Prove-that-in-general-trisection-of-an-angle-is-impossible-with-only-ruler-and-compass-

Question Number 3467 by Rasheed Soomro last updated on 13/Dec/15

P r o v e t h a t i n g e n e r a l t r i s e c t i o n o f a n

a n g l e i s i m p o s s i b l e w i t h o n l y r u l e r

a n d c o m p a s s .

Commented by prakash jain last updated on 13/Dec/15

\sin 3 x = \sin (2 x + x) = \sin 2 x \cos x + \cos 2 x \sin x

= 2 \sin x (1 - \sin^{2} x) + (1 - {2 \sin}^{2} x) \sin x

= 3 \sin x - {4 \sin}^{3} x

If we able to construct trisect any angle

3 x .

If 3 x = 30 °, \sin 10 ° = y

\frac{1}{2} = 3 y - 4 y^{3}

8 y^{3} - 6 y - 1 = 0

2 y = u

u^{3} - 3 u + 1 = 0 \dots (A)

From rational roots theorem, \pm 1 are only

possible rational root which are not possible .

try = a + b \sqrt{c} (a, b, c \in Q), second root will be a - b \sqrt{c}

third root is say α .

(x - (a + b \sqrt{c})) (x - (a - b \sqrt{c}) (x - α) = u^{3} - 3 u + 1

- α (a + b \sqrt{c}) (a - b \sqrt{c}) = 1 \Rightarrow α is rational .

Since the equation has no rational roots,

it has no root which are of the form

a + b \sqrt{c} (a, b, c \in Q) .

Answered by prakash jain last updated on 13/Dec/15

As \sin 10 ° \notin Q and not of form a + b \sqrt{c}, \sin 10 ° cannot

be drawn using ruler and compass .

So 30 ° cannot be trisected .

Commented by prakash jain last updated on 13/Dec/15

However you can trisect 54 ° since \sin 18 ° = \frac{\sqrt{5} - 1}{4}