Take A Break Fellas! ... For Math Enthusiasts!

[Harmonica]

I am surprised than I remember cotangent 30 degrees is √3. I learned that stuff, never used since,  nearly 20 years ago.

Interesting that 7.5 is 1/4 of 30!

Remember the  1,  √2 ,  √3    ,      1 ,2 , √3    etc

Doesn't take much to prove it with this info.

Right on! Look at the 45 & 60 degree triangles.

The "45" has 2 angles that are 45 and one that is 90.

The "60" has a 30, a 60 and a 90.

For the "45" the hypotenuse is root 2 and the sides are ?

For the "60" the hypotenuse is ???? and the sides are ?????

Now what's the formula for Tan 2A?

See what I'm getting at?

By the way, even if you get the right formula to stat with, you've still got to overcome working with square roots in the denominator.

Give it a shot, Neeranam.

Take your time.

[Neeranam]

Come on Harmonica, give us the answer.

[Harmonica]

Come on Harmonica, give us the answer.

Neeranam,

Please review the 2nd para of the origination.

Jai yen krap -- give it time; somebody will come along and take it to another level. I'm not about to give up this stunner so easily and then have it appear on the Net.

You have enough info + 2 hints on a reasonable approach.

Show me what you've done so far!

Goodluck, if you're still interested, that is!

[kayo]

   

Level 13

I interview a person for an accounting position with us and ask only one question.

What is one and one, and the person answered :- ???????????????

So what did the person answer????        

<{POST_SNAPBACK}>

....And the aspiring accountant said,

WHATEVER YOU'd LIKE IT TO BE SIR!!!

And you said, " You're hired!!!"

And we laughed and chok dee'd eachother, and lived happily ever after. !!!!

LOLOLOL

[peder.klockmann]

Harmonica wears only 1 kind of sock, which he buys in 2 colors - black and white. They are all mixed up in his drawer. If he pulls out 3 socks at random, what is the probability that he will have a matched pair?

Bleh

[LivinLOS]

100%

[chownah]

We are not yet at Level 10, but I am posting it early to give enthusiasts enough time to ponder, extrapolate and "figure" this one out. There are no tricks in this question. 

If nobody gets it, then that's that.  I will not be posting the answer.  It would deflate & detract from the power of this "stunner" as an examination " bypass-all-24-questions-but-if-you-attempt-and-get-#25-correct, you-score 100% on-the-exam"  type of power!  Just sheer poetry to me! 

Degree of difficulty 4+ .... out of a maximum of  4       

Level 10:

Without the use of logarithmic and/or trignometric tables, sliderules, calculators, computers etc., and starting from first principles, prove the following:

cot 7½°  =  √2 + √3 + √4 + √6

<{POST_SNAPBACK}>

Start with the fact that a 30-60 right triangle has sides of relative lengths 1,2, and square root of 3....or you could prove this by drawing an equalateral triangle with one altitude and then prove that the altitude bisects the base and the vertex angle and then using the Pythagorean formula (or maybe you'd have to derive this too) you would find the side lengths. Then finding the sine, cosine, tangent secant, cosecant and cotangent of 30 degrees is trivial. Next apply the half angle formulas (or maybe you would have to derive these.....god forbid). Lots of ways you could use the half angle formulas....either do the tangent and then invert for cotangent or do the sine and cosine and then divide them....take your pick. After you do it once you'll have 15 degrees down cold....just do it again starting with 15 degrees and you'll get right down to the real nitty gritty...cot(7.5 degrees). Not bad for a farmer...huh? The puzzle I want to solve is how to keep the crabs from making my rice paddy leak.

Chownah

[BrissyBoy]

there is "history" behind this question.  Starting from first principles is just exactly that -- rock bottom basics in Trigonometry & Geometry & then proceeding into Algebra -- there is unquestionably quite a bit of dealing with square roots in the denominator etc.; but the beauty comes in due to the various approaches to the solution.

Even great men were born little children!   

So start from "basics" and build the great man! 

<{POST_SNAPBACK}>

I have been thinking about this problem for a while now and it is quite a challenge.

Ok then let's start from basics and state the facts in the problem.

Right Hand Triangle

This has to be proved.

cot 7½° = √2 + √3 + √4 + √6

Pythagorean Theorem

a^2 = b^2 + c^2 where a=hypotenuse and b and c are the sides of a right handed triangle.

SOHCAHTOA

Sin(x) = Opposite / Hypotenuse

Cos(x) = Adjacent / Hypotenuse

Tan(x) = Opposite / Adjacent

Cot(x) = Tan(90-x)

so also

tan 82 ½° = √2 + √3 + √4 + √6

there are also interesting relationships with certain size angles.

tan 30° = 1 / √3

cot 30° = √3

cos 30° = √3 / 2

cos 30° = √3 / √4

cot 45° = 1

sin 45° = √2

As to where to go from here I am not sure.

If the adjacent is √2 + √3 + √4 + √6 and opposite is 1 then

hypotenuse^2 = (√2+√3+√4+√6)^2 + 1^2

= 144 + 1

hypotenuse = √145

In reality as chonwah states we will probably have to find the derivative. But doing this with raw math, no calculators and square roots is going to be messy.

Anyone with any thoughts?

[chownah]

[

..........there are also interesting relationships with certain size angles.

tan 30° = 1 / √3

cot 30° = √3

cos 30° = √3 / 2

cos 30° = √3 / √4

cot 45° = 1

sin 45° = √2

As to where to go from here I am not sure..............

<{POST_SNAPBACK}>

BrissyBoy, I'm kind of worried about you. Having "interesting relationships with certain size angles" seems a bit.....strange. I'm not sure if you can talk about this even in the gay forum. And "as to where to go from here," wellllllll maybe Patpong in the back of some almost forgotten soi you could find some certain size angles and............

[Harmonica]

We are not yet at Level 10, but I am posting it early to give enthusiasts enough time to ponder, extrapolate and "figure" this one out. There are no tricks in this question. 

If nobody gets it, then that's that.  I will not be posting the answer.  It would deflate & detract from the power of this "stunner" as an examination " bypass-all-24-questions-but-if-you-attempt-and-get-#25-correct, you-score 100% on-the-exam"  type of power!  Just sheer poetry to me!  

Degree of difficulty 4+ .... out of a maximum of  4       

Level 10:

Without the use of logarithmic and/or trignometric tables, sliderules, calculators, computers etc., and starting from first principles, prove the following:

cot 7½°  =  √2 + √3 + √4 + √6

Start with the fact that a 30-60 right triangle has sides of relative lengths 1,2, and square root of 3....or you could prove this by drawing an equalateral triangle with one altitude and then prove that the altitude bisects the base and the vertex angle and then using the Pythagorean formula (or maybe you'd have to derive this too) you would find the side lengths. Then finding the sine, cosine, tangent secant, cosecant and cotangent of 30 degrees is trivial. Next apply the half angle formulas (or maybe you would have to derive these.....god forbid). Lots of ways you could use the half angle formulas....either do the tangent and then invert for cotangent or do the sine and cosine and then divide them....take your pick. After you do it once you'll have 15 degrees down cold....just do it again starting with 15 degrees and you'll get right down to the real nitty gritty...cot(7.5 degrees). Not bad for a farmer...huh? The puzzle I want to solve is how to keep the crabs from making my rice paddy leak.

Chownah

Thanks for reminding me, chownah -- glad I caught your comment in the "Thai Baht" thread.

Your ideas here are unquestionably in the right direction and of course, were we in a class with me as the teacher, you would score solid marks for this effort.

Take a bow on your progress thus far.

But ....... where is the step-by-step answer? That is what is required.

Read Brissyboy's comments for further insight. I have the feeling that if the 2 of you were to combine your efforts, a systematic, step-by-step proof sequence could be at hand.

Continue!

[Harmonica]

there is "history" behind this question.  Starting from first principles is just exactly that -- rock bottom basics in Trigonometry & Geometry & then proceeding into Algebra -- there is unquestionably quite a bit of dealing with square roots in the denominator etc.; but the beauty comes in due to the various approaches to the solution.

Even great men were born little children!     

So start from "basics" and build the great man!  

I have been thinking about this problem for a while now and it is quite a challenge.

Ok then let's start from basics and state the facts in the problem.

Right Hand Triangle

This has to be proved.

cot 7½° = √2 + √3 + √4 + √6

Pythagorean Theorem

a^2 = b^2 + c^2 where a=hypotenuse and b and c are the sides of a right handed triangle.

SOHCAHTOA

Sin(x) = Opposite / Hypotenuse

Cos(x) = Adjacent / Hypotenuse

Tan(x) = Opposite / Adjacent

Cot(x) = Tan(90-x)

so also

tan 82 ½° = √2 + √3 + √4 + √6

there are also interesting relationships with certain size angles.

tan 30° = 1 / √3

cot 30° = √3

cos 30° = √3 / 2

cos 30° = √3 / √4

cot 45° = 1

sin 45° = √2

As to where to go from here I am not sure.

If the adjacent is √2 + √3 + √4 + √6 and opposite is 1 then

hypotenuse^2 = (√2+√3+√4+√6)^2 + 1^2

= 144 + 1

hypotenuse = √145

In reality as chonwah states we will probably have to find the derivative. But doing this with raw math, no calculators and square roots is going to be messy.

Anyone with any thoughts?

Outstanding effort, Brissyboy. You're currently at the top of the class with chownah -- each receiving 50%.

Continue!

Here's a "hint" question -- what is the formula for Tan(2A) ..... ?

[The_Doctor]

Hmmmm...

Should expand on this...

Tan 75° = tan(45° + 30°)

= tan45° + tan30°

-------------------

1 - tan45° * tan30°

[Harmonica]

Hmmmm...

Should expand on this...

Tan 75° = tan(45° + 30°)

= tan45° + tan30°

  -------------------

  1 - tan45° * tan30°

Nice try, doctor!

Revisit the hint -- Tan(2A)

Perhaps you should consult with Doctor Pat Pong?

[The_Doctor]

cot (n/2) = cot(n) + √ ((cot(n)*cot(n) )+1) (looks better if can do squared!)

cot 30° = 3

cot 15° = √3 + 2

= √3 + √4

cot 7.5° = √3 + √4 + √((√3+√4)*(√3+√4)+1)

so Apparently;

√2 + √3 + √4 + √6 = √3 + √4 + √((√3+√4)*(√3+√4)+1)

√2 + √6 = √((√3+√4)*(√3+√4)+1)

(√2 + √6)(√2 + √6) = (√3+√4)(√3+√4)+1

(2 + 2(√2*√6) + 6) = (3 + 2(√4*√3) + 4)+1

8 + 2(√2*√6) = 8 + 2(√4*√3)

2(√2*√6) = 2(√4*√3)

√2*√6 = √4*√3

Where am I going with this proof! ARGH

It's right god ###### it, but I gotta take a break...

[The_Doctor]

Where was I?

√2*√6 = √4*√3

Ah yes, both sides are;

√2 * √2 * √3

Bingo - where's my prize?

[chownah]

We are not yet at Level 10, but I am posting it early to give enthusiasts enough time to ponder, extrapolate and "figure" this one out. There are no tricks in this question. 

If nobody gets it, then that's that.  I will not be posting the answer.  It would deflate & detract from the power of this "stunner" as an examination " bypass-all-24-questions-but-if-you-attempt-and-get-#25-correct, you-score 100% on-the-exam"  type of power!  Just sheer poetry to me!  

Degree of difficulty 4+ .... out of a maximum of  4       

Level 10:

Without the use of logarithmic and/or trignometric tables, sliderules, calculators, computers etc., and starting from first principles, prove the following:

cot 7½°  =  √2 + √3 + √4 + √6

<{POST_SNAPBACK}>

Start with the fact that a 30-60 right triangle has sides of relative lengths 1,2, and square root of 3....or you could prove this by drawing an equalateral triangle with one altitude and then prove that the altitude bisects the base and the vertex angle and then using the Pythagorean formula (or maybe you'd have to derive this too) you would find the side lengths. Then finding the sine, cosine, tangent secant, cosecant and cotangent of 30 degrees is trivial. Next apply the half angle formulas (or maybe you would have to derive these.....god forbid). Lots of ways you could use the half angle formulas....either do the tangent and then invert for cotangent or do the sine and cosine and then divide them....take your pick. After you do it once you'll have 15 degrees down cold....just do it again starting with 15 degrees and you'll get right down to the real nitty gritty...cot(7.5 degrees). Not bad for a farmer...huh? The puzzle I want to solve is how to keep the crabs from making my rice paddy leak.

Chownah

<{POST_SNAPBACK}>

Thanks for reminding me, chownah -- glad I caught your comment in the "Thai Baht" thread.

Your ideas here are unquestionably in the right direction and of course, were we in a class with me as the teacher, you would score solid marks for this effort.

Take a bow on your progress thus far.

But ....... where is the step-by-step answer? That is what is required.

Read Brissyboy's comments for further insight. I have the feeling that if the 2 of you were to combine your efforts, a systematic, step-by-step proof sequence could be at hand.

Continue!

<{POST_SNAPBACK}>

I'll leave the step by step answer for others......don't want to spoil the fun.

And a question for you: Can this be proven using expansion series?

[Harmonica]

We are not yet at Level 10, but I am posting it early to give enthusiasts enough time to ponder, extrapolate and "figure" this one out. There are no tricks in this question. 

If nobody gets it, then that's that.  I will not be posting the answer.  It would deflate & detract from the power of this "stunner" as an examination " bypass-all-24-questions-but-if-you-attempt-and-get-#25-correct, you-score 100% on-the-exam"  type of power!  Just sheer poetry to me!  

Degree of difficulty 4+ .... out of a maximum of  4       

Level 10:

Without the use of logarithmic and/or trignometric tables, sliderules, calculators, computers etc., and starting from first principles, prove the following:

cot 7½°  =  √2 + √3 + √4 + √6

Start with the fact that a 30-60 right triangle has sides of relative lengths 1,2, and square root of 3....or you could prove this by drawing an equalateral triangle with one altitude and then prove that the altitude bisects the base and the vertex angle and then using the Pythagorean formula (or maybe you'd have to derive this too) you would find the side lengths. Then finding the sine, cosine, tangent secant, cosecant and cotangent of 30 degrees is trivial. Next apply the half angle formulas (or maybe you would have to derive these.....god forbid). Lots of ways you could use the half angle formulas....either do the tangent and then invert for cotangent or do the sine and cosine and then divide them....take your pick. After you do it once you'll have 15 degrees down cold....just do it again starting with 15 degrees and you'll get right down to the real nitty gritty...cot(7.5 degrees). Not bad for a farmer...huh? The puzzle I want to solve is how to keep the crabs from making my rice paddy leak.

Chownah

Thanks for reminding me, chownah -- glad I caught your comment in the "Thai Baht" thread.

Your ideas here are unquestionably in the right direction and of course, were we in a class with me as the teacher, you would score solid marks for this effort.

Take a bow on your progress thus far.

But ....... where is the step-by-step answer? That is what is required.

Read Brissyboy's comments for further insight. I have the feeling that if the 2 of you were to combine your efforts, a systematic, step-by-step proof sequence could be at hand.

Continue!

I'll leave the step by step answer for others......don't want to spoil the fun.

And a question for you: Can this be proven using expansion series?

Not to my knowledge!

[Harmonica]

We are not yet at Level 10, but I am posting it early to give enthusiasts enough time to ponder, extrapolate and "figure" this one out. There are no tricks in this question.

If nobody gets it, then that's that. I will not be posting the answer. It would deflate & detract from the power of this "stunner" as an examination " bypass-all-24-questions-but-if-you-attempt-and-get-#25-correct, you-score 100% on-the-exam" type of power! Just sheer poetry to me!

Degree of difficulty 4+ .... out of a maximum of 4

Level 10:

Without the use of logarithmic and/or trignometric tables, sliderules, calculators, computers etc., and starting from first principles, prove the following:

<span style='font-size:14pt;line-height:100%'>cot 7½° = √2 + √3 + √4 + √6 </span>

4 years later ......

good entertainment with Thais when the conversation drifts to Chess, puzzles, riddles and problems. Several Thai engineering students have taken a stab at this. One in particular got to the 13th line and then stumbled. A girl no less and a beauty at that. Just 27 yrs old. No winner yet. Its ongoing, but I should have kept the answers or had them post here. Would've saved these had I known I'd be paroled in under 4.

A pal(s) will help me post this in Korea, China, Taiwan, Hongkong, Malaysia, Vietnam, Indonesia and India at the yet to-be-decided grad Uni bulletin boards. The posts will be fired off within minutes of each other. Translations (despite prepon of numbers) of the question will have to be done accurately to be fair to all.

Whoever gets it, even if its a long drawn out derivation, I'll post it here and also try to get these folk to post it themselves without me being the middleman, but its possible like with the Thais, they might be shy to plunk down in englishland.

There are many approaches to solve this and no monitor is required once the question is understood because when/if you get, the emotion will fire within like a lightening bolt and you'll be on a high for days.

continuamos .....

[mike44]

Steps to solve are set out below

Start from the identities

Cos(A+=CosACosB-SinASinB

Sin(A+=SinACosB+CosASinB

So that

Cot(A+= [CosACosB-SinASinB]/[sinACosB+CosASinB]

and

Cot(A-= [CosACosB+SinASinB]/[sinACosB-CosASinB] ....(1)

Cot2A = [CosA*CosA-SinA*SinA]/2SinACosA = [CotA*CotA-1]/2CotA ....(2)

Now Cot(60-45) = Cot15

And Cos60=Sqrt(3)/2, Sin60=1/2, Cos45=1/Sqrt(2), Sin45=1/Sqrt(2)

Hence Cot15 = [sqrt(3) + 1]/[sqrt(3) - 1] using (1)

But Cot15 = [Cot 7.5*Cot 7.5 - 1]/2Cot 7.5 using (2)

Hence [Cot 7.5*Cot 7.5 - 1]/2Cot 7.5 = [sqrt(3) + 1] /[sqrt(3) - 1]

And rearranging

Cot 7.5*Cot 7.5 - 2 [sqrt(3) + 1] /[sqrt(3) - 1]Cot 7.5 - 1= 0

This quadratic equation can be solved to yield the solutions

Cot 7.5 = 2+Sqrt(3) +/- [sqrt(6)+Sqrt(2)]

And because Cot 7.5 must be positive

Cot 7.5 = Sqrt(2) + Sqrt(3) + Sqrt(4) + Sqrt(6)