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Prove the following trigonometric identities using the Euler formula ei·φ = cos(φ)+i·sin(φ)
[See lecture on UNIZOR.COM - Math 4 Teens - Trigonometry - Complex Numbers and Trigonometry - Euler's Formula]
Find the sums Ssin(x,y,n)= Σk∈[0,n]sin(x+k·y) Scos(x,y,n)= Σk∈[0,n]cos(x+k·y)
as algebraic expression with arguments x, y and n.
Solution A1
Let's start with a sum of sines.
Recall the formula for a cosine of a sum of two angles cos(a+b) =
= cos(a)·cos(b) − sin(a)·sin(b)
From this formula it's easy to derive another trigonometric identity formula 2·sin(a)·sin(b) =
= cos(a−b)−cos(a+b)
Let's use this formula to transform our series into only a couple of members.
Multiply Ssin (that is, each member of this series) by 2·sin(y/2) and use the trigonometric identity above.
Then after this multiplication by 2·sin(y/2) the kth member of Ssin would look like 2·sin(x+k·y)·sin(y/2) =
= cos(x+k·y−y/2) −
− cos(x+k·y+y/2) =
= cos(x+(2k−1)·y/2) −
− cos(x+(2k+1)·y/2)
Thus, each member of our series is converted into a pair of cosines with a minus in-between: 2·Ssin(x,y,n)·sin(y/2) =
= Σk∈[0,n][cos(x+(2k−1)·y/2) −
− cos(x+(2k+1)·y/2)] =
= [cos(x−y/2)−cos(x+y/2)] +
+ [cos(x+y/2)−cos(x+3y/2)] +
+ [cos(x+3y/2)−cos(x+5y/2)] + etc. and the last two members of a series are
+ [cos(x+(2n−3)/2) −
− cos(x+(2n−1)y/2)] +
+ [cos(x+(2n−1)/2) −
− cos(x+(2n+1)y/2)]
It's easy to see that in this series every members, except the first and the last, cancel each other and the result is 2·Ssin(x,y,n)·sin(y/2) =
= cos(x−y/2) −
− cos(x+(2n+1)·y/2)
The sum of cosines can be treated analogously.
Recall the formula for a sine of a sum of two angles sin(a+b) =
= sin(a)·cos(b) + cos(a)·sin(b)
From this formula it's easy to derive another trigonometric identity formula 2·cos(a)·sin(b) =
= sin(a+b)−sin(a−b)
Let's use this formula to transform our series into only a couple of members.
First, we multiply all members of Scos (that is, each member of this series) by 2·sin(y/2). and use the trigonometric identity above. 2·Scos(x,y,n)·sin(y/2) =
= Σk∈[0,n][sin(x+(2k+1)·y/2) −
− sin(x+(2k−1)·y/2)] =
= sin(x+(2n+1)·y/2) −
− sin(x−y/2)
Of cause, both formulas for Ssin and Scos are correct only if sin(y/2)≠0, that is y≠2πN, where N - any integer number.
If, however, y=2πN, then sin(x+k·y)=sin(x) and cos(x+k·y)=cos(x), so Ssin = (n+1)·sin(x) Scos = (n+1)·cos(x)
Solution A2
A different and, arguably, more elegant approach to solve this problem would be use the Euler formula ei·φ = cos(φ)+i·sin(φ)
[See lecture on UNIZOR.COM - Math 4 Teens - Trigonometry - Complex Numbers and Trigonometry - Euler's Formula]
Using it, we can express cos(x+k·y) and sin(x+k·y) as real and imaginary parts of ei·(x+k·y) = ei·x·(ei·y)k
The sum of these expression by k from 0 to n can be calculated as a geometric series with the first term a=ei·x and factor r=ei·y.
[See lecture on UNIZOR.COM - Math 4 Teens - Algebra - Sequence and Series - Progression Series]
This sum in terms of a and r is S = a·(rn+1−1)/(r−1)
Therefore, our sum is equal to
Σk∈[0,n]ei·x·(ei·y)k =
= ei·x·[(ei·y)n+1−1] /(ei·y−1) =
= [ei·(x+y·(n+1))−ei·x] /(ei·y−1)
What remains is to convert this expression into canonical representation of complex numbers as A+B·i. Then A will represent the sum of cosines, and B will represent a sum of sines.
First, get rid of i in the denominator by multiplying both numerator and denominator by (e−i·y−1).
In the denominator the result will be (ei·y−1)·(e−i·y−1) =
= e0−e−i·y−ei·y+1 =
= 2 − ei·y − e−i·y =
= 2 − cos(y) − i·sin(y) −
− cos(−y) − i·sin(−y) =
= 2 −2cos(y) = 4·sin²(y/2)
As we see, there is only real part in the denominator, an imaginary terms cancel each other.
In the numerator we will have
[ei·(x+y·(n+1))−ei·x]·(e−i·y−1) =
= ei·(x+y·n) − ei·(x+y·(n+1)) −
− ei·(x−y) + ei·x =
= cos(x+y·n)−cos(x+y·(n+1))−
−cos(x−y) + cos(x) +
+ i·[sin(x+y·n)−sin(x+y(n+1))−
−sin(x−y)+sin(x)]
Let's deal with real part that gives the sum of cosines.
We will use another trigonometric identity which can be easily proved using α=½(α+β)+½(α−β) and β=½(α+β)−½(α−β): cos(α) − cos(β) =
= −2sin(½(α+β))·sin(½(α−β))
Using this identity, the numerator for a sum of cosines is cos(x+y·n)−cos(x+y·(n+1))−
−cos(x−y) + cos(x) =
= 2·sin(x+½(2n+1)·y)·sin(½y) −
− 2·sin(x−½y)·sin(½y) =
= 2·sin(½y)·
·[sin(x+½(2n+1)·y) −
− sin(x−½y)]
Combining this numerator and calculated above denominator 4·sin²(y/2), we obtain Scos(x,y,n) =
= [sin(x+(2n+1)·y/2) −
− sin(x−y/2)]·
·2·sin(y/2) / [4·sin²(y/2)] =
= [sin(x+(2n+1)·y/2) −
− sin(x−y/2)] /
/ [2·sin(y/2)]
which corresponds completely to Solution B1.
Similarly, the imaginary part of a numerator can be used to calculate the sum of sines.
We will use yet another trigonometric identity, which can be derived analogously to a difference of cosines above: sin(α) − sin(β) =
= 2sin(½(α−β))·cos(½(α+β))
Using this identity, the numerator for a sum of sines is sin(x+y·n)−sin(x+y(n+1))−
−sin(x−y)+sin(x) =
=−2·sin(½y)·cos(x+½(2n+1)·y)+
+ 2·sin(½y)·cos(x−½y) =
= 2·sin(½y)·
·[cos(x−½y) −
− cos(x+½(2n+1)·y)]
Combining this numerator and calculated above denominator 4·sin²(y/2), we obtain Ssin(x,y,n) =
= [cos(x−y/2) −
− cos(x+(2n+1)·y/2)]·
·2·sin(y/2) / [4·sin²(y/2)] =
= [cos(x−y/2) −
− cos(x+(2n+1)·y/2)] /
/ [2·sin(y/2)]
which corresponds completely to Solution B1.
A traveler has to stay in some city for 7 days.
A restaurant owner near a hotel where a traveler stays does not accept credit cards, and a traveler has no local currency to pay for food.
However, a traveler has a silver chain of 7 links, and a restaurant owner agreed to take one chain link for each dinner.
What is the minimum number of chain links that must be cut open to be able to pay for each dinner separately?
Answer A
Only one link needs to be cut open, link #3.
Problem B
Same as Problem A, but the number of days is N, and the number of links in a chain is the same N.
Answer B
Assume, a traveler cuts open only K links (K must be small relatively to N).
Then he can pay for 1, 2, 3,..., K dinners using only these links.
For the next dinner he needs a segment of chain that contains K+1 links.
With this additional segment of chain he can pay for K+1, K+2,..., 2K+1 dinners.
For the next dinner he needs a segment of 2K+2 links.
Using K individual links that were cut, a segment of a chain with K+1 links and another segment with 2K+2 links a traveler pay for any number of dinners from 1 to K+(K+1)+(2K+2)=4K+3.
For the next dinner he needs a segment of 4K+4 links.
Using K individual links that were cut, a segment of a chain with K+1 links, a segment with 2K+2 links and a segment with 4K+4 links a traveler pay for any number of dinners from 1 to K+(K+1)+(2K+2)+(4K+4)=8K+7.
For the next dinner he needs a segment of 8K+8 links.
Using K individual links that were cut, a segment of a chain with K+1 links, a segment with 2K+2 links, a segment with 4K+4 links and a segment with 8K+8 links a traveler pay for any number of dinners from 1 to K+(K+1)+(2K+2)+(4K+4)+(8K+8)=16K+15.
Let's generalize it.
When K links were cut, it separated a chain into K+1 segments as follows.
Segment #1 should have K+1 links in it.
Segment #2 should have 2K+2=2·(K+1) links in it.
Segment #3 should have 4K+4=4·(K+1) links in it.
Segment #4 should have 8K+8=8·(K+1) links in it.
Segment #5 should have 16K+16=16·(K+1) links in it.
etc.
The last segment #(K+1) should have 2K·(K+1) links in it.
All individual links that were cut splitting a chain into segments and all segments listed above allow to pay for the following number of dinners. K +
+ (K+1) +
+ 2·(K+1) +
+ 4·(K+1)+
+...+
+ 2K·(K+1) =
= K + (2K+1 −1)·(K+1) =
= (K+1)·2K+1 − 1
If N=(K+1)·2K+1−1 then the minimum number of links to cut is K.
If N is not such a number, we have to find K such that (K+1)·2K+1−1 is greater than N. That K is the minimum number of links to cut.
Problem A
Consider two numbers: X=11111111
(digit 1 repeats 8 times) and Y=11111...11111
(digit 1 repeats 100 times).
What is their greatest common factor?
Hint A
Both numbers can be represented as a sum 100+101+...+10k
where k=7 for number X and k=99 for number Y.
From this it's easy to see that Y = X·n + 1111·1096
where n is some natural number.
Answer A
The greatest common factor of numbers X and Y is 1111.
Problem B
Prove that a remainder of the division of some natural number by 9 is the same as a remainder of the division by 9 of the sum of this number's digits.
Example B
Take, for instance, number 2024 and divide it by 9. 2024÷9 = 224 (8)
(remainder is 8)
Now with the sum this number's digits: (2+0+2+4)÷9 = 8÷9 = 0(8)
(the same remainder 8)
The rule of divisibility by 9
Consequently, if the sum of digits of a natural number is divisible by 9, then the number itself is divisible by 9.
Hint B
Represent a number as a sum of its digits, each multiplied by a factor 10n, where n=0 for the right most digit, n=1 for the next digit to the left etc.
Problem C
Prove that if the sum of digits of some natural number N is the same as the sum of digits of the number 2·N then number N is divisible by 9.
Example C
For number 279 the sum of digits is 2+7+9=18.
Double this number: 2·279 = 558
Sum of digits of 558 is also 18.
The number 279 is indeed divisible by 9: 279÷9 = 31 (0)
(remainder is 0)
Hint C
Use the result of the Problem B above and consider the difference between 2·N and N.
Proof C N÷9 = n (r)
(here n is a quotient and r is a remainder with the value from 0 to 8)
Therefore,
N = 9·n + r
Analogously, (2·N)÷9 = m (r)
(same remainder r) 2·N = 9·m + r
Subtract formula for N from formula for 2·N 2·N − N = N =
= (9·m + r) − (9·n + r) =
= 9·(m−n)
Hence N = 9·(m−n)
that is, N is divisible by 9.
Note
The same result (divisibility by 9) can be obtained under weaker conditions.
The theorem can be stated as follows.
Prove that if the sum of digits of some natural number N divided by 9 gives the same remainder as the sum of digits of the number 2·N divided by 9, then number N is divisible by 9.
Obviously, that remainder in both cases will be zero. No other equal remainder can be encountered for sums of digits of N and 2·N divided by 9.
Example CC
For number 144 the sum of digits is 1+4+4=9.
Double this number: 2·144 = 288
Sum of digits of 288 is 18.
Sums of digits are different (though, both are divisible by 9), but the number 144 is indeed divisible by 9: 144÷9 = 16 (0)
(remainder is 0)
You can easily prove it or watch the proof in the lecture.
There are two cities:
CityOfTruth, where all people always tell the truth and
CityOfLies, where all people always lie.
A traveler intends to go to CityOfTruth. He comes to a fork, one road leading to CityOfTruth, another - to CityOfLies and he has to choose which way to go.
Right there he meets a person who, apparently, lives in one of these cities. So a traveler can ask him the direction to CityOfTruth.
The problem is, the person can live in either of these cities and nobody knows whether he tells the truth or lies.
Can a traveler ask this person a question in such a way that there will be no doubts about which way leads to CityOfTruth?
Answer A
The question might be: "Could you show me which way leads to a city where you live?"
If this person lives in CityOfTruth and always tells the truth, he will point to CityOfTruth.
If this person lives in CityOfLies and always lies, he will point to the same CityOfTruth, because pointing to CityOfLies would be the truth, which he never tells..
In any case, he will correctly point to a direction the travel wants to know.
Problem B
A man (M), a wolf (W), a goat (G) and a cabbage (C) have to cross a river from side A to side B.
There is a boat that can hold a man and either one of the others.
The problem is, in the absence of a man a wolf might eat a goat, a goat might eat a cabbage. But in the man's presence nobody eats anything. A wolf does not eat cabbage under any circumstances.
How can they all safely cross the river from side A to side B, so everyone is alive and well at the end?
A person has a revolver designed for 6 bullets, but there is only one loaded and the chamber is spun.
He has two attempts to hit a target.
He shoots once, but no bullet comes out.
He wants to maximize his chances of making a real shot to hit a target.
What would be a better choice for him before the second shot, to spin or not to spin a chamber?
Answer C
Not to spin gives a one in five chances to shoot a bullet.
Spinning gives a one in six chances.
So, not to spin gives a better chance to make a real shot and, hopefully, hit a target.
Problem D
Two players, A and B, are playing against each other.
Each game results in one person winning and another losing.
The loser pays $1 to a winner.
Initially, both players came with $100 each.
At end it appears that player A won 10 games and player B ended with $120.
How many games did they play?
Answer D
They played 40 games.
Problem E
Three wise men (very smart indeed) after discussing some very important subject fell asleep.
Some foolish child was passing along and decided, as a joke, to put some black shoe wax on their foreheads.
All three wise men woke up at the same time and each of them, seeing black spots on two others' foreheads, started to laugh.
But after a short time one of them (a bit wiser than others) stopped laughing, realizing that his own forehead also has a black spot.
What was his logic?
Answer E
Let's call the wise men WiseA (that's the one who stopped laughing first), WiseB and WiseC.
WiseA thinks as follows.
Assume, my forehead is clean.
Then WiseB and WiseC can see only spots on the foreheads of each other.
Considering WiseB is very smart, seeing that WiseC is laughing and seeing no black spot on my forehead, he would immediately realize that WiseC is laughing at him, because he also has a black spot on his forehead. Then WiseB would stop laughing.
Since WiseB still laughs, he must see a black spot on my forehead too. So, I better stop laughing and go to wash my face.
When we switch from n to n+1, function F(n) has changed by f(n+1) = 1/(2n+1) − 1/(2n+2)
So, F(n+1)=F(n)+f(n+1).
Analogously, function G(n) has changed by g(n+1) = −1/(n+1) +
+ 1/(2n+1) + 1/(2n+2)
So, G(n+1)=G(n)+g(n+1).
If f(n+1), an increment of F(n), and g(n+1), an increment of G(n), are equal, assuming F(n)=G(n), we can conclude that F(n+1)=G(n+1), which proves that F(n)=G(n) for any natural number n.
Let's check the equality f(n+1) = g(n+1)
Indeed, transforming an expression to a common denominator, we get f(n+1) = 1/[(2n+1)·(2n+2)]
For g(n+1) the common denominator is (n+1)(2n+1)(2n+2)
The numerator is −(2n+1)(2n+2)+(n+1)(2n+2)+
+(n+1)(2n+1) =
= −4n² − 2n − 4n − 2 +
+ 2n² + 2n + 2n + 2 +
+ 2n² + n + 2n + 1 =
= n+1
Dividing this numerator of g(n+1) by its denominator obtained above and cancelling n+1 from both, we will get g(n+1) = 1/[(2n+1)(2n+2)]
This is exactly what we obtained for f(n+1).
END OF PROOF
Problem B
Two cars are approaching an intersection on two perpendicular roads.
In the beginning at time t=0 car #1 is at distance a from the intersection.
The car #2 (on a perpendicular road) at t=0 is at distance b from the intersection.
Both cars move with the same speed V.
For traffic safety, assume that a≠b, so there will be no collision.
At what time the distance between the cars will be minimal and what that minimal distance will be?
Answer B
Let d(t) be a distance between the cars at any moment t.
Then d²(t) = (a−v·t)² + (b−v·t)²
Minimum of d(t) and d²(t) occur for the same moment in time t.
So, our task is to find minimum of d²(t), which is a quadratic function of t.
The time when the cars are closest to each other is tmin = (a+b)/(2V)
The closest distance between the car at that time is dmin = |a−b| /√2
Construct a quadrilateral ABCD by its 4 sides AB, BC, CD, DA and an angle φ between opposite sides AB and CD.
Hint A
Find point P such that BP is parallel and congruent to CD.
Consider ΔABP.
Problem B
Given a circle of radius R and n-sided regular polygon inscribed into it.
Let P be any point on this circle.
Find a sum of squares of distances from this point P to all vertices of a polygon.
Hint B
(a) Geometrical solution for even number n of vertices of a regular polygon can be obtained by adding pairs of distances from P to ith and (i+½n)th vertices.
(b) General solution can be obtained if using vectors from the center of a circle to all its vertices and to point P.
Answer
Sum of squares of distances from point P to all vertices of a polygon equals to 2nR².
Problem C
Given an equilateral triangle ΔABC.
Extend side AC beyond vertex C to point D and build another equilateral triangle CDE with point E on the same side from AD as point B.
Connect points A and E. Let point M be a midpoint of segment AE.
Connect points B and D. Let point N be a midpoint of segment BD.
Prove that triangle ΔCMN is equilateral.
