1
$\begingroup$

a) Use De Moivre's theorem to express $\frac{\sin 8\theta}{\sin\theta \cos\theta}$ as a polynomial in$ s$, where $s=\sin\theta$

b) Hence solve the equation $x^6-6x^4+10x^2-4=0$

I've been able to do the first question and I worked out the answer to be $8(1-10s^2+24s^4-16s^6)$ but I can't seem to be able to do part b) because I can't see a relationship between the two polynomials.

Any help is appreciated thanks :)

$\endgroup$
1
$\begingroup$

Hint: $\;4\cdot(1-10s^2+24s^4-16s^6) = -\big((2s)^6 - 6 \cdot (2s)^4 + 10 \cdot (2s)^2 - 4\big)\,$.

$\endgroup$
0
$\begingroup$

The key here is to recognise that the equation $x^{6}-6x^{4}+10x^{2}-4=0$ can be changed into $$4(1-10s^{2}+24s^{4}-16s^{6})=0$$

via the substitution $x=2\sin\theta$. Then you can use the identity you proved in the first part to find the roots of the equation in $x$.

$\endgroup$
0
$\begingroup$

Solution

$$x^6-6x^4+10x^2-4=(x^2-2)(x^4-4x^2+2)=(x^2-2)[(x^2-2)^2-2]=0$$

Thus, let $x^2-2=0$ or $x^4-4x^2+2=0$. We obtain $$x=\pm\sqrt{2},$$or$$x=\pm\sqrt{2\pm\sqrt{2}}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.