How to prove (tan12)(tan48)(tan54)(tan72) = 1 without using a calculator?

we have

tanθ tan(60° - θ) tan (60° + θ) = tan3θ.

( refer to http://schoolnotes4u.blogspot.in/search/label/1.3-... for the proof

put θ =12 to get

tan 12 tan(60 - 12) tan (60 + 12) = tan 3 * 12

=> tan 12 tan 48 tan 72 = tan 36

=> tan 12 tan 48 tan 72 = cot (90-36) = cot 54 = 1/ tan 54

=> (tan12)(tan48)(tan54)(tan72) = 1

proved

edit : it is not for my solution but for pinkgreen

8x^3-4x+1=0 is satisfied by sin 18

because

sin 72° = 2 sin 36° cos 36° by the double angle relationship.

sin 72° = 4 sin 18° cos 18° (1 - 2sin2 18°) by the double angle relationship, again.

cos 18° = 4 sin 18° cos 18° (1 - 2sin2 18°) by the cofunction properties: sin 72° = cos 18°.

1 = 4 sin 18° (1 - 2sin2 18°) Let x = sin 18°, this is known as

1 = 4x(1-2x2) substitution, a useful technique in calculus.

8x3-4x+1 = 0

so sin 18 is exact root and why pinkgreen considers approximate is beyond my comprehension.

then the solution become complete

Let x = 12 degrees ==> 3x = 36, 5x = 60

==> cos(3x) = cos(36), cos(5x) = 1/2

Let y = 3x ==> cos(y) = cos36, cos(4y) = cos144

==> cos(4y) + cos(y) = cos144 + cos36 = cos(180 - 36) + cos36

==> cos(4y) + cos(y) = cos144 + cos36 = -cos(36) + cos36

==> cos(4y) + cos(y) = cos144 + cos36 = 0

==> cos(4y) + cos(y) = 0

==> 2cos²(2y) - 1 + cos(y) = 0

==> 2 [ 2cos²(y) - 1 ]² + cos(y) - 1 = 0

==> 8cos⁴(y) - 8cos²(y) + cos(y) + 1 = 0, let z = cos(y)

==> 8z^4 - 8z^2 + z + 1 = 0

Trial and error shows that z = -1, z = 1/2 are the roots to the above equation

==> (z - 1)(2z - 1)(4z^2 - 2z - 1) = 0

By zero product property, solve for z gives z = -1, 1/2, (1/4)(1 - √5), (1/4)(1 + √5)

But since z = cos(y) = cos(36), 0 < z < 1, thus z = (1/4)(1 + √5) is the only solution

Thus cos(36) = (1 + √5)/4 ................. (*)

==> sec(36) = 4/(1 + √5)

==> sec²(36) = 8/(3 + √5) = 2(3 - √5), by rationalizing denominator

==> sec²(36) - 1 = 5 - 2√5

==> tan²(36) = 5 - 2√5

==> 1 - tan²(36) = 2√5 - 4

==> 1/(1 - tan²(36)) = (2 + √5)/2 ................. (**)

tanx * tan4x

= (sin4x * sinx)/(cos4x * cosx)

= (2sin4x * sinx)/(2cos4x * cosx)

= [cos(3x) - cos(5x)] / [cos(5x) + cos(3x) ], by Product to Sum formulae

= [ (1 + √5)/4 - (1/2) ] / [ (1/2) + (1 + √5)/4 ]

= [ (1 + √5) - 2 ] / [ 2 + (1 + √5) ]

= (√5 - 1)/(√5 + 3), then rationalize denominator

= √5 - 2 ................. (***)

Hence

tan12 * tan48 * tan54 * tan72

= tan(x) * tan(4x) * 1/tan(90 - 54) * tan(6x)

= tan(x) * tan(4x) * 1/tan(36) * tan(6x)

= tan(x) * tan(4x) * 1/tan(3x) * tan(6x)

= tan(x) * tan(4x) * 1/tan(3x) * 2tan(3x)/(1 - tan²(3x)), by Double Angle Formula

= 2 * tan(x) * tan(4x) * 1/(1 - tan²(3x))

= 2 * (√5 - 2) * (2 + √5)/2

= 1

@Kali Prasad: ahaha! Nice~

convinced, I loved math in college, till I got here to school this year. we've now a large form of Mathamatica crap, and its all on-line homework. the instructor reads the examples from the e book, so now i have no theory what i'm doing. :/ next semester will be exciting, seeing as I quite managed to bypass Calc I and am now going into Calc II... humorous seeing that i have continuously had As in math.

x=tan12*tan48*tan54*tan72

=tan48*tan12*cot36*cot18

=sin48/cos48*sin12/cos12*cos36/sin36*cos18/sin18

x*cos48*cos12*sin36*sin18=sin48*sin12*cos36*cos18

x*1/2*(cos60+cos36)*sin18*sin36=1/2*(cos36-cos60)*cos18*cos36

x/2*(cos60*sin18+cos36*sin18)*cos54=1/2*(cos36*cos18-cos60*cos18)*sin54

x/2*(1/2*sin18+1/2*sin54-1/2*sin18)*cos54=1/2*(1/2*cos54+1/2*cos18-1/2*cos18)*sin54

x/2*1/2*sin54*cos54=1/2*1/2*cos54*sin54

x/4*sin54*cos54=1/4*sin54*cos54

Thus x=1

LHS=

sin(12*)sin(48*)sin(54*)sin(72*)/

[cos(12*)cos(48*)cos(54*)cos(72*)]=

(-2)sin(12*)sin(48*)(-2)sin(54*)

sin(72*)/[2cos(12*)cos(48*)

2cos(54*)cos(72*)]=

[cos(60*)-cos(36*)][(cos(126*)-

cos(18*)]/{[cos(60*)+cos(36*)]

[cos(126*)+cos(18*)]}=

[0.5-cos(36*)][-sin(36*)-cos(18*)]/

{[0.5+cos(36*)][-sin(36*)+cos(18*)]}=

[0.5-(1-2sin^2(18*)][-2sin(18*)-1]/

{[0.5+(1-2sin^2(18*)][-2sin(18*)+1]}=

[-0.5+2sin^2(18*)][2sin(18*)+1]/

{[1.5-2sin^2(18*)][2sin(18*)-1]}=

1 =

RHS approximately.

The reason is as follows:

Let sin(18*)=x, and let

(-0.5+2x^2)(2x+1)/

[(1.5-2x^2)(2x-1)]=1

then

-x-0.5+4x^3+2x^2=3x-1.5-4x^3+2x^2

=>

8x^3-4x+1=0

=>

The equation has 3 real roots one of

which is x=0.3090169=sin(18*)

approximately.

This shows that there is no error during

the proof, but the given product on the left

of the equation is approximately=1 only.

wow

you probably need to apply double-angle, triple-angle and half-angle formulae to work this out.