Solve Trigonometric Sum: Find S Value

[MarkFL]

It can be shown that the following sum:

\(\displaystyle S=\sum_{k=1}^{89}\left(\sin^6\left(k^{\circ}\right)\right)\)

is rational. Find the value of $S$. (Callme)

 

[anemone]

My solution:

Spoiler

My plan is first to reduce the power of both $\sin^6 x$ and $\cos^6 x$ in their sum, this gives

$\begin{align*}\sin^6 x+\cos^6 x&=\sin^4 x(1-\cos^2 x)+\cos^4 x(1-\sin^2 x)\\&=\sin^4 x+\cos^4 x-(\sin x \cos x)^2(\sin^2 x+\cos^2 x)\\&=\dfrac{(3-4\cos 2x+\cos 4x)+(3+4\cos 2x+\cos 4x)}{8}-\dfrac{\sin^2 2x}{4}\\&=\dfrac{3+\cos 4x}{4}-\dfrac{1-\cos 4x}{8}\\&=\dfrac{5}{8}+\dfrac{3\cos 4x}{8}\end{align*}$

Group the given sum as follows we see that

$\sin^6 1^{\circ}+\sin^6 2^{\circ}+\cdots + \sin^6 88^{\circ}+\sin^6 89^{\circ}$

$=(\sin^6 1^{\circ}+\sin^6 89^{\circ})+(\sin^6 2^{\circ}+\sin^6 88^{\circ})+\cdots+(\sin^6 44^{\circ}+\sin^6 46^{\circ})+(\sin^6 45^{\circ})$

$=(\sin^6 1^{\circ}+\cos^6 1^{\circ})+(\sin^6 2^{\circ}+\cos^6 2^{\circ})+\cdots+(\sin^6 44^{\circ}+\cos^6 44^{\circ})+\left(\dfrac{1}{\sqrt{2}}\right)^6$

$=\left(\dfrac{5}{8}+\dfrac{3\cos 4^{\circ}}{8}\right)+\left(\dfrac{5}{8}+\dfrac{3\cos 8^{\circ}}{8}\right)+\cdots+\left(\dfrac{5}{8}+\dfrac{3\cos 176^{\circ}}{8}\right)+\dfrac{1}{8}$

$=44\left(\dfrac{5}{8}\right)+\dfrac{3}{8}((\cos 4^{\circ}+\cos 176^{\circ}+)+(\cos 8^{\circ}+\cos 172^{\circ})+\cdots+(\cos 88^{\circ}+\cos 92^{\circ}))+\dfrac{1}{8}$

$=44\left(\dfrac{5}{8}\right)+\dfrac{3}{8}(0)+\dfrac{1}{8}$

$=\dfrac{221}{8}$

$\therefore S=\dfrac{221}{8}$

 

[kaliprasad]

It can be shown that the following sum:

\(\displaystyle S=\sum_{k=1}^{89}\left(\sin^6\left(k^{\circ}\right)\right)\)

is rational. Find the value of $S$. (Callme)

Spoiler

we have



$\sin^6 x + \cos^6 x$

= $(\sin ^2x + \cos^2x)^3 - 3 \sin ^2 x \cos^2 x(\sin ^2 x + \cos^2 x)$

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

= $ 1- \dfrac{3}{4}(2 \sin x\, \cos\, x)^2 $

= $1- \dfrac{3}{4}(sin ^2 2x)$

= $1- \dfrac{3}{8}(2 sin ^2 2x)$

= $1- \dfrac{3}{8}(1- cos 4x)$

= $ \dfrac{5}{8}+\dfrac{3}{8}\cos 4x$



as $\sin \,x^{\circ} = \cos \, (90-x)^{\circ}$

now

$S=\sum_{k=1}^{89}\left(\sin^6\left(k^{\circ}\right)\right)$

= $\sin^6 45^{\circ} + \sum_{k=1}^{44}\left(\sin^6\left(k^{\circ}\right)+\cos^6\left(k^{\circ}\right)\right )$

= $\dfrac{1}{8} + \sum_{k=1}^{44}(\dfrac{5}{8}+\dfrac{3}{8}\cos\,4k^{\circ})$

= $\dfrac{221}{8} + \dfrac{3}{8} \sum_{k=1}^{44}(\cos\,4k^{\circ})$

now as $cos 4^{\circ} + cos 4 * 44^{\circ} = 0$ so on so sum of the cosines is zeo so result = $\dfrac{221}{8}$hence $S = \dfrac{221}{8}$

Note: As I was solving Anemone beat me to it.

 

[MarkFL]

Thank you anemone and kaliprasad for participating! (Sun)

My solution is essentially the same:

Spoiler

I first used the co-function identity:

\(\displaystyle \sin\left(90^{\circ}-x\right)=\cos(x)\)

to express the sum as:

\(\displaystyle S=\sum_{k=1}^{44} \sin^6\left(k^{\circ}\right)+\sin^6\left(45^{\circ}\right)+\sum_{k=1}^{44} \cos^6\left(k^{\circ}\right)\)

Hence:

\(\displaystyle S=\sum_{k=1}^{44}\left(\sin^6\left(k^{\circ}\right)+\cos^6\left(k^{\circ}\right)\right)+\frac{1}{8}\)

Now consider the following (sum of 2 cubes and a Pythagorean identity):

\(\displaystyle \sin^6(x)+\cos^6(x)=\sin^4(x)-\sin^2(x)\cos^2(x)+\cos^4(x)\)

Now, if we write everything in terms of sine, we obtain:

\(\displaystyle 3\sin^4(x)-3\sin^2(x)+1\)

Factor and use a Pythagorean identity:

\(\displaystyle 1-3\sin^2(x)\cos^2(x)\)

Apply double-angle identity for cosine:

\(\displaystyle \frac{4-3\left(1-\cos^2(2x)\right)}{4}\)

Pythagorean identity:

\(\displaystyle \frac{4-3\sin^2(2x)}{4}\)

Double-angle identity for cosine:

\(\displaystyle \frac{8-3\left(1-\cos(4x)\right)}{8}\)

\(\displaystyle \frac{3\cos(4x)+5}{8}\)

Hence, we now have:

\(\displaystyle S=\frac{1}{8}\sum_{k=1}^{44}\left(3\cos \left(4k^{\circ}\right)+5\right)+\frac{1}{8}\)

\(\displaystyle S=\frac{1}{8} \left(3\sum_{k=1}^{44} \left(3\cos \left(4k^{\circ} \right) \right)+44\cdot5+1 \right)\)

\(\displaystyle S=\frac{1}{8} \left(3\sum_{k=1}^{44} \left(\cos \left(4k^{\circ} \right) \right)+221 \right)\)

Now, observe that:

\(\displaystyle \cos\left(180^{\circ}-x\right)=-\cos(x)\)

And we may write:

\(\displaystyle S=\frac{1}{8} \left(3\sum_{k=1}^{22} \left(\cos \left(4k^{\circ} \right)-\cos \left(4k^{\circ} \right) \right)+221 \right)\)

The sum goes to zero, and we are left with:

\(\displaystyle S=\frac{221}{8}\)