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This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.

What is the value of the following expression?

$$\left( 1+\cos { \left( \frac { \pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 5\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 7\pi }{ 8 } \right) } \right) $$

Edit: Do I manually put the values of $\cos\frac{\pi}{8}$ and all other cosine terms? Is there a better and shorter way?

N. F. Taussig
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Tanuj
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4 Answers4

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$$\left( 1+\cos { \left( \frac { \pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 5\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 7\pi }{ 8 } \right) } \right) =\\ =\left( 1+\cos { \left( \frac { \pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \pi -\frac { 3\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \pi -\frac { \pi }{ 8 } \right) } \right) =\\ =\left( 1+\cos { \left( \frac { \pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1-\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1-\cos { \left( \frac { \pi }{ 8 } \right) } \right) =\\ =\left( 1-\cos ^{ 2 }{ \frac { \pi }{ 8 } } \right) \left( 1-\cos ^{ 2 }{ \frac { 3\pi }{ 8 } } \right) =\sin ^{ 2 }{ \frac { \pi }{ 8 } } \sin ^{ 2 }{ \frac { 3\pi }{ 8 } } =\frac { \left( 1-\cos { \frac { \pi }{ 4 } } \right) }{ 2 } \cdot \frac { \left( 1-\cos { \frac { 3\pi }{ 4 } } \right) }{ 2 } =\\ =\frac { \left( 1-\cos { \frac { \pi }{ 4 } } \right) \left( 1-\cos { \left( \pi -\frac { \pi }{ 4 } \right) } \right) }{ 4 } =\frac { \left( 1-\cos { \frac { \pi }{ 4 } } \right) \left( 1+\cos { \frac { \pi }{ 4 } } \right) }{ 4 } =\frac { 1-\cos ^{ 2 }{ \frac { \pi }{ 4 } } }{ 4 } =\frac { 1-\frac { 1 }{ 2 } }{ 4 } =\frac { 1 }{ 8 } $$

haqnatural
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The answer is $\dfrac{1}{8}$

As $\cos(\pi-x)=-\cos x$ we can write $$\left(1+\cos\frac{\pi}{8} \right)\left(1+\cos\frac{3\pi}{8} \right)\left(1+\cos\frac{5\pi}{8} \right)\left(1+\cos\frac{7\pi}{8} \right)=\left(1+\cos\frac{\pi}{8} \right)\left(1+\cos\frac{3\pi}{8} \right)\left(1-\cos\frac{3\pi}{8} \right)\left(1-\cos\frac{\pi}{8} \right)= \left(1-\cos^2\frac{\pi}{8} \right)\left(1-\cos^2\frac{3\pi}{8} \right)=\sin^2\frac{\pi}{8}\,\sin^2\frac{3\pi}{8}=\left(\sin\frac{\pi}{8}\,\sin\frac{3\pi}{8}\right)^2$$ For the Werner's formula $$\sin x\sin y=\frac{1}{2} (\cos (x-y)-\cos (x+y))$$ so $$\sin\frac{\pi}{8}\,\sin\frac{3\pi}{8}=\frac{1}{2} \left(\cos \left(\frac{\pi}{8}-\frac{3\pi}{8}\right)-\cos \left(\frac{\pi}{8}+\frac{3\pi}{8}\right)\right)=\frac{1}{2}\left(\cos\frac{-\pi}{4}-\cos\frac{\pi}{2}\right)=\frac{\sqrt{2}}{4}$$ Remember that $\cos\dfrac{-\pi}{4}=\cos\dfrac{\pi}{4}=\dfrac{\sqrt 2}{2}$

So the result is $\left(\dfrac{\sqrt{2}}{4}\right)^2=\dfrac{1}{2}$

Raffaele
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Like The value of $\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$,

if $y=1+c,$ we have $$8(y-1)^4-8(y-1)^2+1=0\iff8y^4+\cdots+1=0$$

$$\implies\prod_{n=0}^3\left(1+\cos\dfrac{(2n+1)\pi}8\right)=\dfrac18$$

lab bhattacharjee
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Recall that $$ 1+\cos\alpha=2\cos^2\frac{\alpha}{2} $$ If $\beta=\pi/16$ (so $8\beta=\pi/2$), you get $$ 16\cos^2\beta\cos^23\beta\cos^25\beta\cos^27\beta= (2\cos\beta\cos7\beta)^2(2\cos3\beta\cos5\beta)^2 $$ Now, with the product-to-sum formula, $$ 2\cos\beta\cos7\beta=\cos8\beta+\cos6\beta=\cos6\beta $$ and $$ 2\cos3\beta\cos5\beta=\cos8\beta+\cos2\beta=\cos2\beta $$ Hence your expression is $$ (\cos6\beta\cos2\beta)^2= \frac{1}{4}(\cos8\beta+\cos4\beta)^2= \frac{1}{4}\cos^2\frac{\pi}{4}=\frac{1}{8} $$

egreg
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