The points A $(0,0)$, B$(\cos(\alpha),\sin(\alpha))$ and C $(\cos(\beta),\sin(\beta))$ are the vertices of a right angled triangle.

Question

$(0,0)$, B$(\cos(\alpha),\sin(\alpha))$ and C $(\cos(\beta),\sin(\beta))$ are the vertices of a right angled triangle. Derive a relation between $\alpha$ and $\beta$." />

I tried using the slope method, basically that $m_{1}\times m_{2}=-1$ and I got the relation that:

$\cos(\alpha-\beta)=0$

Clearly $\alpha-\beta=\pi/2$, i.e. $\sin(\frac{\alpha-\beta}{2})=(\frac{1}{\sqrt{2}})$

But one case which I think should also be considered is $\alpha-\beta=\frac{3\pi}{2}$, which would result in a negative value for $\cos(\frac{\alpha-\beta}{2})$, but the answer to the referenced image doesn't consider it. Why? (The answer given is (A) and (C))

Answer 1

All four answers are correct.

Notice the problem says "if", not "if and only if", so an answer is correct if its inequality implies that the triangle is a right triangle.

You are correct that the triangle is a right triangle if and only if $\cos(\alpha-\beta) = 0$.

We have the double angle formulas

$$ \cos(\alpha-\beta) = 2 \cos^2 \frac{\alpha-\beta}{2} - 1 $$ $$ \cos(\alpha-\beta) = 1 - 2 \sin^2 \frac{\alpha-\beta}{2} $$

By the first,

$$ \cos \frac{\alpha-\beta}{2} = \pm \frac{1}{\sqrt{2}} \implies \cos(\alpha-\beta) = 0 $$

By the second,

$$ \sin \frac{\alpha-\beta}{2} = \pm \frac{1}{\sqrt{2}} \implies \cos(\alpha-\beta) = 0 $$

So each of the four conditions is sufficient to show that the triangle is a right triangle. Of course, only two of them can be true for a single scenario, so these are not necessary conditions.