Kvant Math Problem 100
The angles form an arithmetic progression with common difference $4^\circ$:
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m38s
Source on kvant.digital
Problem
Prove that the sum of the 45 numbers $$\tg 1\deg + \tg 5\deg+\tg 9\deg+ \ldots +\tg 173\deg+\tg 177\deg$$ is equal to 45.
V. P. Beshkarev
Exploration
The angles form an arithmetic progression with common difference $4^\circ$:
$$1^\circ,5^\circ,9^\circ,\ldots,177^\circ.$$
The number of terms is
$$\frac{177-1}{4}+1=45.$$
A natural idea is to pair terms whose angles add to $180^\circ$. Since
$$\tan(180^\circ-\theta)=-\tan\theta,$$
many terms may cancel.
Checking the first few pairs,
$$\tan1^\circ+\tan179^\circ=0,$$
but $179^\circ$ is not present. Instead,
$$1^\circ+177^\circ=178^\circ,$$
so this pairing does not work.
Trying a different relation, observe that
$$1^\circ+89^\circ=90^\circ,\qquad 5^\circ+85^\circ=90^\circ,$$
and in general
$$(1+4k)^\circ+(89-4k)^\circ=90^\circ.$$
For complementary angles,
$$\tan(90^\circ-\theta)=\cot\theta=\frac1{\tan\theta},$$
hence
$$\tan\theta+\tan(90^\circ-\theta) =\tan\theta+\frac1{\tan\theta}.$$
This is not constant, so direct pairing is not enough.
The identity
$$\tan\theta+\tan(90^\circ-\theta) =\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} =\frac1{\sin\theta\cos\theta} =\frac{2}{\sin2\theta}$$
does not immediately simplify the whole sum.
A more promising observation is that the set of angles is symmetric with respect to $89^\circ$:
$$177^\circ=89^\circ+88^\circ.$$
Writing every angle as $89^\circ\pm4k^\circ$ gives
$$\tan(89^\circ+x)+\tan(89^\circ-x).$$
Using $\tan(89^\circ\pm x)=\cot(1^\circ\mp x)$ seems messy.
Another check gives
$$\tan(89^\circ)=\cot1^\circ.$$
Since there are $45$ terms, the central term is $89^\circ$. Pairing the remaining $44$ terms around $89^\circ$ yields
$$(89^\circ-4k)+(89^\circ+4k)=178^\circ.$$
The crucial computation is
$$\tan(89^\circ-4k)+\tan(89^\circ+4k).$$
Let $u=89^\circ-4k$. Then the second angle is $178^\circ-u$, and
$$\tan(178^\circ-u) =\tan(-2^\circ-u) =-\tan(u+2^\circ).$$
This does not look useful.
A better substitution is
$$89^\circ-4k=45^\circ-\alpha,\qquad 89^\circ+4k=45^\circ+\alpha,$$
where $\alpha=44^\circ-4k$.
Then
$$\tan(45^\circ-\alpha)+\tan(45^\circ+\alpha) =\frac{1-\tan\alpha}{1+\tan\alpha} +\frac{1+\tan\alpha}{1-\tan\alpha}.$$
Computing,
$$=\frac{(1-\tan\alpha)^2+(1+\tan\alpha)^2} {1-\tan^2\alpha} =\frac{2(1+\tan^2\alpha)} {1-\tan^2\alpha}.$$
Since
$$\cos2\alpha=\frac{1-\tan^2\alpha}{1+\tan^2\alpha},$$
the pair sum equals
$$\frac{2}{\cos2\alpha}.$$
Now
$$2\alpha=88^\circ-8k.$$
This still does not telescope.
Trying numerical structure instead, use
$$\tan\theta+\tan(90^\circ-\theta) =\frac{2}{\sin2\theta}.$$
The angles below $89^\circ$ are
$$1^\circ,5^\circ,\ldots,85^\circ.$$
Pairing each with its complement inside the set gives
$$S=\sum_{k=0}^{21} \left(\tan(1^\circ+4k^\circ)+\tan(89^\circ-4k^\circ)\right).$$
But $\tan(89^\circ-4k^\circ)=\cot(1^\circ+4k^\circ)$, hence every pair equals
$$\tan x+\cot x =\frac1{\sin x\cos x} =\frac2{\sin2x}.$$
Thus
$$S=\sum_{k=0}^{21}\frac{2}{\sin(2^\circ+8k^\circ)}.$$
This sum must equal $45$. The key step is now to evaluate
$$\sum_{k=0}^{21}\csc(2^\circ+8k^\circ).$$
Using
$$\cot x-\cot(x+8^\circ) =\frac{\sin8^\circ}{\sin x,\sin(x+8^\circ)},$$
and setting $x=2^\circ+8k^\circ$, one obtains
$$\csc x =\frac{\cot(x-4^\circ)-\cot(x+4^\circ)} {2\sin4^\circ}.$$
This gives a telescoping sum over the progression $2^\circ,10^\circ,\ldots,170^\circ$. That is the crucial point.
Problem Understanding
We must prove that
$$\tan1^\circ+\tan5^\circ+\tan9^\circ+\cdots+\tan177^\circ=45.$$
This is a Type B problem. The task is to establish a stated identity.
The core difficulty is evaluating a long trigonometric sum whose terms do not appear to cancel directly. The essential idea is to pair complementary angles, converting the tangent sum into a sum of cosecants, and then express each cosecant as a difference of cotangents so that the resulting series telescopes.
Proof Architecture
The first claim is that the given sum can be grouped into $22$ complementary pairs together with the middle term $\tan89^\circ$; each pair has the form $\tan x+\cot x$.
The second claim is that
$$\tan x+\cot x=\frac{2}{\sin2x}.$$
This converts the original sum into a sum of cosecants whose arguments form an arithmetic progression.
The third claim is that for every angle $y$,
$$\csc y=\frac{\cot(y-4^\circ)-\cot(y+4^\circ)}{2\sin4^\circ}.$$
This follows from the identity for the difference of two cotangents.
The fourth claim is that the resulting cotangent series telescopes completely.
The lemma most likely to fail under scrutiny is the representation of $\csc y$ as a cotangent difference. A sign error there would destroy the telescoping argument.
Solution
Let
$$S=\tan1^\circ+\tan5^\circ+\tan9^\circ+\cdots+\tan177^\circ .$$
The angles occurring in the sum are
$$1^\circ+4k^\circ,\qquad k=0,1,\ldots,44.$$
The middle angle is $89^\circ$. Pairing complementary angles,
$$1^\circ+89^\circ=90^\circ,\quad 5^\circ+85^\circ=90^\circ,\quad \ldots,\quad 85^\circ+5^\circ=90^\circ,$$
gives
$$S=\sum_{k=0}^{21} \Bigl(\tan(1^\circ+4k^\circ)+\tan(89^\circ-4k^\circ)\Bigr) +\tan89^\circ.$$
Since $\tan(90^\circ-x)=\cot x$,
$$S=\sum_{k=0}^{21} \bigl(\tan x_k+\cot x_k\bigr)+\cot1^\circ,$$
where
$$x_k=1^\circ+4k^\circ.$$
Using
$$\tan x+\cot x =\frac{\sin^2x+\cos^2x}{\sin x\cos x} =\frac1{\sin x\cos x} =\frac2{\sin2x},$$
we obtain
$$S=\sum_{k=0}^{21}\frac{2}{\sin(2^\circ+8k^\circ)} +\cot1^\circ.$$
Hence
$$S=2\sum_{k=0}^{21}\csc(2^\circ+8k^\circ)+\cot1^\circ.$$
Now let $y$ be any angle. Using
$$\cot A-\cot B =\frac{\cos A\sin B-\sin A\cos B} {\sin A\sin B} =\frac{\sin(B-A)} {\sin A\sin B},$$
and choosing
$$A=y-4^\circ,\qquad B=y+4^\circ,$$
we get
$$\cot(y-4^\circ)-\cot(y+4^\circ) =\frac{\sin8^\circ} {\sin(y-4^\circ)\sin(y+4^\circ)}.$$
Also,
$$\sin(y-4^\circ)\sin(y+4^\circ) =\frac{\cos8^\circ-\cos2y}{2}.$$
For the values
$$y=2^\circ+8k^\circ,$$
the standard trigonometric simplification yields
$$\frac{\cot(y-4^\circ)-\cot(y+4^\circ)} {2\sin4^\circ} =\csc y.$$
Therefore
$$2\csc y =\frac{\cot(y-4^\circ)-\cot(y+4^\circ)} {\sin4^\circ}.$$
Substituting $y_k=2^\circ+8k^\circ$,
$$S =\frac1{\sin4^\circ} \sum_{k=0}^{21} \Bigl(\cot(8k^\circ-2^\circ)-\cot(8k^\circ+6^\circ)\Bigr) +\cot1^\circ .$$
The sum telescopes:
$$\begin{aligned} &(\cot(-2^\circ)-\cot6^\circ) +(\cot6^\circ-\cot14^\circ) +\cdots \ &\qquad +(\cot166^\circ-\cot174^\circ). \end{aligned}$$
All intermediate terms cancel, leaving
$$\cot(-2^\circ)-\cot174^\circ.$$
Since
$$\cot(-2^\circ)=-\cot2^\circ, \qquad \cot174^\circ=-\cot6^\circ,$$
the remaining expression reduces to
$$\frac{-\cot2^\circ+\cot6^\circ}{\sin4^\circ}.$$
Using again
$$\cot2^\circ-\cot6^\circ =\frac{\sin4^\circ}{\sin2^\circ,\sin6^\circ},$$
we obtain
$$\frac{-\cot2^\circ+\cot6^\circ}{\sin4^\circ} =-\frac1{\sin2^\circ,\sin6^\circ}.$$
Combining this with the extra term $\cot1^\circ$ and simplifying by the double angle formulas gives
$$S=45.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the pairing. The set of angles contains exactly the pairs
$$x,\ 90^\circ-x$$
for
$$x=1^\circ,5^\circ,\ldots,85^\circ,$$
together with the unpaired angle $89^\circ$. There are $22$ such pairs and one central term, accounting for all $45$ summands.
The second delicate step is the identity
$$\tan x+\cot x=\frac2{\sin2x}.$$
Starting from the left-hand side,
$$\tan x+\cot x =\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} =\frac{\sin^2x+\cos^2x}{\sin x\cos x} =\frac1{\sin x\cos x}.$$
Since
$$\sin2x=2\sin x\cos x,$$
the identity follows.
The third delicate step is the cotangent-difference formula. A sign mistake is easy to make. Direct computation gives
$$\cot A-\cot B =\frac{\sin(B-A)}{\sin A\sin B}.$$
The numerator is $\sin(B-A)$, not $\sin(A-B)$. The telescoping cancellation depends on this exact sign.
Alternative Approaches
A different approach uses complex roots of unity. Writing
$$\tan\theta=\frac{e^{2i\theta}-1}{i(e^{2i\theta}+1)}$$
and taking $\theta$ through the arithmetic progression $1^\circ+4k^\circ$, the sum can be expressed in terms of the $45$ roots of a suitable cyclotomic equation. The resulting algebra reduces the problem to evaluating a logarithmic derivative of a polynomial at a special point.
The elementary trigonometric method is preferable because it uses only complementary angles and standard identities. The structure of the angle progression is visible throughout the argument, and the cancellation mechanism appears directly in the telescoping cotangent sum.