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Evaluate $$(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots(2-\sec^2{44^{\circ}})(2-\sec^2{46^{\circ}})\cdots(2-\sec^2{89^{\circ}})$$

This same problem come from Problem 21:But that problem is very very easy $$(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots\cdots(2-\sec^2{89^{\circ}})$$

and since $sec^2{45^{\circ}}=2$ so $$(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots(2-\sec^2{89^{\circ}})=0$$

so I ask this follow problem have value?:$$(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots(2-\sec^2{44^{\circ}})(2-\sec^2{46^{\circ}})\cdots(2-\sec^2{89^{\circ}})$$ Thank you, and this problem is interesting. we kown that $$2-\sec^2{x}=\dfrac{\sec^2{x}}{\sec{2x}}$$ and we kown $$\sin{1^{0}}\sin{2^{0}}\cdots\sin{89^{0}}=\dfrac{6}{4^{45}}\sqrt{10}$$ This result use this $$\sin{3x}=4\sin{x}\sin{(x+60^{0})}\sin{(60^{0}-x)}$$

so $$\cos{1^{0}}\cos{2^{0}}\cdots\cos{89^{0}}=\dfrac{6}{4^{45}}\sqrt{10}$$ so \begin{align} &(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots(2-\sec^2{44^{\circ}})(2-\sec^2{46^{\circ}})\cdots(2-\sec^2{89^{0}})\\ &=\dfrac{cos{2^{0}}\cos{4^{0}}\cdots\cos{88^{0}}(-\sin{2^{0}})(-\sin{4^{0}})\cdots(-\sin{88^{0}})}{\cos^2{1^{0}}\cos^2{2^{0}}\cdots\cos^2{89^{0}}}\\ &=\dfrac{(\sin{4^{0}}\sin{8^{0}}\cdots\sin{88^{0}})^2}{A^2} \end{align}

where $A=\dfrac{12}{4^{45}}\sqrt{5}$

so we must find $\sin{4^{0}}\sin{8^{0}}\cdots\sin{88^{0}}$ I think must use $$\sin{3x}=4\sin{x}\sin{(60-x)}\sin{(60+x)}$$

Thank you, I have solve it. I have this result $$(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots(2-\sec^2{44^{\circ}})(2-\sec^2{46^{\circ}})\cdots(2-\sec^2{89^{\circ}})=2^{88}$$

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up vote 10 down vote accepted

$$ \begin{align} \prod_{\substack{k=1\\k\ne45}}^{89}\left(2-\sec^2(k\pi/180)\right) &=\prod_{\substack{k=1\\k\ne45}}^{89}\left(1-\tan^2(k\pi/180)\right)\tag{1}\\ &=\prod_{\substack{k=1\\k\ne45}}^{89}\left(\frac{\cos^2(k\pi/180)-\sin^2(k\pi/180)}{\cos^2(k\pi/180)}\right)\tag{2}\\ &=\prod_{k=1}^{44}\frac{\cos(k\pi/90)}{\cos^2(k\pi/180)} \prod_{k=46}^{89}\frac{\cos(k\pi/90)}{\cos^2(k\pi/180)}\tag{3}\\ &=\prod_{k=1}^{44}\frac{\cos(k\pi/90)}{\cos^2(k\pi/180)} \prod_{k=1}^{44}\frac{-\cos(k\pi/90)}{\sin^2(k\pi/180)}\tag{4}\\ &=(-1)^{44}\left(\prod_{k=1}^{44}\frac{2\cos(k\pi/90)}{\sin(k\pi/90)}\right)^2\tag{5}\\ &=2^{88}\prod_{k=1}^{44}\cot^2(k\pi/90)\tag{6}\\ &=2^{88}\prod_{k=1}^{22}\cot^2(k\pi/90)\prod_{k=23}^{44}\cot^2(k\pi/90)\tag{7}\\ &=2^{88}\prod_{k=1}^{22}\cot^2(k\pi/90)\prod_{k=1}^{22}\tan^2(k\pi/90)\tag{8}\\[6pt] &=2^{88}\tag{9} \end{align} $$ Justification:

$(1):$ $\sec^2(x)=1+\tan^2(x)$

$(2):$ $\tan(x)=\dfrac{\sin(x)}{\cos(x)}$

$(3):$ $\cos(2x)=\cos^2(x)-\sin^2(x)$

$(4):$ $\cos(\pi-x)=-\cos(x)$ and $\cos(\pi/2-x)=\sin(x)$

$(5):$ $\sin(2x)=2\sin(x)\cos(x)$

$(6):$ $\cot(x)=\dfrac{\cos(x)}{\sin(x)}$

$(7):$ splitting the product

$(8):$ $\cot(\pi/2-x)=\tan(x)$

$(9):$ $\cot(x)\tan(x)=1$


N[Product[2-Sec[k Pi/180]^2,{k,1,44}]Product[2-Sec[k Pi/180]^2,{k,46,89}],20]


N[2^88, 20]


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nice! you alway give me a nice methods,Thank you ,my frend,and someone awalys say my problem is very bad,such… – math110 Jun 1 '13 at 5:31
I think the post is a very beautiful item, as a result, they are believed to be in a mess – math110 Jun 1 '13 at 5:32
I believe that you must not think like that, – math110 Jun 1 '13 at 5:32
This is an interesting problem. Did it originate from Problem 21, or somewhere else? Although Problem 21 is easily solvable because there is a factor of $0$, it is amazing that taking out the factor of $0$ leaves such a nice answer. – robjohn Jun 1 '13 at 5:57

Actual problem is to find the value of the Product $P$ and $Q$ where

$$ P= \left(1-{Tan^2{1^\circ}}\right) \left(1-{Tan^2{2^\circ}}\right)\cdots \left(1-{Tan^2{44^\circ}}\right) $$ $$ Q= \left(1-{Tan^2{46^\circ}}\right) \left(1-{Tan^2{47^\circ}}\right)\cdots \left(1-{Tan^2{89^\circ}}\right) $$ $\implies$ $$ Q= \left(1-{Cot^2{44^\circ}}\right) \left(1-{Cot^2{43^\circ}}\right)\cdots \left(1-{Cot^2{1^\circ}}\right) $$ Let $$ R= Tan{1^\circ} \:Tan{2^\circ}\cdots Tan{44^\circ} $$ So

$$ Q=\frac{P}{R^2}$$ Also

$$ S=\left(1+Tan{1^\circ}\right)\left(1+Tan{2^\circ}\right)\cdots \left(1+Tan{44^\circ}\right)=2^{22} $$ and

$$ T=\left(1-Tan{1^\circ}\right)\left(1-Tan{2^\circ}\right)\cdots \left(1-Tan{44^\circ}\right) $$ so

$$ T=\left(\frac{2Tan{44^\circ}}{1+Tan{44^\circ}}\right) \left(\frac{2Tan{43^\circ}}{1+Tan{43^\circ}}\right) \cdots \left(\frac{2Tan{1^\circ}}{1+Tan{1^\circ}}\right)$$ So

$$ T=\frac{2^{44}R}{S}=2^{22}R$$ But

$$P=ST=2^{44}R$$ So

$$Q=\frac{P}{R^2}=\frac{2^{44}}{R}$$ Finally


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Sadly, the answer won't be anything nice. You've removed essentially the only trick to this problem and now you just have the product of a bunch of almost unrelated numbers. Note that the later terms will start getting quite large so the actual value will be very large (of the order of $10^{26}$)

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No,I don't think so,But thank you all the same – math110 Jun 1 '13 at 2:01
There's really no reason at all for this to be something significant though. Cosine functions don't change in a nice way (e.g. linearly) and cos(1) certainly isn't a nice value (I don't believe we even have an exact value for it in involving only roots, powers, etc. in a finite representation). – john Jun 1 '13 at 3:29
$2^{88}$ may not be nice, but it is nicer than the original. – robjohn Jun 1 '13 at 5:03
oh wow. This is quite surprising to me. It uses a result I've never seen before but it's really quite cool. Thanks @robjohn – john Jun 1 '13 at 5:06
@john: which result is uncommon? I tried to break everything down to pretty basic steps. – robjohn Jun 1 '13 at 5:07

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