Today we started out class with another proof by induction example.  The problem was:

Prove n is greater than or equal to 2, a(x1+x2+x3+…+xn)=ax1+ax2+ax3+…+axn 

n=2,   a(x1+ax2) =ax1=ax2 (distributive property)  TRUE!

Now we have to continue the problem by assuming n=k,

a(x1+x2+x3+..+xk)+ax1+ax2+ax3+…axk

The next step is to assume k+1,

a(x1+x2+x3+…+xk+xk+1)=ax1+ax2+ax3+…+axk+axk+1

The left part of the equation can also be writtten as a((x1+x2+x3+…+xk)+xk+1)

By rewriting the left side of the equation like this we can subsititute in when we assumed n=k.

ax1+ax2+ax3+…+axk+axk+1 = ax1+ax2+ax3+…+axk+axk+1 q.e.d.

I apologize for not being abke to figure out how to insert subscripts.  All of the numbers and k’s are supposed to be in subscript form.

After the warm-up proof by induction problem we moved to the main focus of the class which was to prove identities.  The previous day in class we went over all of the diffferent properties which included pythagorean, quotient, and reciprocal.

The proof we did today in class was sinx/1+cosx —>1-cosx/sinx    

To start the problem we multiplied sinx/1+cosx by 1-cosx/1-cosx.  We did this to obtain 1-cosx in the numerator just like the other one.

The result of the multiplication was sinx(1-cosx)/1-cos(squared)x

The 1-cos(squared)x translates to sin(squared)x

It now looks like sinx(1-cosx)/sin(squared)x. Since the demoninator is squared, written out it looks like sinx(1-cosx)/sinx times sinx.  SInce there is both a sinx in the numerator and in the denominator we can elminate one of them in the deonminator.

Therefore, the final is 1-cosx/sinx=1-cosx/sinx q.e.d.

If you need more help a great website for information is http://www.purplemath.com/modules/proving.htm. 

Most of us are familiar with the Pythagorean Theorem of a²+b²=c², well on Monday we learned how that can apply to our unit circle and the 8 different Pythagorean Properties.If you look at the unit circle and create a right triangle with a point on the center of the circle and on the border, then you can use cosine and sine in place of a and b in the Pythagorean Theorem. With cos(x) being the short side, and sin(x) being the long side, we know the hypotenuse is 1. When you plug in cos(x) and sin(x) for a and b, you get the equation sin²x + cos²x = 1. This is the first property! (All the properties will be written in purple.)

1. sin²x + cos²x = 1
You can also write property #1 in terms of sin²x or cos²x like so:
1a) sin²x = 1 – cos²x
1b) cos²x = 1 – sin²x

After we learned the first property, we then skipped to properties #4 and 5.  These two properties are called quotient properties.
4. tan(x) = sin(x)/cos(x)
5. cot(x) = cos(x)/(sin(x)

The last three properties are called the reciprocal properties because it is 1 over a number.
6. csc(x) = 1/sin(x), which can also be written as  sin(x) = 1/csc (x)
7. sec(x) = 1/cos(x), which can also be written as cos(x) = 1/sec(x)
8. cot(x) = 1/tan(x), which can also be written as tan(x) = 1/cot(x)

As you can see, we used all 6 of the trig functions within those properties. These properties are the ones that are very important and you should begin to memorize. We then went back to properties #2 and #3, which you do not need to memorize because they are the first equation, just divided by either cosine or sine.

2. (sin²x + cos²x)/(sin²x) = 1/sin²x
 We then simplified this by
sin²x/sin²x + cos²x/sin²x = 1/sin²x
the first term sin²x/sin²x is simply, and becomes 1
the second term seems more complex, but if you look at it, it is the same as cot²x
the third term, on the right side of the parentheses is the same as csc²x
so the simplified equation of this is:
2) 1 + cot²x = scs²x
This can also be written as:
2a) 1 = csc²x – cot²x
2b) cot²x = csc²x – 1

The third property is the same as #2, except you divide by cos²x.
3. (sin²x + cos²x)/(cos²x) = 1/cos²x
Again, we could simplify this equation by writing it out as:

sin²x/cos²x + cos²x/cos²x = 1/cos²x
The first term, again, looks complex, but it is actually just tan²x
The second term, is simple, and simplifies to 1
The third term is the same as sec²x
So the simplified equation for property #3 is,
tan²x + 1 = sec²x
this can also be written as:
3a) tan²x = sec²x -1
3b) 1 = sec²x – tan²x

All the these properties can be tied in with proof by induction, so we then did a problem at the end of the class.
The problem we were given was:
cotA + tan A ? scsA · secA
the first step we took was getting rid of cotangent and tangent and turning into something we are more familiar with:
cotA  ? cosA/sinA   and    tanA ? sinA/cosA    so the new equation is:
cosA/sinA + sinA/cosA
next, to be able to add the two terms, we got a common denominator of sinAcosA. so the new equation is:
cos² A/sinAcosA + sin²A/sinAcosA
to get the first term you had to multiple cosA/sinA by cosA/cosA
to get the second term you had to multiple sinA/cosA bu sinA/sinA
the next step is to add the two terms together to get:
cos²A + sin²A/sinAcosA
The numerator of the equation is something we know! It is property #1, so therefore we know it equals 1! This helps us out because then we can simplify that equation to get:
1/sinAcosA
This is also the same as:
1/sinA · 1/cosA
And!
1/sinA = cscA
And!
1/cosA = secA
Therefore we know cotA + tanA can turn into cscA · secA
Tombstone!

<<WE STARTED OFF THE CLASS WITH A BASIC PROOF BY INDUCTION PROBLEM THAT WAS…

2+2²+3²+4²+…. 2^K = 2^(K+1) +2

2+2²+3²+4²… 2^K+2^(K+1) = 2^(K+2) – 2

2^(K+1) + 2^(K+1) = 2^(K+2)

2(2^K+1) = 2^(K+2) Q.E.D. #TOMBSTONE

#HAVINFUNYET ?

NEXT WE LEARNED THE #PYTHAGOREAN, #QUOTIENT, AND #RECIPROCAL PROPERTIES…

#PYTHAGOREAN…

1) SIN²X+COS²X=1   1A) SIN²X=1-COS²X       1B) COS²X=1-SIN²X

2)1+COT²X=CSC²X   2A) COT²X=CSC²X-1      2B) 1=CSC²X-COT²X

3)TAN²X+1=SEC²X    3A) TAN²X=SEC²X-1      3B) 1=SEC²X-TAN²X

#QUOTIENT …

4) TAN(X)=(SIN(X)) / (COS(X))

5) COT(X)= (COS(X)) / (SIN(X))

#RECIPROCAL …

6) SIN(X)=1 / (CSC(X))       6A) CSC(X)=1 /SIN(X)

7) COS(X)=1 / (SEC(X))       7A) SEC(X)=1 / COS(X)

TAN(X)=1 / COT(X)         8A) COT(X)=1 / TAN(X)

THANNNNNNNN…..

WE DID A PRACTICE PROBLEM! HERE IT IS AMIGOS!

COT(A) + TAN(A) =                  CSC(A) + SEC(A)

( COS(A)/SIN(A) ) + ( SIN(A)/COS(A) ) = ^^

( COS²A/(SIN(A)COS(A)) ) + ( SIN²A/(SIN(A)COS(A)) ) = ^^

1/( SIN (A)COS(A) )= ^^

1/SIN(A) x 1/COS(A) = CSC(A)SEC(A)

#RIGHTRIGHTRIGHT #YOUSTILLDOIT

Today in #bienprecal, we did some more solving of proof by induction equations. The first one we did was 2+4+6+8+…+2n=. The steps we used to solve this are as follows:

1. Find what the equation equals.

2+4+6+8+…+2n=n²+n

2. Assume Pk (do this by changing all the n’s to k’s)

2+4+6+8+…+2k=k²+k

3. Prove Pk+1 is true (do this by adding this to the left side and changing all the k’s on the right to k+1)

2+4+6+8+…+2k+2(k+1)=(k+1)²+(k+1)

4 Use the asumption from step 2 to substitute for the “2+4+6+8+…+2k”

k²+k+2(k+1)=(k+1)²+(k+1)

5. Simplify using #algebra

k²+k+2(k+1)=(k+1)²+(k+1) equals

k²+3k+2=k²+3k+2 q.e.d.  Therefore, we know it is TRUE!

The second one we did was 1²+2²+3²+…+n²=.  You solve this one the same way we solved the first.

1. Find out what the equation equals (In this case it was given to us since it’s hard to find by just guess and check)

1²+2²+3²+…+n²=n(n+1)((2n+1)/6)

2. Assume Pk

1²+2²+3²+…+k²=k(k+1)((2k+1)/6)

3. Show that Pk+1 is true

1²+2²+3²+…+k²+(k+1)²=(k+1)((k+1)+1)((2(k+1)+1)/6)

4. Insert the assumption “n(n+1)((2n+1)/6)” from step 2 in for “1²+2²+3²+…+k²”

n(n+1)((2n+1)/6)+(k+1)²=(k+1)((k+1)+1)((2(k+1)+1)/6)

5. Simplify using #algebra

n(n+1)((2n+1)/6)+(k+1)²=(k+1)((k+1)+1)((2(k+1)+1)/6) equals

(k+1)((2k²+7k+6)/6)=(k+1)((2k²+7k+6)/6)

If you still need more help, you can check out the #KhanAcademy video about Proof by Induction

We dove right into class today. We review how to do the proof by induction. After we tried a problem we tried something a little more difficult.

 Proving 1^2+2^2+3^2+…+2n=n(n+1)(2n+1/6)

First: Prove That the First Domino Falls

n=1

1^2=1(1+1)(2*1+1/6)

simplify

4=4 The first domino falls!

Second: Assume another domino falls

 1^2+2^2+3^2+…+2k=k(k+1)(2k+1/6)

 1^2+2^2+3^2+…+2(k+1)=(k+1)((k+1)+1)(2(k+1)+1/6)

Sence  1^2+2^2+3^2+…+2n=k(k+1)(2k+1/6) We can subsitute in the problem above

 (k)(k+1)(2k+1/6)+2(k+1)=(k+1)((k+1)+1)(2(k+1)+1/6)

(k)(k+1)(2K+1/6)+2(K+1)=(K+1)(k+2)(2k+3/6)

(k+1)(2k^2+k/6)+2k+2=(k+1)(2k^2+7k+5/6)

(k+1)(2k^2+7k+5/6)=(k+1)(2k^2+7k+5/6)QED Or Tombstone

What your favorite QED or Tombstone?

 Jets and Sharks you say? In future it will be the QED’s and Tombstone’s

http://www.youtube.com/embed/3YT9riTfzNM