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Prove that

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

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

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

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

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  • $\begingroup$ @orangeskid: thanks for the edit! that was a silly mistype. $\endgroup$ – Fabian Jan 14 '15 at 13:30
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    $\begingroup$ no worries; couldn't let it gather those "hater votes" $\endgroup$ – Orest Bucicovschi Jan 14 '15 at 13:39
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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 $$n \not= 2$$

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

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