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$$\cos 2 \alpha = \frac{c}{mc} = \frac{1}{m} = c^2 - s^2 = cos^2 _\alpha - sin^2 \alpha$$

$$\sin 2 \alpha = \frac{s+ms}{mc} = ( 1 + \frac{1}{m} ) \frac{s}{c} = ( c^2+s^2+c^2-s^2 ) \frac{s}{c}$$

$$ = 2 s c = 2 \sin \alpha \cos$$

$$\frac{n}{1} = \frac{ms+mns}{2s} ,{2n} = {m+mn}$$

$$= 4c^2 - 2c^2 - 2s^2 - 1 = 4c^2 - 3.$$

$$\cos 3\alpha = \frac{c}{n} = 4c^3 - 3c = 4\cos^3\alpha - 3\cos\alpha$$

$$\sin 3\alpha = \frac{s(1+m+mn)}{n} ,$$

$$= s \left \{ \frac{1}{n} + m ( {1} + \frac{1}{n} ) \right \} = s \left ( \frac{1}{n} + m*\frac{2}{m} \right )$$

$$= s \left ( \frac{1}{n} + m*\frac{2}{m} \right )

= s \left ( \frac{1}{n} + 2 \right ) = s (4c^2 - 1),$$

$$ = s (3 - 4s^2),$$

$$ = 3 \sin \alpha - 4 \sin^3$$

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$$\frac{mo}{1} = \frac {mn+mno}{1+m},

o ( 1 + m - n ) = n, \frac{1}{o} = \frac{1}{n} + \frac{m}{n} - 1 = $$

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If the vertical angle $$\mathrm{A}{O}{E},$$ of a triangle be quadrisected by lines which meet the base in $$\mathrm{B},\mathrm{C},$$ and $$\mathrm{D}$$, so that $$\mathrm{A},\mathrm{B},\mathrm{C},\mathrm{D},$$and $$\mathrm{E}$$ are in alfabetical order, then the segments of the base satisfy the following relation:

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Since $$\nabla POP^2$$ is right angled at $$ P $$ $$m^2 c^2 = c^2 + ( s + ms )^2 = 0$$ $$m^2 ( c^2 - s^2 ) - 2ms^2 - 1 = 0$$ $${m ( c^2 - s^2 ) - 1 } { m + 1 } = 0$$

$$\left \{ m ( c^2 - s^2 ) - 1 \right \}$$

$$\left \{ m ( c^2 - s^2 ) - 1 \right \} \left \{ m + 1 \right \} = 0 $$

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$$OP$$ is $$\perp QP $$ and angles $$ QOP, POP_1, P_1OP_2, P_2OP_3$$ and so on are each equal to $$\alpha$$

-- Choose $$OQ$$ as unit lenght.

Then $$OP = \cos \alpha \equiv c,$$

and $$PQ = PP_1 = \sin \alpha \equiv s,$$ and $$ c^2 + s^2 = 1 $$

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Another method for deriving $$\sin 2\alpha \sin 3\alpha$$ and so on.

We assume the theorem that the bisector of the vertical angle of a triangle divides the base into segments that are proportional to the adjacent sides.

In the figure $$OP$$ is $$\perp QP $$ and angles $$ QOP, POP_1, P_1OP_2, P_2OP_3$$ and so on are each equal to $$\alpha$$

Choose $$OQ$$ as unit length.

Then $$OP = \cos \alpha \equiv c,$$

and $$PQ = PP_1 = \sin alpha \equiv s,$$ and $$ c^2 + s^2 = 1 $$

Since $$ OP_1$$ bisects angle $$POP_2,$$ we may, using the above theorem,

denote $$OP_2$$ and $$P_1P_2$$

by $$mc$$ and $$ms$$ where $$m$$ remains to be determined.

Using the same theorem, we may denote,

$$OP_3$$ and $$ P_2P_3$$ by $$n$$ and $$mns,$$

$$OP_4$$ and $$P^3P^4$$ by $$moc$$ and $$ mnos$$, and so on.