Proof by Bygone Experience

7/28/2019 Proof by Bygone Experience

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PROOF BY BYGONE EXPERIENCE

(MY RECENT WRITING)

Mr. SANONG HAUYRERAI

MATHEMATICS TEACHER

RATTANAKOSINSOMPHOT BOWONNIWET

SALAYA SCHOOL

FROM JUNE TO JULY, 2012

EDDUCATION YEAR, 2012


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Proof by creating according to bygone experience

That bygone experience has to be excellent experience that be arisen

By our continuously distinguished practice, that is important we ought to have

The distinct goal as follows. we will do what only we like and we will like

What only we do we have to accept that every-one have same time but

The excellent experience make we have the time greater than the others.

1.1 Several example

Example 1 Prove that< + + +...+ 1

Proof Let P(n):< + + +...+ 1

And

n):

1

Since < + = so (2) is to be true.And since

< + We will get

+ + +...+ < + +...+ That is (k+1) is to be true for all positive integer more one.By mathematical induction we conclude that (n) is to be trueAnd let

(n):

+

+

+...+

1

It can be proved by mathematical induction and be left the proof as an exercise.

Example 2 Prove that there is not an integer lying between n and n+1 for all n Proof Assume that there exists x such that n


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That is


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Example 8 Prove that 1) k+(k+1)+(k+2)+(k+3)+...+n= = 2) 2k+2(k+1)+2(k+2)+...+2n= =(n-k+1)(n+k)3) 2k-1+2(k+1)-1+2(k+2)-1+...+2n-1=

-1)=(n-k+1)(n+k-1)

Proof 1) k+(k+1)+(k+2)+(k+3)+...+n= k+(k+1)+(k+2)+(k+3)+...+k+(n-k)

= [k+k+k+k+...+k (n-k)+1 term] + [1+2+3+...+(n-k)]

= k[(n-k)+1]+

= [(n-k)+1][k+ ]

= (n-k+1)

=

=

Point 2 and 3 are left the proof as an exercise.

Example 9 Prove that 1) (x, x+4)=1, 2, 4 and 2) (x, x+6)=1, 2, 3, 6; x Proof 1) Let x is integer, where x=4n, 4n+1, 4n+2, 4n+3;n

1.1) Let (x, x+4)=(4n+1, 4n+5)=d

We will get d/(4n+5-4n-1) and then d/4 (1, 2, 4)

But d/4n+1, so d=1

1.2) Let (x, x+4)=(4n+2, 4n+6)=d

We will get d/(4n+6-4n-2) and then d/4 (1, 2, 4)

But d/4n+2, so d=2

1.3) Let (x, x+4)=(4n, 4n+4)=d

We will get d/(4n+4-4n) and then d/4

But d/4n, so d=4

Point 2 is left the proof as an exercise.

Example 10 Define

=

,

=

and

=

, Prove that

1) (, )=n 2) (, )=n+1 3) (, )=1, 4, 123) (, )=1, 4, 12 4) (, )=nProof All points are left as an exercise.

Example 11 Let a, b, c, d are sequence of positive integer, where a is odd number

And more than or equal to three


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Prove that [a, c]>[b, d]

Proof The proof is to be an exercise.

Example 12 Let sequences 1+1!, 2+2!, 3+3!,...,n+n!

Find all n

that make n+n! is perfect square.

Show showing is left an exercise.

Example 13 Let N=1.1!+2.2!+3.3!+...+n.n! Prove that N+1=(n+1)!

Proof Hint: (n+1)!=n!+n!+n!+...+n!...n-term +n! The proof is to be an exercise.

Or by using mathematical Induction

1.2 THE SUM AND THE PRODUCT

As positive real number have closure property for addition and multiplication

So, we can find the sum and the product when defined two all positive real number

But what we require knowing is when the sum is lesser than the product, when the sum is

Equal to the product and when the sum is more than the product. Please look at the

Following property.

Theorem 1 a+b=aba= , a,b Proof The proof is an exercise.

Theorem 2 a+b=ab

a=b=2, a, b are natural number.

Proof according to theorem 1 b=2So, a=b=2Theorem 3 If a


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Proof Its proof is an exercise.

Lemma 1 If a3 and b3 then 2ab3a+3b, a,b (the proof is an exercise.)Theorem 7 If a2 then ab(2a, 2b) and a+b(a+2,b+2), a,b Proof Its proof is an exercise.

1.3 TRIANGULAR NUMBER

Definition The positive integer n is called that triangular number if and only if n can be

Written in pattern n=1+2+3+4+5+...+k, k According to definition we let : nth number of triangular numberSo, =1+2+3+4+5+...n=

The following expressions are to be true that have to prove which arise from definition

Above.

1) If n and m are odd number in order that is continuous then -=4k+1, k 2) If n and m are even number in order that is continuous then -=4k+3, k 3) When n is a triangular number then 9n+1 is the triangular number.

4) When E is odd number then the quotient of-1 by 8 is a triangular number.5) The difference of square of triangular number in order continuous is perfect cube.

6) The sum of reciprocal of triangular number is equal to 2.

7) When m and n are a triangular number will get that 3m+n is triangular number.

8) When m and n are a triangular number will get that m+3n is triangular number.

9) When m and n are a triangular number will get that + is triangular number.10) When m , n will get that +


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6) 2 + n is the triangular number.7) +2n is the triangular number.

Proof 1) If n and m are odd number in order that is continuous then -=4k+1, k According to designed expression we find that

-= =4k+1,k 6) The sum of reciprocal of triangular number is equal to 2.

According to designed expression we find that The sum of reciprocal of

Triangular number is =2 = 2((1-

)+(-)+(-)+...+( - )+( -))=2(1-) and

=2

=2.1=2

10) When m , n will get that +


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(b-a)-1 when a and b are natural number that is a lesser than b

So [-]-1 is the number of natural number lying between and Considering

-

]-1=

-

-1=

=nThe following we will show that is to be true all n by mathematical inductionLet P(n): -]-1=n

1) P(1) is to be true because -]-1=(3-1)-1=1 (true)2) Assume that P(k) is to be true

That is -]-1=k,k Considering P(k+1) as follows, -]-1= - -1=k+1, Hence P(k+1) is to be true.

Therefore P(n) is to be true for all n We conclude that the number of natural number lying between the triangularNumber from 1 to n+1 is sequence of natural number from 1 to n

Assignment the remaining properties are left as an exercise.

1.4 ABOUT TRIANGLES

When we say to a triangle, we have to recall to its sides and angles and connect to

bearing of sides and angles together. The following we will say to concise theorem and

An example that involve

Theorem 8 Let ABC is a triangle having and are opposite side of and Respectively. We will obtain that= AC=ABProof It is left the proof an exercise (by creating the figure of triangle)

Example 14 Let ABC is an triangle with and are opposite side of and Respectively prove that n(

) > n(

)

n(

) > n(

Proof The proof is left an exercise (by theorem above)

Example 15 Let ABC is all triangles prove that the sum length of two sides is greater than

Remaining side.

Proof The proof is left as an exercise.(by Pythagorean theorem)

1.5 INVERSE FUNCTION OF TRIGONOMETRIC FUNCTION


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In this part we will say to merely inverse function of trigonometric function and want to

Introduce the proof which be written from my bygone experience

Definition 1 Arcsine function is a set of order pairs (x, y) when x[-1,1] and y[-,]Where x= call y that arcsine of x, written by y=arcsinxThat is y=arcsinx x=

Definition 2 Arccosine function is a set of order pairs (x,y) when x[-1,1] and y[0,]Where x= call y that arccosine of x, written by y=arccosxThat is y=arccosx x=

Definition 3 Arctangent function is a set of order pairs (x,y) when x(-,) and y(-,)Where x=

call y that arctangent of x, written by y=arctanx

That is y=arctanx x=Definition 4 Arccotangent function is a set of order pairs (x,y) when x(-,) and y(0,)Where x= call y that arccotant of x, written by y=arccotxThat is y=arccotx x=

Definition 5 Arcsecant function is a set of order pairs (x,y) when x(-,1][1,) andY[0,)(,] where x= call y that arcsecant of x, written byY=arcsecx , that is y=arcsecx

x=

Definition 6 Arccosecant function is a set of order pairs (x,y) when x(-,1][1,) andY[-,0)(0,] where x=, call y that arccosecant of x, written byY=arccscx, that is y=arccscx x=

Some property of inverse function of trigonometric function

1 Arcsin : [-1,1][-,] is one to one onto increasing function.2 Arccos : [-1,1][0,] is one to one onto decreasing function.3 Arctan : (-,)(-,) is one to one onto increasing function.4 Arccot : (-,)(0,) is one to one onto decreasing function.5 Arccsc : (-,-1][1,)[-,0)(0,] is one to one onto decreasing function on sub domain.6 Arcsec : (-,-1][1,)[0,)(,] is one to one onto decreasing function on sub domain.7 Arcsin(-x) = - arcsinx , x[-1,1] 8 Arcsin(siny) = y, y[-,]


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9 Sin(arcsinx) = x, x[-1,1] 10 Arccos(-x) = - arccosx, x[-1,1]11 Arccos(cosy) = y, y[0,] 12 Cos(arccosx) = x, x[-1,1]13 Arcsinx + arccosx =

, x[-1,1] 14 Arctan(-x) = - arctanx, x(-,)15 Arctan(tany) = y, y(-,) 16 Tan(arctanx) = x,x(-,)17 Cot(arccotx) = x, x 18 Arccot(coty) = y, y(0,)19 Sec(arcsecx) = x, x(-,-1][1,) 20 Arcsec(secy) = y, y[0,)(,]21 Csc(arccscx) = x, x(-,-1][1,) 22 Arccsc(cscy) = y, y[-,0)(0,]23 If arctanx + arctany (-,) then arctanx + arctany = arctan( )24 If arctanx + arctany then arctanx + arctany = + arctan( )26 Arccot(-x) = -arccotx, x(-,)27 Arcsec(-x) = -arcsecx, x(-,-1][1,)28 Arccsc(-x) = -arccscx, x(-,-1][1,)Proof 7 Arcsin(-x) = - arcsinx , x[-1,1]

Since Arcsin(-x)= = -x, y[- ]

-

= x

= x arcsinx = -y -arcsinx = yTherefore Arcsin(-x) = - arcsinx , x[-1,1]

10 Arccos(-x) = - arccosx, x[-1,1]Since Arccos(-x)=y = -x, y[0,]

-= x = x arccosx = -y- arccosx = yTherefore Arccos(-x) = - arccosx, x[-1,1]


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23 If arctanx + arctany (-,) then arctanx + arctany = arctan( )Let arctanx + arctany = , arctanx = A and arctany = B, where

arctanx = A

= x and arctany = B

= y

Now, we have (A + B) = , (-,) and arctan( )(-,) = = ==n+) arctan( )= n+=0+=, (-,)Therefore If arctanx + arctany (-,) then arctanx + arctany = arctan( )

And remaining properties are assigned the proof as an exercise.

Theorem 1 1

=

, x

[-1,1]

2 = , x[-1,1]3 = , x04 = , x05 = , x 6 = , x 7

=

, x

8 = , x 9 = 2x , x

10 = , x Proof 1 = , x[-1,1]

Since, arcsinx = y = x, y[-,]And

=

=

So, = = , x[-1,1]2 = , x[-1,1]

Since, arccosx = y = x, y[0,]And = =


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So, = , x[-1,1]3 = , x0

Since, arccotx = y

= x, y

(0,

)

= So, = = , 04 = , x0

Since, arctanx = y = x, y(-,0)(0,) = , y(-,0)(0,)So,

=

, x

0

5 = , x Since, arctanx = y = x, y(-,)And = = = So, = and then = = , x

6 = , x Since, arctanx = y = x, y(-,)And = = = = = So, = = , y(-,0) and [0,] respectivelyThat is == , x

7 = , x Since, arccotx = y

= x, y

(0,

)

And = = = = So, = = , y(0,] and (,) respectivelyThat is == , x


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8 = , x Since, arccotx = y = x, y(0,)And

=

=

=

So, = and then = , x 9 = 2x , x

Since, 2arcsinx = y arcsinx = , y(-,) = x = x = = 1-2= 1 - 4+4

So, = 1-1 + 4- 4=4- 4Therefore, =2(-x)(-) =2x , y(-,0) and [0,) respectivelyThat is = 2x , x

10

=

, x

Since, arctanx = y , y(-,)And = 1+=+1So, = Finally, we have = , x

Theorem 2 1 Arcsin(-x) = - arcsinx , x[-1,1]2 Arccos(-x) = - arccosx, x[-1,1]3 Arctan(-x) = - arctanx, x(-,)4 Arccot(-x) = -arccotx, x(-,)5 Arcsec(-x) = -arcsecx, x(-,-1][1,)6 Arccsc(-x) = -arccscx, x(-,-1][1,)7 Arcsinx + arccosx =

, x[-1,1]


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8 Arctanx + arccotx =, x

9 Arcsecx + arccscx =, x(-,-1][1,)

10 Arccosx + arcos(-x) =

, x

[-1,1]

11 Arctanx + arctan() = , x0Proof 7 Arcsinx + arccosx = , x[-1,1]According to definition, we have

-


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So, - arcsecx-


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Proof Since, arctanx, arctany(-,), we will imply thatarctanx + arctany(-,).....(1) and arctan( (-,).....(2)Let arctanx+arctany=

;

(-

,

)

Now, we have = arctanx+arctany)= =n+arctan( ))And it make arctanx+arctany = n+arctan( ).....(3)And from (1), (2) we find that n{ -1, 0, 1 }We can consider n as follows.

From (3), sin(arctanx+arctany)=sin(n+arctan( )), 1-1 ontoSin(arctanx)

arctany)+

arctany)

)=

Sin(n)arctan( ))+arctan( ))n)So, we obtain

+ = Rearrange term,

=

()= ||

Finally, we have

=

||and if

xy0 which make ||==1 and, then n=0Therefore, arctanx + arctany = arctan(

xy>1 then 1-xy


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So,

,y> and


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Hence, = >0, see a graph of arccosx=y, y(0,]But arcos( )=kk(0,),where y(0,)Therefore, arcos(

)=k=y=arcsinx

Finally, we can write arcsinx=( ) () or arcsinx=-arccosx2 Arcsinx=arctan(

Let arcsinx=y =x, where x[-1,1]=, y[-,]=And == = then = ||

1) -1


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We are obviously that k=-y

Therefore, arcsec()=k=-y=-arcsinx, -1


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Since, = = = so, = || 1) When -1


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Let arctanx=y =x, x =, y(-,)=We have = = = = , = ||

1) Since, x

0

-


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Finally, we can write arctanx= 3 Arctanx=

(

) ( )

Let arctanx=y =x, x =, y(-,)=Consider =1+=1+ so, =

1) x


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2) x>00


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Theorem 6 formula of arccosx into term of arcsin, arctan, arcsec, arccsc

And arccot, when -1x11 Arccosx=

or arccosx=

-arcsinx

2 Arccosx={ 3 Arccosx=arcsec(

, x04 Arccosx= 5 Arccosx=arccot( ), -1


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Finally, we conclude arccosx= or arccosx=-arcsinx2 Arccosx=

{

Let arccosx=y =x, x[-1,1]=, y[0,]=Considering, = -1= -1 so, = ||

1) -1x


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Let arccosx=y =x, x[-1,1]=, y[0,]=Considering = = = , =

1) Since -10, y[,)And, as arccsc(

)=kk(0,], while y[,)We find that -y= k(0,], therefore arccosx=-arccsc( )

2) Since 0


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1.6 EQUATION OF TRIGONOMETRIC INVERSE FUNCTIOIN

Real numbers that is obtained from solving that equation of trigonometric inverse

Function always have to be tested by substitution of variable in that equation.

As the real numbers that is obtained from solving that equation did not require

To be its answer.

Example 1 solve the following equation

1) arcsinx=2arctanx 3) arcsinx +arcsin(1-x)=arccosx

2) arccos(2x)-arccosx= 4) arctan()+arctan()=

Solution 1) arcsinx=2arctanx

Let arctanx=A

=x, A

(-

,)

Consider, )=): sine is a fuctionSo, x=2) ): 2x)=2=

: theorem 1=

Thence, x }Test : 1) -1-10- arcsinx0 and - 2arctanx0

2) 0 is to be true.

3) 0


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2x= - : theorem 1

4x=x- 9

=3(1-

)

4=1 and, then x{-,}Test: 1) since, arccos(-1)-arccos(-

)=-(-)=And, arccos(1)-arccos(

)=0-=-So, {-

} is solution set3) arcsinx +arcsin(1-x)=arccosx

Let arcsinx=A

=X

=

, A

[-

,]

And arcsin(1-x)=B =1-x = , B[-,]Now, we see (A+B)=arccosx, A+B[0,]And, =arccosx)Hence, -=XSubstituting, -x(1-x)=xRearranging,

=2x-

Rearranging again, -2-+2x=4-4+Rearranging again, 2-5+2x=0Factorizing, x(2-5x+2)=x(2x-1)(x-2)=0We have x{0,1, } which have to test as followsWhen x=0, arcsin0+arcsin(1-0)=0+

=arccos0=: trueWhen x=

, arcsin+arcsin(1-)=+=arccos=: trueWhen x=2, it is to be false, as arcsin2 is not defined.

Finally, we obtain solution set is {0, }4) Since, arctan(

)+arctan()=So,)=


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And make=1

Now following,

=1

Rearrange term, =1We see that x{ , } and can test as followsWhen x=-

, arctan( +arctan(

=arctan( )+arctan( )=arctan(

)=arctan1

=X=

is left the test as an exercise.EXERCISE

1 Find value of the following functions.

1) arcsin 2) arccsc2

3) arctan 4) arccot5) arccos(- ) 6) arcsin)7) arctan(-)) 8) arccot )9) arctan2) 10) arccos)

11) arcsin( ) 12) arccos(- ))13) arctan(

) 14)

arcsin(-

))

15) arcsin( )) 16) arcsin)2 Find the solution of each of the following.

1) arcsin)+arccos(-)) 1) arcsin+arcsin)3) arctan -arcsin ) 4) arctan(- )+arccos(-))


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5) arcsin +arcsin) 6) arcsin(-)-arccos )7) 2arcsin+arccos) 8) arctan(-)+arcsin)9)

2arctan

-arctan

10)

2arctan(-

))+

arctan(-

))

11) arcsin( - 12) arctan+arctan13) arctan2+arctan3 14) arctan

+arctan+arctan+arct6an3 Show the following expressions

1) arcsin+arcsin = 2) arcsin+arccos+arcsin=

3) 2arctan=arctan 4) arctan+arctan+arctan=

5) arctan-arctan1=arctan 6) arctan+arctan=arctan+arctan 7) 2arctan

+arctan+2arctan= 8) arccot-arctan-arccos=09) arctan1+arctan2+arctan3==2(arctan1+arctan+arctan)10) If A=arctan

and B=arctan then =4 Prove the following expressions.

1) arcsinx=arccsc, |

|

1

2) arccosx=arcsec, || 13) arctanx=arccot, x and x04) 2arctanx)=2arctanx+arctan)5) arctanx+arctan

=, x>0 and arctanx+arctan=-, x17)

A=

B, A=arctan

, B=arctan

5 Find a solution set of each of the following.

1) arcsinx=2arctanx 2) arctan2x+arctan3x=

3) arctan(x+1)+arctan(x-1)= 4) arcsinx+arccosx=

5) arctanx+arctan= 6) arccos2x-arccosx=


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7) 2arcsinx+arcsin(2x )=6 Show that the function

(x)=arcsin(

)-2arcsin

is constant for x

[0,4]

1.7 SYMMETRY AND APPLICATION

In this part we want to introduce symmetry of graph being basic type its definition and test

Lastly we want to lead this knowledge for using to prove basic theorem about trigonometry

Function that concern angles (-), (+), (-), (+), ( -), ( +) and, where is length

Of circumference of the unit circle beginning from a point (1, 0) to anti-clockwise and 0


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Definition1 (Symmetry)

1 A graph is said to be symmetric with respect to the X-axis if, whenever (x, y) is a point

on the graph, (x, -y) is also a point on the graph.

2 A graph is said to be symmetric with respect to the Y-axis if, whenever (x, y) is a point

on the graph, (-x, y) is also a point on the graph.

3 A graph is said to be symmetric with respect to the origin if, whenever (x, y) is a point

on the graph, (-x, -y) is also a point on the graph.

4 A graph is said to be symmetric with respect to the identity (f(x)=x) if, whenever (x, y)

is a point on the graph, (y, x) is also a point on the graph.

5 A graph is said to be symmetric with respect to the inverse of identity ( g(x)=-x

if, whenever (x, -y) is appoint on the graph, (-y, x) is also a point on the graph.

Or, that is to say.

Definition2 let is an equation1 A graph of =0 is said to be symmetric with respect to the Y-axis

if, whenever

are points lying on its graph.

That is, (x, y)p2 A graph of =0 is said to be symmetric with respect to the X-axis

if, whenever are points lying on its graph.That is, (x, y)p

3 A graph of =0 is said to be symmetric with respect to the originif, whenever are points lying on its graph.That is,

(x, y)

p

4 A graph of p =0 is said to be symmetric with respect to the identity functionif, whenever are points lying on its graph.That is, (x, y)p

5 A graph of p =0 is said to be symmetric with respect to the inverse ofIdentity function if, whenever (x, y), (-y , -x) are points lying on its graph.


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That is, (x, y)p(-y, -x)p, (x, y)pTEST FOR SYMMETRY

1 The graph of an equation in x and y is symmetric with respect to X-axis if replacing

y by -y yields an equivalent equation.

2 The graph of an equation in x and y is symmetric with respect to Y-axis if replacing

x by -x yields an equivalent equation.

3 The graph of an equation in x and y is symmetric with respect to the origin if replacing

x by -x and y by -y yields an equivalent equation

4 The graph of an equation in x and y is symmetric with respect to the identity function

if replacing x by y and y by x yields an equivalent equation.

5 The graph of an equation in x and y is symmetric with respect to the inverse function

of the identity function if replacing x by -y and y by -x yields an equivalent equation.

Example1 Show that

1 An equation y= has symmetry property along the Y-axis.2 An equation x= has symmetry property along the X-axis.3 An equation xy=1 has symmetry property along the identity function.

4 An equation xy=-1 has symmetry property along the inverse function of identity function.

5 An equation y= has symmetry property along the origin6 An equation +=1 has all symmetry property (the five types)Solution 1 When replacing x by -x in its original will get

y=== which see obviously yields an equivalent equation.So, an equation y= has symmetry property along the Y-axis.2 When replacing y by -y in its original will get

x=

=

=

which see obviously yields an equivalent equation.

So, an equation x= has symmetry property along the X-axis.3 When replacing x by y and y by x in its original will get1=xy=yx=xy which see obviously yields an equivalent equation.

So, an equation xy=1 has symmetry property along the identity function.

4 When replacing x by -y and y by -x in its original will get


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-1=xy=(-y)(-x)=yx=xy which see obviously yields an equivalent equation.

So, an equation xy=-1 has symmetry property along the inverse

Function of identity function.

5 When replacing x by -x and y by -y in its original will get

0=-y=-(-y)=-+y=-y which see obviously yields an equivalent equation.So, an equation y= has symmetry property along the origin.6 Since, =, = and +=+So will obtain that an equation +=1 has all symmetry property (the five types).

Symmetry about straight line

Its one property is its symmetric axis has to halves and be perpendicular with line segment

That is connected by two points that is symmetrical.Symmetry about an origin point

Its one property is its symmetric point has to halves line segment that is connected by two

Points that is symmetrical.

The following we will use two properties above and 6th

item in an example1 for leading to

Prove theorem of trigonometric function that concern an angle or length of a circumference

as follows.

, (

-

), (

+

), (

-

), (

+

), (

-

), (

+

) and -

, where 0

Proof by Bygone Experience

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