Prove the following using trigonometric identities

Prove the following

using trigonometric identities:

$\sin^4 x + \cos^{15} x = 1$

how do I resolve cosine

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If that's supposed to be an identity, then it's false. –  Alraxite Jul 10 at 14:46
Are you trying to solve the equation? That's a different thing than proving the relation (which is decidedly not true in general ... try, for instance, $x = \pi$). –  Blue Jul 10 at 14:46

As $0\le \cos^2x\le1,\cos^{15}x-\cos^4x=\cos^4x(\cos^{11}x-1)\le0$

$\implies \sin^4x+\cos^{15}x\le \sin^4x+\cos^4x$

Now, $\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x\le 1$

The equality occurs if one of $\sin x,\cos x$ is $0$

If $\cos x=0,\sin x=\pm1,\implies \sin^4x=1$

If $\sin x=0,\cos x=\pm1\implies \cos^{15}x=\pm1$

So, we can clearly identify when the given proposition holds.

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Doesn't that just show that the sum is less than or equal to 1, not equal to 1? –  William Ballinger Jul 10 at 14:45
@William, please refer to the last statement –  lab bhattacharjee Jul 10 at 14:46
@labbhattacharjee +1 for the work, and solution. :-D –  amWhy Jul 10 at 14:57
@amWhy, have you noticed the latest version. Anyway thanks for the reminding –  lab bhattacharjee Jul 10 at 14:59
Yes, and I changed my comment, accordingly! ;-) –  amWhy Jul 10 at 14:59

$\cos^{15}x=1-\sin^4x=(1-\sin^2x)(1+\sin^2x)=\cos^2 x(1+\sin^2 x)$

$\implies \cos^2 x(\cos^{13}x-\sin^2 x-1)=0$

$\implies \cos^2 x=0,$ or $\cos^{13}x=1+\sin^2 x$

For second case:, $\cos^{13}x=1+\sin^2 x$, $L.H.S=\cos^{13}x\leq 1$ and $R.H.S=1+\sin^2 x\geq 1$, so $L.H.S=R.H.S\implies \cos^{13}x=1$ and $1+\sin^2 x=1$

Solving further is easy.

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Writing $c$ for $\cos x$, and recalling $\sin^2x = 1- \cos^2 x$, we have
\begin{align} 0 &= (1-c^2)^2 + c^{15} - 1 \\ &= c \; \left( c - 1 \right)\left( 2 + 2 c + c^2 + c^3 + c^4 + c^5 + c^6 + c^7+c^8 + c^9+c^{10}+c^{11}+c^{12} \right) \\ &= c \; \left( c - 1 \right )\left( c^{12} + ( 1 + c )( 2 + c^2 + c^4 + c^6 + c^8 + c^{10} ) \right) \end{align}
In the last factor, $2+c^2+c^4+c^6+c^8+c^{10}$ is strictly positive, while $1+c$ is non-negative, so that their product is non-negative, and that product vanishes only when $c=-1$. On the other hand, $c^{12}$ is non-negative, and it vanishes only when $c=0$. Consequently, the last factor as a whole never vanishes.
Therefore, we must have that $\cos x = 0$ or $\cos x = 1$.