# $\sin^n a + \cos^n a = 1$ is only true when $n=2$

Prove that

$$\forall a\in\mathbb R:\quad\sin^n a + \cos^n a = 1$$ is only true when $n=2$

• Welcome to the site. What have you tried so far? – mattos Jan 14 '15 at 13:22
• for $a=\pi/2$ it is true for all $n.$ – Leox Jan 14 '15 at 13:23
• Come on, it is obvious he meant that it must hold for every $a$ :) – Ant Jan 14 '15 at 13:24
• It should probably be specified that if it holds for all $a\in\mathbb{R}$, then $n=2$. – Eff Jan 14 '15 at 13:24
• @Ant: Yes, but still, OP can learn from this. It's important to say what you mean in mathematics. – MPW Jan 14 '15 at 13:29

Simply because, if $a\not\equiv0 \mod\dfrac\pi 2$, we have $\,\,0< \lvert\sin a\rvert<1$ and $\,\,0< \lvert\cos a\rvert<1$, so that: $$\sin^n a\le \lvert\sin a\rvert^n<\sin^2 a\quad\text{and}\quad\cos^n a\le \lvert\cos a\rvert^n<\cos^2 a$$ for all $n>2$.

• If ur logic is correct in this instance,then why is sina+cosa not equal to 1. – Dhruv Sawhney Jan 14 '15 at 14:01
• Because $\sin^2 a<\lvert\sin a \rvert$, and the analog for $\cos$, so that $\lvert\sin a \rvert +\lvert\cos a \rvert>1$. – Bernard Jan 14 '15 at 14:05

Hint:

Set $a=\pi/4$ and then see where you end up with...

• @orangeskid: thanks for the edit! that was a silly mistype. – Fabian Jan 14 '15 at 13:30
• no worries; couldn't let it gather those "hater votes" – Orest Bucicovschi Jan 14 '15 at 13:39

Rewrite equation in the form:

$$\cos^n(a) + sin^n(a) = cos^2(a) + sin^2(a)$$ $$\cos^2(a)(cos^{n-2}(a) - 1) + sin^2(a)(sin^{n-2}(a) - 1) = 0$$

Left part is negative for $$n \not= 2$$

• nice way to do it :) – Ant Jan 14 '15 at 14:52