# 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 '13 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 '13 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 '13 at 14:45
@William, please refer to the last statement – lab bhattacharjee Jul 10 '13 at 14:46
@labbhattacharjee +1 for the work, and solution. :-D – amWhy Jul 10 '13 at 14:57
@amWhy, have you noticed the latest version. Anyway thanks for the reminding – lab bhattacharjee Jul 10 '13 at 14:59
Yes, and I changed my comment, accordingly! ;-) – amWhy Jul 10 '13 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 $\cos x = 1$.

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I will take an approach based on the properties of the trigonometric functions involved. First off, $\ f(x) \ = \ \cos^{15} x \ + \ \sin^4 x \$ has a period of $\ 2 \pi \ ,$ since the sine and cosine functions do. So we can restrict the basic search for solution angles to the "principal circle", $\ 0 \ \le \ x \ < \ 2 \pi \ .$

Next, we can use the Pythagorean Identity to re-write the function as

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

$$= \ \cos^{15} x \ + \ \cos^4 x \ - \ 2 \cos^2 x \ + \ 1 \ \ .$$

Since cosine is an even function (even symmetry about $\ x = 0 \$), each term in this "polynomial in cosines" is even, and hence $\ f(x) \$ is an even function. Further, because $\ \cos(\pi - x )$ $= \ \cos(\pi + x ) \ ,$ cosine also has even symmetry about $\ x = \pi \ ,$ and so $\ f(x) \$ does as well. We can thus further limit our discussion to $\ 0 \ \le \ x \ \le \ \pi \ .$

If we simply set $\ f(x) \ = \ 1 \ ,$ we have

$$\cos^{15} x \ + \ \cos^4 x \ - \ 2 \cos^2 x \ + \ 1 \ = \ 1 \ \ \Rightarrow \ \ \cos^2 x \ ( \cos^{13} x \ + \ \cos^2 x \ - \ 2 ) \ = \ 0 \ ,$$

from which we see immediately that there are solutions from $\ \cos x = 0 \ ,$ one at $\ x = \frac{\pi}{2} \$ and the other, by symmetry, at $\ x = \frac{3 \pi}{2} \ .$ A cursory "inspection" of the factor in parentheses indicates that there is also a solution from $\ \cos x = 1 \ ,$ giving $\ x = 0 \ .$

We have found three solutions then in the principal circle; if we want all solutions to the equation of the problem, we may write $\ x \ = \ 0 + 2k \pi \ , \ \frac{\pi}{2} + 2k \pi \ , \ \frac{3\pi}{2} + 2k \pi \$ , or just $\ x \ = 2k \pi , \ \frac{\pi}{2} + k \pi \ .$ It still remains to determine whether we have found all of the possible solutions.

For this, we can look at the behavior of the terms of $\ f(x) \ .$ Of course, $0 \ \le \ \sin^4 x \ \le \ 1$ everywhere; $\ 0 \ \le \ |\cos^{15} x| \ \le \ 1 \ ,$ but $\ \cos^{15} x \$ itself is only positive over $\ 0 \ \le \ x \ < \frac{\pi}{2} \ ,$ so there are no solutions to $\ f(x) \ = \ 1 \$ on the interval $\ \frac{\pi}{2} < x \le \pi \ .$ By application of the symmetries we've mentioned, principal circle solutions must therefore lie in $\ 0 \ \le \ x \ \le \ \frac{\pi}{2} \ \text{and} \ \ \frac{3 \pi}{2} \ \le \ x \ < \ 2 \pi \ .$

$$\ | \cos^{15} x | \ \le \ \left( \frac{\sqrt{3}}{2} \right)^{15} \ \approx \ 0.116 \ \ \text{for} \ \ \frac{\pi}{6} \le x \le \frac{\pi}{2} \$$

and

$$\ | \sin^4 x | \ \le \ \left( \frac{\sqrt{2}}{2} \right)^4 \ = \ 0.25 \ \ \text{for} \ \ 0 \le x \le \frac{\pi}{6} \ \ .$$

So the roots of $\ f(x) \ = \ 1$ must lie in fairly narrow "windows" about $\ x = 0 \$ and $\ x = \frac{\pi}{2} \ .$

For the last portion of the argument, we will need to invoke the "binomial approximation" (since we are eschewing the tools of calculus, which would have made a lot of this argument much easier). In the neighborhood of $\ x = 0 \ ,$ we will write

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

$$\approx \ (1 - \frac{15}{2} \sin^2 x) \ + \ \sin^4 x \ \approx \ 1 \ - \ \frac{15}{2} x^2 \ + \ x^4 \ ,$$

this last result from the "small-angle approximation". Since $\ x^4 \$ is much smaller than $\ x^2 \$ in this neighborhood, it is clear that $\ f(x) \ = \ 1 \$ only at $\ x = 0 \$ , so there are no other solutions to be found. In the neighborhood of $\ x = \frac{\pi}{2} \ ,$ we will use the co-relation $\cos x \ = \ \sin(\frac{\pi}{2} - x )$ . Writing $\ \xi \ = \ \frac{\pi}{2} - x \ ,$ our function becomes

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

$$= \ 1 \ - \ 2 \sin^2 \xi \ + \ \sin^4 \xi \ + \ \sin^{15} \xi \ \approx \ 1 \ - \ 2 \xi^2 \ + \ \xi^4 \ + \ \xi^{15} \ \ ,$$

which, by reasoning similar to that for the previous approximation, tells us there is only one solution to $\ f(x) \ = \ 1 \$ exactly at $\ x = \frac{\pi}{2} \ .$ This also permits us to "pick up" the solution at $\ x = \frac{3 \pi}{2} \ ,$ by the earlier symmetry argument. Thus, we have confirmed that the principal circle solutions we proposed above are the only ones. Below is a graph of $\ f(x) \ .$

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