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Finding value of $\sin(2^\circ)\cdot \sin(4^\circ)\cdot\cdot \cdot \sin(88^\circ)$

Try: Assuming $X=\sin(2^\circ)\cdot \sin(4^\circ)\cdot\cdot\cdot \sin(88^\circ)$

and $Y=\cos(2^\circ)\cdot \cos(4^\circ)\cdot\cdot\cdot \cos(88^\circ)$

So $2^{22}XY=\sin(4^\circ)\cdot \sin(8\circ)\cdot\cdot \sin(176^\circ)$

Could some help me to solve it, thanks in advance

DXT
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    https://math.stackexchange.com/questions/1893810/what-is-the-value-of-sin26-sin212-sin218-cdots-texttill-15?rq=1 – Guy Fsone Jan 23 '18 at 09:33

1 Answers1

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As $\sin(180^\circ-x)=\sin x$

$$\prod_{r=1}^{44}\sin(2r^\circ)=+\sqrt{\prod_{r=1}^{89}\sin(2r^\circ)}$$ as $\sin90^\circ=?$

Using this,

$$\sin90x=90\sin x\cos x+\cdots+2^{90-1}\sin^{89}x\cos x$$

Now if $\sin90x=0,90x=180^\circ n$ where $n$ is any integer

$\implies$ the roots of $\cos x(2^{90-1}\sin^{89}x+\cdots+90\sin x)=0$ are $\sin2r^\circ$ where $1\le r\le90$

$\implies$ the roots of $2^{90-1}\sin^{89}x+\cdots+90\sin x=0$ are $\sin2r^\circ$ where $1\le r\le90,r\ne45$

$\implies$ the roots of $2^{90-1}\sin^{88}x+\cdots+90=0$ are $\sin2r^\circ$ where $1\le r<90,r\ne45$

$\implies\prod_{r=1}^{89}\sin(2r^\circ)=\dfrac{90}{2^{90-1}}$

lab bhattacharjee
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