Finding value of product of Cosines

Question

Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$

My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\sin \frac{\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$

So we have $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$

Could some help me to solve it, Thanks in Advanced

Answer 1

Let $\xi=\exp\left(\frac{2\pi i}{40}\right)$. The given product equals

$$ \frac{1}{16}\left(1+\xi+\xi^{-1}\right)\left(1+\xi^3+\xi^{-3}\right)\left(1+\xi^9+\xi^{-9}\right)\left(1+\xi^{27}+\xi^{-27}\right)$$ or $$ \frac{1}{16\xi\xi^3\xi^9\xi^{27}}\cdot\frac{\xi^3-1}{\xi-1}\cdot\frac{\xi^9-1}{\xi^3-1}\cdot\frac{\xi^{27}-1}{\xi^9-1}\cdot\frac{\xi^{81}-1}{\xi^{27}-1}$$ or (by telescopic property and the fact that $\xi^{81}=\xi$) $$ \frac{1}{16 \xi^{1+3+9+27}}=\frac{1}{16\xi^{40}}=\color{red}{\frac{1}{16}}.$$

Answer 2

Hint:

Using $\cos2x=1-2\sin^2x,$

$$\sin3x=3\sin x-4\sin^3x=\cdots=\sin x(1+2\cos2x)$$

Put $2x=\dfrac\pi{20},\dfrac{3\pi}{20},\dfrac{9\pi}{20},\dfrac{27\pi}{20}$ one by one to recognize the Telescoping nature.

Finally use $\sin(2m\pi+y)=\sin y$ where $m$ is any integer.

Answer 3

Put $$f(x)=\left(x+\cos \frac{\pi}{20}\right)\left(x+\cos \frac{3\pi}{20}\right)\left(x+\cos \frac{5\pi}{20}\right)\left(x+\cos \frac{7\pi}{20}\right)\left(x+\cos \frac{9\pi}{20}\right)$$

Then you have to find $f({1\over 2})/({1\over 2}+{\sqrt{2}\over2})$

Answer 4

You can end it by the following way: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)=$$ $$=\left(\frac{1}{2}+\cos9^{\circ}\right)\left(\frac{1}{2}+\sin9^{\circ}\right)\left(\frac{1}{2}+\cos27^{\circ}\right)\left(\frac{1}{2}-\sin27^{\circ}\right)=$$ $$=\left(\frac{1}{4}+\frac{1}{2}(\cos9^{\circ}+\sin9^{\circ})+\frac{1}{2}\sin18^{\circ}\right)\left(\frac{1}{4}+\frac{1}{2}(\cos27^{\circ}-\sin27^{\circ})-\frac{1}{2}\sin54^{\circ}\right)=$$ $$=\left(\frac{1}{4}+\frac{1}{2}(\sin81^{\circ}+\sin9^{\circ})+\frac{1}{2}\sin18^{\circ}\right)\left(\frac{1}{4}+\frac{1}{2}(\sin63^{\circ}-\sin27^{\circ})-\frac{1}{2}\cos36^{\circ}\right)=$$ $$=\left(\frac{1}{4}+\sin45^{\circ}\cos36^{\circ}+\frac{1}{2}\sin18^{\circ}\right)\left(\frac{1}{4}+\sin18^{\circ}\cos45^{\circ}-\frac{1}{2}\cos36^{\circ}\right)=$$ $$=\left(\frac{1}{4}+\frac{\sqrt5+1}{4\sqrt2}+\frac{\sqrt5-1}{8}\right)\left(\frac{1}{4}+\frac{\sqrt5-1}{4\sqrt2}-\frac{\sqrt5+1}{8}\right)=$$ $$=\left(\frac{1}{8}+\frac{\sqrt5}{4\sqrt2}+\frac{1}{4\sqrt2}+\frac{\sqrt5}{8}\right)\left(\frac{1}{8}+\frac{\sqrt5}{4\sqrt2}-\frac{1}{4\sqrt2}-\frac{\sqrt5}{8}\right)=$$ $$=\left(\frac{1}{8}+\frac{\sqrt5}{4\sqrt2}\right)^2-\left(\frac{1}{4\sqrt2}+\frac{\sqrt5}{8}\right)^2=$$ $$=\frac{1}{64}+\frac{\sqrt5}{16\sqrt2}+\frac{5}{32}-\frac{5}{64}-\frac{\sqrt5}{16\sqrt2}-\frac{1}{32}=\frac{1}{8}-\frac{1}{16}=\frac{1}{16}.$$

Answer 5

Durgesh: to complement Jack D'Aurizio fantastic answer above.

If you use Euler's identity

$$e^{ix} = cos x + i\cdot sin x$$

and then apply for $-x$

$$e^{-ix} = cos (-x) + i\cdot sin(-x) = cos x - i\cdot sinx$$

and sum both equations, you get the well known

$$cos x={e^{ix}+e^{-ix}\over2}$$

so

$$\frac12+cos x={1+e^{ix}+e^{-ix}\over2}$$

that is

$$\frac12+cos x={1+e^{ix}+e^{2ix}\over{2e^{ix}}}={1\over{2e^{ix}}}{{e^{3ix}-1}\over{e^{ix}-1}}$$

Substituting for $3x$, $9x$ and $27x$ and multiplying all 4 equations, you get Jack's magical leap :-)