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While solving trigonometric problems I've noticed possible pattern that reminded me of Fibonacci numbers and Pascal triangle. So I tried to find next "element" of this pattern (8 degree exponent ) and failed to do so. So can someone please tell me is there a pattern or just "coincidence". Possible pattern :

$\sin^2x + \cos^2x = 1 $

$\sin^4x + \cos^4x = 1 - 2\sin^2x\cos^2x$

$\sin^6x + \cos^6x = 1 - 3\sin^2x\cos^2x$

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I can tell you that there is a pattern :) – Ahmed Masud Dec 31 '11 at 14:23
up vote 8 down vote accepted

There is a pattern but I do not think it is as simple as what the first three terms might suggest.

Let me use the abbreviation $s := \sin x$ and $c := \cos x$. In this notation, we have $$ (s^6 + c^6) = (s^6 + c^6)(s^2 + c^2) = (s^8 + c^8) + s^2c^2(s^4+c^4). $$ Now plugging in the expressions you obtained for $s^6+c^6$ and $s^4+c^4$, we get $$ 1 - 3s^2c^2 = (s^8+c^8)+s^2c^2(1-2s^2c^2), $$ or: $$ s^8+ c^8 = 1 - 4s^2c^2 + 2s^4c^4. $$

Alternate derivation: Since $s^2 + c^2 = 1$, let $s^2 = \frac{1}{2} + u$ and $c^2 = \frac12 - u$ for some $u$. Then $$ s^2 c^2 = \frac14 - u^2. \tag{$\dagger$} $$ Now calculating $s^8+c^8$ is straightforward, if a bit involved: $$ \begin{align*} s^8 + c^8 &= \left(\frac12 + u\right)^4 + \left(\frac12 - u\right)^4 \\\\ &= 2 \left( \frac{1}{2^4} + u^4 + 6 \cdot \frac{1}{2^2} \cdot u^2 \right) \\\\ &= \frac{1}{8} +2 u^4 + 3 u^2 \\\\ &\stackrel{(\dagger)}{=} \frac{1}{8} +2 \left( \frac14 - s^2 c^2 \right)^2 + 3 \left( \frac14 - s^2 c^2 \right) \\\\ &= \frac18 + 2\cdot \frac{1}{4^2} + \frac34 +2 s^4 c^4 - 2 \cdot 2 \cdot \frac14 \cdot s^2 c^2 - 3 s^2 c^2 \\\\ &= 1 + 2 s^4 c^4 - 4s^2 c^2 . \end{align*} $$

Although this seems longer, the second method generalises easily for $s^{2n}+c^{2n}$ for all $n$.

What does $\sin^{2n} x + \cos^{2n} x$ look like?

It is a standard fact that any symmetric polynomial in $\lambda_1$ and $\lambda_2$ can be expressed in terms of the elementary symmetric polynomials, which, in this case, is just $\lambda_1 + \lambda_2$ and $\lambda_1 \lambda_2$. Now for $\lambda_1 = s^2 = \sin^2 x$ and $\lambda_2 = c^2 = \cos^2 x$, we have $\lambda_1 + \lambda_2 = 1$. That is, every symmetric polynomial in $s = \sin x$ and $c = \cos x$ with the restriction that every term has an even degree in both variables can be expressed as a polynomial in $\sin x\cos x$.

For the special case of $s^{2n} + c^{2n}$, I already explained how to get such a representation: $$ s^{2n} + c^{2n} = \left( \frac12 + u \right)^n + \left( \frac12 - u \right)^n. $$ Expanding using the binomial theorem, we find that all the odd powers of $x$ cancel and the even powers add up, so that we have an even polynomial in $x$. The degree of this polynomial is exactly $2 \lfloor n/2 \rfloor$.

At the next step, we replace $u^2$ by $\frac14 - s^2c^2$, so that we again get an even polynomial -- this time in $s \cdot c$ -- of the same degree, namely $2 \lfloor n/2 \rfloor$. In other words, we get a polynomial in $s^2 c^2$ of degree $\lfloor n / 2 \rfloor$. (However, the actual polynomials themselves do not seem very interesting.)

In general, one should not expect much more simplification. As Henning notes below, for $n =2$ and $n=3$, the expression $\lfloor n / 2 \rfloor$ is just $1$, which is why we got a polynomial of containing only a constant term and a $\sin^2 x \cos^2 x$ term. This pretty pattern disappears even for $n=4$.

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So the true pattern is that $\sin^{2n}+\cos^{2n}$ is a polynomial of degree $\lfloor n/2\rfloor$ in $\sin^2\cos^2$ with constant term $1$. (Odd powers of $x$ always cancel out after the binomial expansion). For $n=2$ and $n=3$ this polynomial is of degree $1$ with fixed constant term, so it looks like the beginning of a prettier pattern than it really is. – Henning Makholm Dec 31 '11 at 15:10
Note: Thanks to a poor choice of notation, I had used $x$ in a different sense than the question; that has subsequently been changed to $u$ in my answer. The $x$ in @Henning's comment should be read as "$u$" for compatibility with my current answer. – Srivatsan Dec 31 '11 at 16:02

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