0

I have a hint that we need to use an identity that $\sin^4x+\cos^4x=1-\frac{1}{2}\sin^22x$ in order to find the value of $\sin {^4 \frac{\pi}{16}} +\sin {^4 \frac{3\pi}{16}} +\sin {^4 \frac{5\pi}{16}} +\sin {^4 \frac{7\pi}{16}} $ .

I tried and error for different values of $x$ but I still could not eliminate the $\cos$-term

Anson NG
  • 965

4 Answers4

5

We have $$\sin\frac{\pi}{16}=\cos(\frac{\pi}{2}-\frac{\pi}{16})=\cos\frac{7\pi}{16} $$ and $$\sin\frac{3\pi}{16}=\cos(\frac{\pi}{2}-\frac{3\pi}{16})=\cos\frac{5\pi}{16}. $$ Thus $$A=\sin^4\frac{\pi}{16}+\sin^4\frac{3\pi}{16}+\sin^4\frac{5\pi}{16}+\sin^4\frac{7\pi}{16} $$ $$=\sin^4\frac{\pi}{16}+\sin^4\frac{3\pi}{16}+\cos^4\frac{3\pi}{16}+\cos^4\frac{\pi}{16}. $$ Using the hint, we get: $$A=1-\frac{1}{2}\sin^2\frac{2\pi}{16}+1-\frac{1}{2} \sin^2\frac{6\pi}{16}$$ $$=2-\frac{1}{2}(\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8})$$ $$=2-\frac{1}{2}(\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8})=2-\frac{1}{2}=\frac{3}{2}. $$

Qurultay
  • 5,224
  • I believe it is the experience that makes a difference. I would not be able to produce this solution after another hour of thinking. Many thanks! – Anson NG May 21 '19 at 15:24
3

Note that $\cos (\frac{\pi}{2}-x) = \sin (x)$. Subbing $x= \frac{\pi}{16}$ and $x= \frac{3\pi}{16}$ into the identity yields $\sin^4 \left( \frac{\pi}{16} \right) + \sin^4 \left( \frac{7 \pi}{16} \right) = 1 - \frac{1}{2} \sin^2 \left( \frac{\pi}{8} \right)$, and $\sin^4 \left( \frac{3\pi}{16} \right) + \sin^4 \left( \frac{5 \pi}{16} \right) = 1 - \frac{1}{2} \sin^2 \left( \frac{3\pi}{8} \right)$. You can now use the values of $\sin\left( \frac{\pi}{8} \right)$ and $\sin\left( \frac{3\pi}{8} \right)$ to compute values (finding these with half-angle formulas if need be.)

auscrypt
  • 8,186
1

Another way:

Convert sine ratios into cosine using $\sin\left(\dfrac\pi2-x\right)=\cos x$

As $\cos\dfrac{8(2n+1)\pi}{16}=0$

Using multiple Angles formula,

$$f(c)=\cos8x=2^7c^8-\binom812^5c^6+160c^4-32c^2+1$$ where $c=\cos x$

The roots of $f(c)=0$ are $\cos\dfrac{(2n+1)\pi}{16},0\le n\le7$

The roots of $$128t^4-256t^3+160t^2-32t+1=0$$ are $t_n=\cos^2\dfrac{(2n+1)\pi}{16},0\le n\le3$

$$\sum_{n=0}^3t_n^2=\left(\dfrac{256}{132}\right)^2-2\left(\dfrac{160}{128}\right)=?$$

Generalization :

In fact, if $\cos8x=\cos8A, 8x=2m\pi\pm8A$

$x=\dfrac{2m\pi}8+A$ where $0\le m\le7$

So, the roots of $128t^4-256t^3+160t^2-32t+1-\cos8c=0$

will be $c_m=\cos^2\left(\dfrac{2m\pi}8+A\right),0\le m\le3$

$$\implies\sum_{m=0}^3c_m^2=?$$

Here $8A=\dfrac\pi2$

lab bhattacharjee
  • 274,582
0

Hint: $$\sin\left(\frac{\pi}{16}\right)=\frac{1}{2} \sqrt{2-\sqrt{2+\sqrt{2}}}$$

Dr. Sonnhard Graubner
  • 95,283