Trigonometry Problem?

Question

Well I've done most of a trig problem, now it boils down to proving that cos36*cos72=1/4 (in degrees). Help anyone?

Answer 1

I assume you want to prove it, not just calculate it with your calculator. This boils down to finding the exact values of the angles.

Create isosceles triangle BAC. Angles B and C are 72° and angle A is 36°. A bisector of angle B runs meets the line AC at point D. Now we have two additional isosceles triangles.
Triangle CBD is 72°, 36°, 72°.
Triangle BAD is 36°, 108°, 36°

Then we have:
BC = BD = AD = 1
DC = x
AB = AC = x + 1

Using similar triangles BAC and CBD we have

DC/BC = BC/AC
x/1 = 1/(x + 1)
x(x + 1) = 1
x² + x - 1 = 0
x = [-1 ± √(1 + 4*1*1)] /2 = (-1 + √5)/2
The solution must be positive.

Using the Law of Cosines and triangle CBD we have:

(CD)² = (BD)² + (BC)² - 2*BD*BC*cos36°
x² = 1 + 1 - 2*1*1*cos36°
x² = 2 - 2cos36°
2cos36° = 2 - x²
cos36° = (2 - x²)/2 = 1 - x²/2

cos36° = 1 - [(-1 + √5)/2]²/2 = 1 - [(6 - 2√5)/4]/2
cos36° = 1 - (6 - 2√5)/8 = (2 + 2√5)/8 = (√5 + 1)/4

Use the double angle formula.

cos72° = 2cos²36° - 1 = 2*[(√5 + 1)/4]² - 1
cos72° = (6 + 2√5)/8 - 1 = (2√5 - 2)/8 = (√5 - 1)/4
____________

(cos36°)*(cos72°) = [(√5 + 1)/4] * [(√5 - 1)/4]

(cos36°)*(cos72°) = (5 - 1)/16 = 4/16 = 1/4

Answer 2

Cos36 = 0.809017
Cos72 = 0.309017
0.809017 x 0.309017 = 0.25 = 1/4

Answer is a number not measured in degrees.

Answer 3

i believe he means
cos 36° cos 72° = 1/4

you can search on google for
cos(36/180*pi) * cos(72/180*pi) ... it returns 0.25
____________________________________________

Let approach it this way ...
say 18° = x
so we want to find the value of
y = cos 2x * cos 4x

now 5x = 5*18° = 90°
we know,
sin 5x = 1
cos 5x = 0

cos 4x = cos (5x - x) = cos (90° - x) = sin x

so y = cos 2x sin x
= (1 - 2 sin² x ) *sin x

now we need to do is to find value of sin(x)

note: cos 5x = 0
cos 5x = cos(2x + 3x)
= cos 2x cos 3x - sin 2x sin 3x
= (2 cos² x -1)*(4 cos^3 x - 3 cos x) - 2 sin x cos x (3 sin x - 4 sin^3 x)
= cos x (2 cos² x -1)*( 4 cos² x - 3) - 2 sin² x cos x (3 - 4 sin² x)
= cos x* [ (2 cos² x - 1)*( 4 cos² x - 3) + 2 (cos² x -1)*(4 cos²x -1) ]
= cos x * [16 cos^4 x - 20 cos² x + 5]

Thus,
cos 5x = cos x * [16 cos^4 x - 20 cos² x + 5] = 0
now since cos x != 0
=> 16 cos^4 x - 20 cos² x + 5 = 0
put cos x = α
so we have,
16 α^4 - 20 α² + 5 = 0
α² = [20 ± √(400 - 320)]/32
= [5 ±√5]/8

now note that
α = cos x
we are trying to find sin x
sin x = √(1- cos² x) = √(1 - α²) = √[ (3 -√5)/8 ]

i neglected sin x = √[ (3 +√5)/8 ]. (why ?)

now,
y = (1 - 2 sin² x ) *sin x
substitute for sin x ... and get the desired result ... you should be able to do that ... and don't forget to answer the "why ?"

just for fun, try using sin 5x = 1 to arrive at the same result. it's not difficult if you know how to factorize ...

Answer 4

wtf is 1/4 in degrees? is it 80 degrees?
multibply cos36 with cos 72, thenit sohuld be 1/4=1/4