# Homework Help: Product=1 homework question

1. May 4, 2008

### xphloem

1. The problem statement, all variables and given/known data

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

2. Relevant equations

tan (90-x)=cot x

3. The attempt at a solution
but these angles are not supplementary. Ho is this possible?

2. May 4, 2008

### symbolipoint

What were you expecting?

$$$(.1051)(.9004)(2.246)(4.7046) = 0.99993...$$$

That seems close enough to call "1".

3. May 4, 2008

### tiny-tim

Welcome to PF!

Hi xphloem! Welcome to PF!

How did you come across this?

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!

4. May 5, 2008

### kamerling

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. May 5, 2008

### xphloem

thanks all. I got this in a book called brain teasers. I have solved it myself. thanks for all the help!

6. May 5, 2008

### kamerling

so how did you solve it? any hints?

7. May 6, 2008

### tiny-tim

ooh, kamerling, that's clever!

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?

8. May 6, 2008

### xphloem

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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