Find the length of the perpendicular from the origin to the straight line joining the two points whose coordinates are $(a c o s α, a s i n α)$ and $(a c o s β, a s i n β)$

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Solution

Line formed from joining $(a c o s α, a s i n α)$ and $(a c o s β, a s i n β)$

$\Rightarrow y - α s i n β$

$= \frac{a s i n β - a s i n α}{a c o s β - a c o s α} \times x - a c o s β$

$y - α s i n β$

$= \frac{2 s i n (\frac{β - α}{2}) c o s (\frac{β + α}{2})}{- 2 s i n (\frac{β - α}{2}) s i n (\frac{β + α}{2})} \times (x - a) c o s β$

$\Rightarrow y - α s i n β = - c o t (\frac{β + α}{2}) (x - a c o s β)$

$\Rightarrow y + c o t (\frac{β + α}{2}) x - a c o s β c o t (\frac{β + α}{2})$

$- a s i n β = 0$

Then, the length of perpendicular

$\Rightarrow ∣ ∣ ∣ ∣ ∣ \frac{0 (y) + 0 - a c o s β c o t (\frac{β + α}{2}) - a s i n β}{\sqrt{1 + c o t^{2} (\frac{α + β}{2})}} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow \frac{a c o s β c o t (\frac{α + β}{2}) + a s i n β}{c o s e c (\frac{α + β}{2})}$

$\Rightarrow a c o s β c o s (\frac{α + β}{2}) + a s i n β (\frac{α + β}{2})$

$\Rightarrow a c o s (\frac{α - β}{2})$

[Using $c o s A c o s B + s i n A s i n B = c o s (A - B)$ ]

Hence, proved.

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