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We are given that $\sin\theta + \sin^3\theta + \sin^2\theta = 1$

Find the value of $\cos^{12}\theta + 3\cos^{10}\theta + 3\cos^{8}\theta + \cos^6\theta + 2\cos^4\theta + 2\cos^2\theta - 2$

Now, I was able to establish the following from the first equation:

$\sin\theta + \sin^3\theta + \sin^2\theta = 1 = \sin^2\theta + \cos^2\theta \implies \sin\theta + \sin^3\theta = \cos^2\theta$

The next obvious step was to simplify the second expression. I let $\cos^2\theta = x$:

$f(x) = x^6 + 3x^5 + 3x^4 + x^3 + 2x^2 + 2x - 2$

$f(-1) = 0 \implies f(x) = (x + 1)(x^5 + 2x^4 + x^3 + 2x - 2)$

I was stuck after this.

Ben Grossmann
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Gerard
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    It seems to me that "the next obvious step" would've been to rewrite the cosine polynomial in terms of sines (via $\cos^2\theta = \sin\theta + \sin^3\theta$). The result is an unattractive degree-$18$ polynomial. The given sine relation lets you whittle that down to a quadratic, but the actual value of the expression isn't immediately clear. – Blue Jul 15 '13 at 15:00
  • BTW: $f(-1)\neq 0$. (Expanding your factorized version misses the $2x$ term.) – Blue Jul 15 '13 at 15:04
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    I don't think $f(-1) = 0$. Can you check? – Calvin Lin Jul 15 '13 at 15:04

3 Answers3

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Observe that the powers of $\cos \theta$ are all even, suggesting that we should use the conversion $\cos^2 \theta = 1 - \sin ^2 \theta$. For simplicity, let $x = \sin \theta$. We are given that

$$(x + x^2 + x^3) = 1$$

and want to find

$$(1-x^2)^6 + 3(1-x^2)^5 + 3(1-x^2)^4 + (1-x^2)^3 + 2(1-x^2)^2 + 2(1-x^2) - 2 $$

By long division, we could factor out $x^3 + x^2 + x -1 $ to get $214x^2 + 20x - 68$, but then it's not clear what to do after that.

You could use the fact that $x^3 + x^2 + x - 1 = 0 $ has 1 real root; but I doubt that is how they want you to proceed.

Calvin Lin
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  • And it has only one real root, by Fermat's test (There is only one change of sign in the coefficients). – OR. Jul 15 '13 at 15:12
  • @RGB When I learned this in school the test was attributed to Descartes. – Ovi Jul 16 '13 at 16:11
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Use that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ and $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.$$

Then replace $e^{ix}$ by $z$ and clear denominators. The first condition becomes an equality of the form $f(z)=0$, the second a polynomial $g(z)$. Use long division to get $g=fq+r$. The value of $f$ will be the same as that of $r$. If you are lucky $r=0$. If not, you will have to solve $f(z)=0$ and evaluate $r(z)$.

I did the computations ... not pretty. $r\neq0$.

OR.
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  • Elegant solution! But a little out of my league. – Gerard Jul 15 '13 at 15:02
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    It is just a standard thing. The idea in Blue's comment also goes along these lines. Turn $\sin$ in to $Z$ and $\cos^2$ in the second polynomial into $1-\sin^2=1-Z^2$. The idea is to then use long division to bring down the degree of the polynomial you want to evaluate. If the remainder in the long division is zero you are very lucky. Otherwise you need to solve for $Z$ in the first polynomial and plug into the second. In Calvin Lin's answer he is giving you the remainder of that division. Fortunately the first polynomial is cubic. You can find its roots easily. Do you know how? – OR. Jul 15 '13 at 15:09
  • Yes, I can proceed from there. Thanks. – Gerard Jul 15 '13 at 15:12
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$\cos^{12}\theta + 3\cos^{10}\theta + 3\cos^{8}\theta + \cos^6\theta + 2\cos^4\theta + 2\cos^2\theta - 2$

By taking $cos^6\theta$ common this can be written as :

$\cos^6\theta [\cos^6\theta +3\cos^4\theta +3\cos^2\theta +1] +2\cos^2\theta ( \cos^2\theta +1)-2$

= $\cos^6\theta (\cos^2\theta +1)^3 +2\cos^2\theta ( \cos^2\theta +1)-2$

Now taking $\cos^2\theta ( \cos^2\theta +1)$ as common you get :

= $\cos^2\theta ( \cos^2\theta +1) [ \cos^4\theta (\cos^2\theta +1)^2 +2]-2$

I hope this will help you further...

Styles
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Sachin
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