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Product=1 homework question

  • Thread starter xphloem
  • Start date
  • #1
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Homework Statement



My calculator shows tan 6. tan 42. tan 66. tan 78 =1
Ho is this possible?

Homework Equations



tan (90-x)=cot x

The Attempt at a Solution


but these angles are not supplementary. Ho is this possible?
 

Answers and Replies

  • #2
symbolipoint
Homework Helper
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What were you expecting?

[tex] \[
(.1051)(.9004)(2.246)(4.7046) = 0.99993...
\]
[/tex]

That seems close enough to call "1".
 
  • #3
tiny-tim
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Welcome to PF!

Hi xphloem! Welcome to PF! :smile:

How did you come across this? :smile:

If we take inverses, we can rewrite this as tan 12 tan 24 tan 48 tan 96 = 1.

So I suspect it has something to do with a regular pentagon, whose interior angles are 108º, and whose exterior angles are 72º.

But that's as far as I've got! :rolleyes:
 
  • #4
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It's actually 12 tan 24 tan 48 tan 96 = -1

I've been able to prove cos 12 cos 24 cos 48 cos 96 = -1/16

sin 12 sin 24 sin 48 sin 96 is equal to 1/16 but I can only prove this with a calculator :grumpy:

sin(24) = 2sin(12)cos(12)

sin(48) = 2sin(24)cos(24) = 4sin(12)cos(12)cos(24)

sin(96) = 2sin(48)cos(48) = 8sin(12)cos(12)cos(24)cos(48)

sin(192) = -sin(12) = 2sin(96)cos(96) = 16sin(12)cos(12)cos(24)cos(48)cos(96)

divide by 16 sin(12) to get:

cos 12 cos 24 cos 48 cos 96 = -1/16
 
  • #5
10
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thanks all. I got this in a book called brain teasers. I have solved it myself. thanks for all the help!
 
  • #6
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thanks all. I got this in a book called brain teasers. I have solved it myself. thanks for all the help!
so how did you solve it? any hints?
 
  • #7
tiny-tim
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cos 12 cos 24 cos 48 cos 96 = -1/16
ooh, kamerling, that's clever! :smile:

Your splitting the problem into cos and sin products has given me the following idea:

tanAtanBtanCtanD = -1
iff cosAcosBcosCcosD + sinAsinBsinCsinD = 0​

But (4cosAcosBcosCcosD + sinAsinBsinCsinD)

= [cos(A+B) + cos(A-B)][cos(C+D) + cos(C-D)] + [cos(A+B) - cos(A-B)][cos(C+D) - cos(C-D)]

= 2[cos(A+B)cos(C+D) + cos(A-B)cos(C-D)]

= cos(A+B+C+D) + cos(-A-B+C+D) + cos(A-B-C+D) + cos(-A+B-C+D)

(putting E = A + B + C + D)

= cosE + cos(E - 2(A+B)) + cos(E - 2(B+C)) + cos(E - 2(C+A)).

So, in particular, tanAtanBtanCtanD = -1 if A+B+C+D = 180º and:
cos2(B+C) + cos2(C+A) + cos2(A+B) = -1;​

or if A+B+C+D = 90º and:
sin2(B+C) + sin2(C+A) + sin2(A+B)) = 0.​

(For example, 12º 24º 48º and 96º, because cos72º + cos120º + cos144º = cos72º - cos60º - cos36º = -1)​

can anyone find a simpler proof
or some geometric explanation for this? :smile:
 
  • #8
10
0
so how did you solve it? any hints?
I got this from the following equations:
2 sin A sin B= cos (A-B)-cos(A+B)
2 cos A cos B=cos (A-B)+cos(A+B)

Divide the (i) by (ii)
to obtain the result :)
 

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