one participant outlining a method to find Sin(2x) based on the initial condition. Overall, the thread encourages mathematical exploration and problem-solving rather than straightforward answers.
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if sin(x)+cos(x)=1/3 then Sin(x)^5+Cos(x)^5=?
 
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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)
 
HallsofIvy said:
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
 
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