MHB Solve Trigonometric Sum: Find S Value

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Sum Trigonometric
AI Thread Summary
The discussion centers around finding the rational value of the sum S, defined as S = ∑(sin^6(k°)) for k from 1 to 89. Participants, including Callme, Anemone, and kaliprasad, share their approaches to solving this problem. Anemone's solution is acknowledged as being reached before Callme's, indicating a collaborative effort. The conversation highlights the mathematical reasoning behind the sum's rationality. Ultimately, the focus remains on determining the exact value of S.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
It can be shown that the following sum:

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

is rational. Find the value of $S$. (Callme)
 
Mathematics news on Phys.org
My solution:

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

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

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

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

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

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

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

$=\dfrac{221}{8}$

$\therefore S=\dfrac{221}{8}$
 
MarkFL said:
It can be shown that the following sum:

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

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

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.
 
Thank you anemone and kaliprasad for participating! (Sun)

My solution is essentially the same:

I first used the co-function identity:

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

to express the sum as:

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

Hence:

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

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

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

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

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

Factor and use a Pythagorean identity:

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

Apply double-angle identity for cosine:

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

Pythagorean identity:

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

Double-angle identity for cosine:

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

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

Hence, we now have:

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

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

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

Now, observe that:

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

And we may write:

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

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

$$S=\frac{221}{8}$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top