# General Solution of the equation $\sin^{2015}(\phi)+\cos^{2015}(\phi) = 1$

(1) Calculation of General Solution of the equation $\sin^{2015}(\phi)+\cos^{2015}(\phi) = 1$

(2) Calculation of General Solution of the equation $\sin^{3}(\phi)+\cos^{5}(\phi) = 1$

$\bf{My\; Try::}$ For $(1)$ one:: We Can write $\sin^{2015}(\phi)\leq \sin^{2}(\phi)\;\;,\cos^{2015}(\phi)\leq 1\;\forall \phi \in \mathbb{R}$

So $\sin^{2015}(\phi)+\cos^{2015}(\phi)\leq \sin^{2}(\phi)+\cos^{2}(\phi) = 1$

Now How can I solve after that, Help me

thanks

## 3 Answers

Hint: $\sin^{2015}\phi - \sin^2\phi = 0 \Rightarrow \sin^2\phi\left(\sin^{2013}\phi - 1\right) = 0 \to \sin\phi = 0,1$. Same for cosine.

• This is cute... – Eric Stucky Nov 16 '14 at 2:49
• From the given, we have $(\sin^{2015}(\phi)- \sin^2(x))+(\cos^{2015}(\phi) - \cos^2(x))= 0$. How can we conclude $\sin^{2015}(\phi)- \sin^2(x) = 0$ and $\cos^{2015}(\phi) -\cos^2(x)= 0$? – Mick Nov 16 '14 at 5:06
• Thanks OC-Sansoo, But i have a doubt same as Mick which is written above, plz clearfy me, Thanks – juantheron Nov 16 '14 at 11:44

Note that for $m,m\geq 2$, we have $\sin^{m}(x) \leq \sin^2(x)$, $\cos^{n}(x) \leq \cos^2(x)$ and $\sin^2(x) + \cos^2(x) = 1$.

Clearly, $s,c\geq0.$ If $s,c>0,$ we have both $s^{2015}<s^2,c^{2015}<c^2.$ This is impossible since $s^{2015}+c^{2015}=s^2+c^2.$ Thus $s=0,c=1$ or viceversa, e.g. $\phi=0,\pi/2$ modulo $2\pi$