If `sin(x) + cos(x) = 1,` find `sin^3(x) + cos^3(x).`

Hello!

1. It is simple to solve the equation `sin(x)+cos(x)=1` and substitute the result into the expression `sin^3(x)+cos^3(x).` But we may solve the problem without solving the equation.

First,

`sin^3(x)+cos^3(x)=`

`=(sin(x)+cos(x))^3-3sin(x)cos^2(x)-3sin^2(x)cos(x)=`

`=(sin(x)+cos(x))^3-3sin(x)cos(x)(sin(x)+cos(x))=`

`=1-3sin(x)cos(x).`

Also,

`1=sin^2(x)+cos^2(x)=(sin(x)+cos(x))^2-2sin(x)cos(x)=`

`=1-2sin(x)cos(x),` therefore `sin(x)cos(x)=0.`

Finally, `sin^3(x)+cos^3(x)=1-3sin(x)cos(x)=1,` which is the answer.

2. If we want to solve the equation first, we divide and multiply by `sqrt(2)` and recall that `1/sqrt(2)=sin(pi/4)=cos(pi/4).` Therefore

`sin(x)+cos(x)=sqrt(2)*(cos(x)*cos(pi/4)+sin(x)*sin(pi/4))=`

`=sqrt(2)*cos(x-pi/4)=1.`

So `cos(x-pi/4)=cos(pi/4),`

`x-pi/4=+-pi/4+2k pi,` where `k`...

Hello!

1. It is simple to solve the equation `sin(x)+cos(x)=1` and substitute the result into the expression `sin^3(x)+cos^3(x).` But we may solve the problem without solving the equation.

First,

`sin^3(x)+cos^3(x)=`

`=(sin(x)+cos(x))^3-3sin(x)cos^2(x)-3sin^2(x)cos(x)=`

`=(sin(x)+cos(x))^3-3sin(x)cos(x)(sin(x)+cos(x))=`

`=1-3sin(x)cos(x).`

Also,

`1=sin^2(x)+cos^2(x)=(sin(x)+cos(x))^2-2sin(x)cos(x)=`

`=1-2sin(x)cos(x),` therefore `sin(x)cos(x)=0.`

Finally, `sin^3(x)+cos^3(x)=1-3sin(x)cos(x)=1,` which is the answer.

2. If we want to solve the equation first, we divide and multiply by `sqrt(2)` and recall that `1/sqrt(2)=sin(pi/4)=cos(pi/4).` Therefore

`sin(x)+cos(x)=sqrt(2)*(cos(x)*cos(pi/4)+sin(x)*sin(pi/4))=`

`=sqrt(2)*cos(x-pi/4)=1.`

So `cos(x-pi/4)=cos(pi/4),`

`x-pi/4=+-pi/4+2k pi,` where `k` is any integer.

Therefore `x=2k pi` or `x=pi/2+2k pi.` Then `sin(x)=0,` `cos(x)=1` or `sin(x)=1,` `cos(x)=0.` And the expression in question always = 1.