Trigonometry problem: $\sin^4 \frac {\pi}{7} + \sin^4 \frac {3\pi}{7} + \sin^4 \frac {5\pi}{7} = \frac {a}{b}$

Question

The question:

$$\sin^4 \frac {\pi}{7} + \sin^4 \frac {3\pi}{7} + \sin^4 \frac {5\pi}{7} = \frac {a}{b} $$

Find $a+b$.

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My attempt:

$$\sin^4 \theta = (\sin^2 \theta)^2 = \left(\frac{1-\cos2\theta}{2}\right)^2 \\ = \frac {\cos4\theta - 4\cos2\theta+3}{8}$$

So, using this result, the question simplifies to:

$$\frac {\cos \frac{4\pi}{7} + \cos \frac{12\pi}{7} + \cos \frac{20\pi}{7} - 4 (\cos \frac{2\pi}{7} + \cos \frac{6\pi}{7} + \cos \frac{10\pi}{7})+9}{8}$$

I tried using the identity $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$ but it just doesn't get shorter. I assume that this question is easily solvable using trigonometric identities, but I would also really appreciate alternate solutions not using pure trigonometry.

Answer 1

Using the identity $2\sin A\cos B = \sin(A+B) + \sin(A - B)$, we have $$ \begin{align*} 2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7} &= \sin\frac{4\pi}{7} + \sin(0) = \sin\frac{4\pi}{7}\\ 2\sin\frac{2\pi}{7}\cos\frac{6\pi}{7} &= \sin\frac{8\pi}{7} + \sin\frac{-4\pi}{7} = -\sin\frac{\pi}{7} - \sin\frac{4\pi}{7} \\ 2\sin\frac{2\pi}{7}\cos\frac{10\pi}{7} &= \sin\frac{12\pi}{7} + \sin\frac{-8\pi}{7} = -\sin\frac{5\pi}{7} + \sin\frac{\pi}{7}, \\ \end{align*} $$ so their sum is $-\sin\tfrac{5\pi}{7} = -\sin\tfrac{2\pi}{7}$. Hence multiplying and dividing the sum $\cos\tfrac{2π}{7}+\cos\tfrac{6π}{7}+\cos\tfrac{10π}{7}$ by $2\sin\tfrac{2\pi}{7}$ gives $-\frac{1}{2}$. Now do the same with $\cos\tfrac{4π}{7}+\cos\tfrac{12π}{7}+\cos\tfrac{20π}{7}$ via $2\sin\tfrac{4\pi}{7}$.

Answer 2

Hint:

Like Prove that $\cos(\pi/7)$ is root of equation $8x^3-4x^2-4x+1=0$,

$\cos\dfrac{2r\pi}7; r=1,2,3$ are the roots of $$8c^3+4c^2-4c-1=0$$

Using $\cos2x=1-2\sin^2x, s_r=\sin^2\dfrac{r\pi}7; r=1,3,5$ are the roots of $$8(1-2s)^3+4(1-2s)^2-4(1-2s)-1=0\iff64s^3+(\cdots)s^2+(\cdots)s+(\cdots)=0$$

Use Vieta's formula $$s_1^2+s_3^2+s_5^2=(s_1+s_3+s_5)^2-2(s_1s_3+s_3s_5+s_5s_1)+?$$

Answer 3

In the numerator of your last step, for the first part, convert $\cos x$ to $\frac12\left(e^{ix}+e^{-ix}\right)$:

$$\begin{align*} \cos \frac{4\pi}{7} + \cos \frac{12\pi}{7} + \cos\frac{20\pi}7 &= \cos \frac{4\pi}{7} + \cos \frac{-2\pi}{7} + \cos\frac{6\pi}7\\ &= \frac{e^{i4\pi/7}+e^{i(-4)\pi/7}}2+\frac{e^{i2\pi/7}+e^{i(-2)\pi/7}}2+\frac{e^{i6\pi/7}+e^{i(-6)\pi/7}}2\\ &= \frac12 \left(e^{i4\pi/7}+e^{i(-4)\pi/7}+e^{i2\pi/7}+e^{i(-2)\pi/7}+e^{i6\pi/7}+e^{i(-6)\pi/7}+1\right)-\frac12\\ &= \frac12 \left(e^{i(-6)\pi/7}+e^{i(-4)\pi/7}+e^{i(-2)\pi/7}+e^{i0\pi/7}+e^{i2\pi/7}+e^{i4\pi/7}+e^{i6\pi/7}\right)-\frac12\\ &= \frac{e^{i(-6)\pi/7}}2 \left(\frac{e^{7\cdot i2\pi/7}-1}{e^{i2\pi/7}-1}\right)-\frac12\\ &= \frac{e^{i(-6)\pi/7}}2 \left(\frac{0}{e^{i2\pi/7}-1}\right)-\frac12\\ &= -\frac12 \end{align*}$$

Similarly, inside the $-4(\ldots)$:

$$\begin{align*} \cos \frac{2\pi}{7} + \cos \frac{6\pi}{7} + \cos\frac{10\pi}7 &= \cos \frac{12\pi}{7} + \cos \frac{20\pi}{7} + \cos\frac{4\pi}7\\ &= -\frac12 \end{align*}$$

Answer 4

Let $$f(k,r) \doteq \sum_{j=0}^{k-1} \sin^{2r}\frac{(2j+1)\pi}{2k+1}.$$ We show that $$f(k,r) = \frac{2k+1}{2^{2r+1}} \binom{2r}{r}$$ for positive integers $k$ and $r$. In particular, this yields the requested result $f(3,2) = 21/16$.

Substituting $\sin \theta = (e^{i\theta}-e^{-i\theta})/(2i)$, we write $f(k,r)$ as $$f(k,r) = f(\omega,k,r) \quad\text{with}\quad \omega \doteq \exp\left(\frac{2 \pi i}{2k+1}\right),$$ where $$f(x,k,r) \doteq \frac{1}{(2i)^{2r}}\sum_{j=0}^{k-1} \left(x^{2j+1}-x^{-(2j+1)}\right)^{2r}.$$ A binomial expansion yields $$f(x,k,r) = \frac{1}{(-4)^r}\sum_{a=0}^{2r}{(-1)^a \binom{2r}{a}\sum_{j=0}^{k-1} x^{2(2j+1)(a-r)}}.$$ We break the sum up into terms $a < r$, $a > r$, and $a = r$ to write this as $$f(x,k,r) = g(x,k,r) + g(x^{-1},k,r) + \frac{k}{4^r}\binom{2r}{r},$$ where $$g(x,k,r) \doteq \frac{1}{4^r}\sum_{a=0}^{r-1}{(-1)^{a-r} \binom{2r}{a}\sum_{j=0}^{k-1} \left(x^{2(a-r)}\right)^{2j+1}}.$$ Evaluating the geometric sum yields $$g(x,k,r) = \frac{1}{4^r}\sum_{a=0}^{r-1}{(-1)^{a-r} \binom{2r}{a} \frac{x^{2(2k+1)(a-r)}-x^{2(a-r)}}{x^{4(a-r)}-1}}.$$ For the special case $x = \omega$, the fact that $\omega^{2k+1}=1$ implies $$g(\omega,k,r) = -\frac{1}{4^r}\sum_{a=0}^{r-1}{(-1)^{a-r} \binom{2r}{a} \frac{1}{\omega^{2(a-r)}+1}}.$$ The identity $1/(z+1) + 1/(z^{-1}+1) = 1$ yields the further simplification $$f(\omega,k,r) =\frac{k}{4^r}\binom{2r}{r} - \frac{1}{4^r}\sum_{a=0}^{r-1}{(-1)^{a-r} \binom{2r}{a}}.$$ Finally, expanding $(1-1)^{2r}$ with the binomial theorem simplifies the second term above and we arrive at the result $$f(k,r) = \frac{2k+1}{2^{2r+1}} \binom{2r}{r}.$$