Prove \[ \frac{1}{\cos 0^\circ \cos 1^\circ} + \frac{1}{\cos 1^\circ \cos 2^\circ} + \cdots + \frac{1}{\cos 88^\circ \cos 89^\circ} = \frac{\cos 1^\circ}{\sin^2 1^\circ}. \]
Problem
Source: USAMO 1992
Tags: trigonometry, induction, algebra, telescoping sums
Shogia
29.03.2006 22:17
I was hoping to check my answer but since no one has posted one here; I will propose what I came up with.
First change the LHS to summation notation so we have:
$\sum\limits_{n=0}^{88} \frac{1}{{\cos(n)}{\cos(n+1)}}=\frac{\cos 1}{\sin^2 1}.$
Seeing $\cos(n)$ and $\cos(n+1)$ suggests an approach using a telescoping series so lets find a partial fraction decomposition for the summation on the LHS.
$\frac{A}{\cos(n)}+\frac{B}{\cos(n+1)}=\frac{1}{{\cos(n)}{\cos(n+1)}}$
Finding the common denominator and setting the numerators equal provides ${A}{\cos(n+1)}+{B}{\cos(n)}=1$ which is the same as $A\cos(n)\cos(1)-A\sin(1)\sin(n)+B\cos(n)=1$ by the cosine sum identity. Now, remembering that $\sin^2(x)+\cos^2(x)=1$ group the terms with cosine and the terms with sine and factor each out respectively producing $(A\cos(1)+B)(\cos(n))+(-A\sin(1))(\sin(n))=1$
Now we need to find values for A and B such that we obtain our familiar trig identity so
$-A\sin(1)=\sin(n)$ or $A=\frac{-\sin(n)}{\sin(1)}$ Substituting this value for A into $A\cos(1)+B=\cos(n)$ we can solve to find $B=\frac{\cos(n)\sin(1)+\cos(1)\sin(n)}{\sin(1)}$ which we recognize as $B=\frac{\sin(n+1)}{\sin(1)}$ by the sum relation for sine.
Now replacing these values for A and B into the partial fraction decomposition we find that the LHS is equivalent to:
$\sum\limits_{n=0}^{88} \frac{-\sin(n)}{\cos(n)\sin(1)}+\frac{\sin(n+1)}{\cos(n+1)\sin(1)}$
From this we can factor a $\frac{1}{\sin(1)}$ from the summation and use $\tan(x)=\frac{\sin(x)}{\cos(x)}$ to get:
$\frac{1}{\sin(1)}\sum\limits_{n=0}^{88}\tan(n+1)-\tan(n)$
The summation part of this telescopes to $\tan(89)-\tan(0)=\tan(89)$ so the total is $\frac{\sin(89)}{\sin(1)\cos(89)}$ Which is the same as $\frac{cos(1)}{\sin^2(1)}$ since $\sin(90-x)=\cos(x)$ and likewise for sine.
calc rulz
19.08.2007 19:48
We prove using induction that $ \sum_{n = 0}^{k}\frac{1}{{\cos(n)}{\cos(n+1)}}=\frac{\tan (k+1)}{\sin 1}.$
For $ k = 0$ we have $ \sum_{n = 0}^{0}\frac{1}{{\cos(n)}{\cos(n+1)}}=\frac{1}{\cos0\cos1}=\frac{\tan 1}{\sin 1}$
Assume it holds for $ k-1$. Then
\begin{eqnarray*}\sum_{n=0}^{k}\frac{1}{{\cos(n)}{\cos(n+1)}}&=&\frac{\tan k}{\sin 1}+\frac{1}{{\cos(k)}{\cos(k+1)}}\\ &=&\frac{\sin{k}\cos(k+1)+\sin 1}{\cos(k)\cos (k+1)\sin 1}\\ &=&\frac{\cos{k}\sin(k+1)}{\cos(k)\cos(k+1)\sin 1}\\ &=&\frac{\tan(k+1)}{\sin 1}\end{eqnarray*}
Note the use of $ \sin 1 =\sin (k+1-k) =\cos{k}\sin(k+1)-\sin{k}\cos(k+1)$ from line 2 to line 3.
Thus $ \sum_{n = 0}^{88}\frac{1}{{\cos(n)}{\cos(n+1)}}=\frac{\tan{89}}{\sin1}=\frac{\cos{1}}{\sin^{2}{1}}$
VANCHANH123
20.08.2007 07:01
MithsApprentice wrote: Prove \[ \frac{1}{\cos 0\dg\cos 1\dg}+\frac{1}{\cos 1\dg\cos 2\dg}+\cdots+\frac{1}{\cos 88\dg\cos 89\dg}=\frac{\cos 1\dg}{\sin^{2}1\dg}.\] We use : $ \frac{sin1}{cosn.cos(n+1)}=\frac{sin[(n+1)-n]}{cosn.cos(n+1)}=cot(n)-cot(n+1)$ there for: $ \frac{1}{\cos 0\dg\cos 1\dg}+\frac{1}{\cos 1\dg\cos 2\dg}+\cdots+\frac{1}{\cos 88\dg\cos 89\dg}=\frac{1}{sin1}(cot1)=\frac{cos1}{sin^{2}1}$
Boy Soprano II
19.01.2008 18:01
Let $ M_k$ denote the point $ (1, \tan k^\circ)$. [asy][asy]size(200); defaultpen(1); pair O=(0,0), a=expi(0), b=expi(1/6), c=expi(2/6), d=expi(3/6), y=expi(32/30), z= expi(34/30); pair A=a, B=b/b.x, C= c/c.x, D=d/d.x, Y=y/y.x, Z=z/z.x, E=(D+Y)/2; pair X=(O+E)/2; draw(O--A--Z); draw(B--O--C--D--O--Y--Z--O); label("\(O\)",O,SW); label("\(M_0\)",A,ESE); label("\(M_1\)",B,ESE); label("\(M_2\)",C,ESE); label("\(M_3\)",D,ESE); label("\(\vdots\)",E,ESE); label("\(M_{88}\)",Y,ESE); label("\(M_{89}\)",Z,ESE); label("\(\ddots\)",X);[/asy][/asy] Then : \begin{align*} OM_k &= 1/\cos k^\circ \\ \text{m} \angle M_a O M_b &= (b-a)^\circ \\ [M_a O M_b ] &= \tfrac{1}{2} \sin (b-a)^\circ \cdot OM_a \cdot OM_b = \frac{\tfrac{1}{2} \sin (b-a)^\circ}{\cos a^\circ \cos b^\circ} , \end{align*} so \begin{align*} \sum_{k=0}^{88} \frac{\tfrac{1}{2} \sin 1^{\circ}}{\cos k^\circ \cos (k+1)^\circ} &= \sum_{k=0}^{88} [M_k O M_{k+1} ] \\ &= [M_0 O M_{89} ] = \frac{\tfrac{1}{2} \sin 89^\circ}{ \cos 0^\circ \cos 89^\circ} = \frac{\tfrac{1}{2} \cos 1^\circ}{\sin 1^\circ}, \end{align*} which proves the desired identity.
Math4tots
11.10.2008 09:10
This might be a newbie type question, but how did calc rulz know to induct on such an outlandish expression? What was the motivation?
Binomial-theorem
15.01.2013 02:12
Note that multiplying both sides by $\sin(1^\circ)$ gives us that we desire:
$\frac{\sin(1^\circ)}{\cos(0^\circ)\cos(1^\circ)}+\frac{\sin(1^\circ)}{\cos(1^\circ)\cos(2^\circ)}+\cdots \frac{\sin(1^\circ)}{\cos(88^\circ)\cos(89^\circ)}=\frac{\cos(1^\circ)}{\sin(1^\circ)}$.
Writing all the $\sin(1^\circ)$ as $\sin(n-(n-1))^\circ)$, we get $\sin(n)\cos(n-1)-\sin(n-1)\cos(n)$ which when divided by $\cos(n)*\cos(n-1)$ gives us $\tan(n)-\tan(n-1)$.
Therefore, the sum telescopes as follows: $\sum_{i=1}^{89}(\frac{\sin(1^\circ)}{\cos(i^\circ)\cos(i+1^\circ)}=\sum_{i=1}^{89}(\tan(n)-\tan(n-1))=\tan(89^\circ)-\tan(0^\circ)$ which is $\tan(89^\circ)$. Then use $\tan(89^\circ)=\cot(1^\circ)=\frac{\cos(1^\circ)}{\sin(1^\circ)}$ and we are done.
mathmax12
06.12.2023 08:25
seriously? 0 mohs Multiply both sides by $\sin(1^{\circ})$, then we have $\frac{\sin(1-0)}{\cos(0)\cos(1)}+\frac{\sin(2-1)}{\cos(1)\cos(2)}+....+\frac{\sin(89-88}{\cos(88)\cos(89)},$ since $\tan(a)-\tan(b)=\frac{\sin(a-b)}{\cos(a)\cos(b)},$ we can telescope this to get $\tan(89)-\tan(0)=\cot(1),$ as desired $\blacksquare$ (everyhting in degrees)