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If $$\sin^8\theta+\cos^8\theta=\frac{17}{32}$$ find the value of $\theta$ using de Moivre's theorem. I tried a lot exapanding using Binomial theorem and taking real part and equating it given value.

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    $\begingroup$ Have you tried adapting the methods in the answers to your previous question? $\endgroup$ – Arnaud D. Sep 21 '18 at 12:09
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    $\begingroup$ $$8\cdot17=(e^{ix}-e^{ix})^8+(e^{ix}+e^{ix})^8$$ $$\implies68=(e^{i4x})^2+\binom82e^{i4x}+\binom84$$ $\endgroup$ – lab bhattacharjee Sep 21 '18 at 12:35
  • $\begingroup$ I have tried using previous method suggested to me but I my question paper asked me to solve using De-moviere's theorem. $\endgroup$ – priyanka kumari Sep 21 '18 at 12:58
  • $\begingroup$ Not able to understand $\endgroup$ – priyanka kumari Sep 21 '18 at 12:59
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    $\begingroup$ @Krzysztof Myśliwiec In case of "Solve this using l'Hôspital" - as i was tortured in the school - i can still show mot-a-mot that $x/x\to 1$ by using l'Hôspital then implicitly or explicitly use the factor $1$ in the limit i have to compute. The author of the problem will have to make contortions in the problem request to avoid this solution. All this discussion about constraints is not even part of didactics of mathematics. In my subjective opinion, also not part of mathematics, people imposing constraints of such kind also act outside the mathematical spirit. $\endgroup$ – dan_fulea Sep 21 '18 at 22:37
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De Moivre's Theorem: $e^{ix} = cos(x) + i*sin(x)$.

From this famous equation follow the well-known expressions for the cosine: $ cos(x) = (e^{ix}+e^{-ix})/2$ and the sine: $sin(x) = (e^{ix} -e^{-ix})/(2i)$.

We seek the solutions to the equation $sin^8(x) + cos^8(x) = 17/32$. Substitute the cosine and sine expressions. After some calculation we obtain: $e^{8ix} + 28*e^{4ix} + 70 + 28*e^{-4ix} + e^{-8ix} = 68$.

Now use the cosine formula to rewrite this result in terms of $cos(4x)$ and $cos^2(4x)$:

$cos(4x)*[cos(4x) + 14] = 0$

The second term is positive for all values of $x$. Hence we must solve $cos(4x) = 0$. The result is: $x = (1 + 2k) \pi/8$ where $k = 0, 1, 2,3 ….$ With a pocket calculator you can check numerically that this result is correct.

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HINT

According to the formula \begin{align*} \sin(2\theta) = 2\sin(\theta)\cos(\theta) \end{align*}

We can rewrite the given expresion as

\begin{align*} \sin^{8}(\theta) + \cos^{8}(\theta) & = [\sin^{4}(\theta) + \cos^{4}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\ & = [(\sin^{2}(\theta)+\cos^{2}(\theta))^{2} - 2\sin^{2}(\theta)\cos^{2}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\ & = [1 - 2\sin^{2}(\theta)\cos^{2}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\ & = \left[1 - \frac{\sin^{2}(2\theta)}{2}\right]^{2} - \frac{\sin^{4}(2\theta)}{8} = 1 -\sin^{2}(2\theta) + \frac{\sin^{4}(2\theta)}{8} = \frac{17}{32} \end{align*}

Therefore, if we make the substitution $y = \sin(2\theta)$, the given problem is equivalent to \begin{align*} & 4y^{4} - 32y^{2} + 15 = 0 \Longleftrightarrow 4(y^{4} - 8y^{2}) + 15 = 0 \Longleftrightarrow 4(y^{4} - 8y^{2} + 16) - 49 = 0 \Longleftrightarrow\\ & 4(y^{2}-4)^{2} = 49 \Longleftrightarrow y^{2} - 4 = \pm\frac{7}{2} \Longleftrightarrow y^{2} = 4\pm\frac{7}{2} \Longleftrightarrow y = \pm\frac{1}{\sqrt{2}} = \pm\sin\left(\frac{\pi}{4}\right) \end{align*}

Can you proceed from here?

EDIT

According to the Newton's binomial theorem as well as the De Moivre's formula, we have \begin{align*} \cos(8\theta) + i\sin(8\theta) = [\cos(\theta) + i\sin(\theta)]^{8} = \sum_{k=0}^{8}\binom{8}{k}[\cos(\theta)]^{k}[i\sin(\theta)]^{8-k} \end{align*} As a consequence, you can obtain the value of the expression $\sin^{8}(\theta) + \cos^{8}(\theta)$ in terms of sines and cosines and solve the corresponding equation.

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  • $\begingroup$ Yes thanks for your answer even I tried this in my paper but I was asked to solve using De-moviere's theorem. $\endgroup$ – priyanka kumari Sep 23 '18 at 0:26
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I try to give a solution with the given constraint. Using de Moivre's formula we will show $$ \begin{aligned} \cos^4 t +\sin^4t &=\frac 1{4}\Big[\ \cos 4t+3\ \Big]\ , \\ \cos^8 t +\sin^8t &=\frac 1{64}\Big[\ \cos 8t+28\cos 4t +35\ \Big]\ . \end{aligned} $$ There is no prize of beauty we can win for the proof, so let us use $c:=\cos t$, $s=\sin t$ as shortcuts for an easy typing. The strategy is to write $\cos 4t$ and $\cos 8t$ in terms of an even polynomial in $c,s$, with no mixed monomials. Let us get explicit relations: $$ \begin{aligned} \cos 4t &=\text{Real part of }(\cos 4t +i\sin 4t)\\ &=\text{Real part of }(\cos t +i\sin t)^4\quad\text{(de Moivre)}\\ &=c^4-6c^2s^2+s^4\\ &=c^4-3(\underbrace{(c^2+s^2)^2}_{=1}-c^4-s^4)+s^4\\ &=4(c^4+s^4)-3\ ,\\ &\qquad\text{showing the first formula.}\\[2mm] 2c^2s^2 &=(c^2+s^2)^2-c^4-s^4=1-c^4-s^4\ ,\\ 4c^4s^4 &=(1-c^4-s^4)^2\\ &=1+c^8+s^8-2c^4-2s^4+2c^4s^4\ ,\qquad\text{ so}\\ 2c^4s^4 &=1+c^8+s^8-2c^4-2s^4\ ,\qquad\text{ so}\\[2mm] \cos 8t &=\text{Real part of }(\cos 8t +i\sin 8t)\\ &=\text{Real part of }(\cos 4t +i\sin 4t)^2\qquad\text{(de Moivre)}\\ &=\cos^24t-\sin^24t\\ &=2\cos^24t-1\qquad\text{(or going straightforward here...)}\\ &=2(4c^4+4s^4-3)^2 \\ &=\dots \ . \end{aligned} $$ Some further lines are showing the claimed formula for $\cos 8t$. (The term in $c^4s^4$ can be split as above to have only monomials in $c$, or in $s$, or constant ones.)

From the given relation we have to solve the equation in $t$ (instead of $\theta$, which is too hard to type): $$ \frac {17}{32} =\frac 1{64}\Big[\ \cos 8t+28\cos 4t +35\ \Big]\ . $$ Possibly using de Moivre, we write again $\cos 8t = 2\cos^2 4t-1$ and obtain an equation of second degree in $u:=\cos 4t\in[-1,1]$, $$ \frac {17}{32} =\frac 1{64}\Big[\ 2u^2+28u +34\ \Big]\ . $$ Now $17/32$ cancels on both sides, we get $2(u^2+17u)=0$, equivalently (for $|u|\le 1$) $u=0$, i.e. $4t$ is an odd multiple of $\pi/2$, i.e. $t$ is an odd multiple of $\pi/8$.

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