# Math Problem - 10



## nitrocan (Aug 14, 2008)

Hopefully hard enough to stay unsolved for a day 

Find sin^3(18) + sin^2(18). (WITHOUT a calculator.)

Edit: Try finding the solution with finding and not finding sin18.


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## 3.14159265358979323846264 (Aug 14, 2008)

This is a nice question, lots of ways to do it. I did a quick calculation but in a not so elegant way. If I had some more time, I'd find something nice. 

Keep the questions coming btw, they're fun even though I don't have much time for them now.


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## nitrocan (Aug 14, 2008)

i will post one of many solutions tomorrow


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## fanwuq (Aug 14, 2008)

I should have really paid attention in trig. Time to pull out that long list of formulas...


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## brunson (Aug 14, 2008)

Is that in radians or degrees? ;-)


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## nitrocan (Aug 14, 2008)

degrees .


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## cmhardw (Aug 14, 2008)

Here is my solution by finding sin(18). I am still trying to find out how to solve it without finding sin(18) explicitly - I'm stumped on this.



Spoiler



Draw a regular pentagon ABCDE, with side length s. Start at the lower left vertex of the base and label clockwise. If you draw segments AC and CE you will form an isosceles triangle. Now connect point C with the midpoint of segment AE, call this point M. Triangle ACM is a right triangle with angle ACM = 18. The length of segment AM is s/2, the length of segment AC is [(1+sqrt(5))/2]*s. Now sin(ACM) = (s/2) / [[(1+sqrt(5))/2]*s] = (sqrt(5)-1)/4

So sin^3(18) + sin^2(18) =
= [(sqrt(5)-1)/4]^3 + [(sqrt(5)-1)/4]^2
= 1/8

I'm still looking at how to do it without finding sin(18), but in my opinion finding sin(18) is not very difficult.



Chris


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## nitrocan (Aug 14, 2008)

here's my way of finding sin18:


Spoiler



18 * 5 = 90, so sin18*2 = cos 18*3
lets say 18 = x
sin2x = cos3x
2sinxcosx = 4(cosx)^3 - 3cosx
2sinx = 4(cosx)^2 - 3
2sinx = 4(1-(sinx)^2) - 3
then we say a to sinx
2a = 4 - 4a^2 - 3
4a^2 + 2a - 1 = 0
4a^2 + 2a + (1/4) - (1/4) - 1 = 0
(2a + 1/2)^2 - 5/4 = 0
2a + 1/2 = sqrt(5)/2
2a = (sqrt(5) - 1) / 2
sin18 = (sqrt(5) - 1) / 4


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## Brett (Aug 15, 2008)

Lol. Maybe those 45 minutes of naptime every day were a little important .

Although I think I just learned more about trig in your post then all year...


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## 3.14159265358979323846264 (Aug 15, 2008)

nitrocan said:


> here's my way of finding sin18:
> 
> 
> Spoiler
> ...





Spoiler



> ((sqrt(5) - 1) / 4)**3 + ((sqrt(5) - 1) / 4)**2;

/ 1/2 \3 / 1/2 \2
|5 | |5 |
|---- - 1/4| + |---- - 1/4|
\ 4 / \ 4 /

> evalf(%);

0.1249999998

> sin(18)**2+sin(18)**3;

3 2
sin(18) + sin(18)

> evalf(%);

0.1404386720
> evalf(sin(18)); evalf((sqrt(5)-1)/4);

-0.7509872468


0.3090169942

Also, I don't have time to look at your whole solution but one thing I did quickly see: solving the equation for a, why would you discard the negative root?


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## Lucas Garron (Aug 15, 2008)

3.14159265358979323846264 said:


> Spoiler
> 
> 
> 
> solving the equation for a, why would you discard the negative root?


Because he knows what he's doing?
(Think about it... where's argument?  )


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## cmhardw (Aug 15, 2008)

Lucas Garron said:


> 3.14159265358979323846264 said:
> 
> 
> > Spoiler
> ...



The negative root is valid for the quadratic equation, yes. But for the trig....

*cough* think unit circle *cough* ;-)

Chris


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## nitrocan (Aug 15, 2008)

3.14159265358979323846264 said:


> Also, I don't have time to look at your whole solution but one thing I did quickly see: solving the equation for a, why would you discard the negative root?[/spoiler]


sin18 can't be negative.
here is the solution without finding sin18:


Spoiler



sin^3(18) + sin^2(18)
=sin^2(18)(sin18 + 1)
=sin^2(18)(sin18 + sin90)
=sin^2(18)(2.sin54.cos36)
=2sin^18.cos^36
=2[1/2(sin54 - sin18)]^2
=1/2(sin54 - sin18)^2
=1/2(cos36-cos72)^2
now we will solve cos 36 - cos 72
cos36 - cos72 = -2sin(54).sin(-18)
=2sin54.sin18
=2.cos36.cos72
=2.(sin36.cos36.cos72)/sin36
=2.(sin72/2 . cos72) /sin36
=sin72.cos72 / sin36
=sin144/2sin36
=1/2
so our answer is 1/2(1/2)^2 = 1/2*1/4 = 1/8


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## badmephisto (Aug 15, 2008)

thank god i did not attempt to find the solution to this (looking at Nitrocan's)
Its Trigiculus


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## cmhardw (Aug 15, 2008)

nitrocan I'm impressed. I was still stumped on how to find this without finding sin(18), that was a nice proof. I have to admit, I don't remember my sum-product formulas :-( I remember nearly all of the other "basic" formulas from first semester trig, except those :-( I chose on purpose not to remember the formulas involving tan, and just to re-derive them using the cos and sin versions. You've inspired me to study my trig formulas again lol.

Chris


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## badmephisto (Aug 16, 2008)

cmhardw said:


> nitrocan I'm impressed. I was still stumped on how to find this without finding sin(18), that was a nice proof. I have to admit, I don't remember my sum-product formulas :-( I remember nearly all of the other "basic" formulas from first semester trig, except those :-( I chose on purpose not to remember the formulas involving tan, and just to re-derive them using the cos and sin versions. You've inspired me to study my trig formulas again lol.
> 
> Chris



I hate trig, so I always end up converting everything into e^i(theta) versions, and working with those. It can get a little messier but its much more intuitive


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## JBCM627 (Aug 16, 2008)

badmephisto said:


> I hate trig, so I always end up converting everything into e^i(theta) versions, and working with those. It can get a little messier but its much more intuitive



And easier! You don't have to remember or look up all those trig ids...


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