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

Monday, February 9, 2009

If $\displaystyle{\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}={1},$ find $\displaystyle{{\sin}}^{{3}}{\left({x}\right)}+{{\cos}}^{{3}}{\left({x}\right)}.$

Hello!

1. It is simple to solve the equation $\displaystyle{\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}={1}$ and substitute the result into the expression $\displaystyle{{\sin}}^{{3}}{\left({x}\right)}+{{\cos}}^{{3}}{\left({x}\right)}.$ But we may solve the problem without solving the equation.

First,

$\displaystyle{{\sin}}^{{3}}{\left({x}\right)}+{{\cos}}^{{3}}{\left({x}\right)}=$

$\displaystyle={{\left({\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}\right)}}^{{3}}-{3}{\sin{{\left({x}\right)}}}{{\cos}}^{{2}}{\left({x}\right)}-{3}{{\sin}}^{{2}}{\left({x}\right)}{\cos{{\left({x}\right)}}}=$

$\displaystyle={{\left({\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}\right)}}^{{3}}-{3}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}{\left({\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}\right)}=$

$\displaystyle={1}-{3}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}.$

Also,

$\displaystyle{1}={{\sin}}^{{2}}{\left({x}\right)}+{{\cos}}^{{2}}{\left({x}\right)}={{\left({\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}\right)}}^{{2}}-{2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}=$

$\displaystyle={1}-{2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}},$ therefore $\displaystyle{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}={0}.$

Finally, $\displaystyle{{\sin}}^{{3}}{\left({x}\right)}+{{\cos}}^{{3}}{\left({x}\right)}={1}-{3}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}={1},$ which is the answer.

2. If we want to solve the equation first, we divide and multiply by $\displaystyle\sqrt{{{2}}}$ and recall that $\displaystyle\frac{{1}}{\sqrt{{{2}}}}={\sin{{\left(\frac{\pi}{{4}}\right)}}}={\cos{{\left(\frac{\pi}{{4}}\right)}}}.$ Therefore

$\displaystyle{\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}=\sqrt{{{2}}}\cdot{\left({\cos{{\left({x}\right)}}}\cdot{\cos{{\left(\frac{\pi}{{4}}\right)}}}+{\sin{{\left({x}\right)}}}\cdot{\sin{{\left(\frac{\pi}{{4}}\right)}}}\right)}=$

$\displaystyle=\sqrt{{{2}}}\cdot{\cos{{\left({x}-\frac{\pi}{{4}}\right)}}}={1}.$

So $\displaystyle{\cos{{\left({x}-\frac{\pi}{{4}}\right)}}}={\cos{{\left(\frac{\pi}{{4}}\right)}}},$

$\displaystyle{x}-\frac{\pi}{{4}}=\pm\frac{\pi}{{4}}+{2}{k}\pi,$ where $\displaystyle{k}$ is any integer.

Therefore $\displaystyle{x}={2}{k}\pi$ or $\displaystyle{x}=\frac{\pi}{{2}}+{2}{k}\pi.$ Then $\displaystyle{\sin{{\left({x}\right)}}}={0},$ $\displaystyle{\cos{{\left({x}\right)}}}={1}$ or $\displaystyle{\sin{{\left({x}\right)}}}={1},$ $\displaystyle{\cos{{\left({x}\right)}}}={0}.$ And the expression in question always = 1.