Consider three points P≡$$($$−$$sin$$$$($$β−α$$)$$,−$$cos$$β$$)$$,Q≡$$($$$$cos$$$$($$β−α$$)$$,$$sin$$β$$)$$, and R≡$$($$$$cos$$$$($$β−α$$+$$θ$$)$$,$$sin$$$$($$β−θ$$))$$, where $$0$$

Question

Consider three points P≡$$($$−$$sin$$$$($$β−α$$)$$,−$$cos$$β$$)$$,Q≡$$($$$$cos$$$$($$β−α$$)$$,$$sin$$β$$)$$,  and  R≡$$($$$$cos$$$$($$β−α$$+$$θ$$)$$,$$sin$$$$($$β−θ$$))$$, where $$0$$<α,β,θ<π$$4$$. Then

Infinity Learn · Accepted Answer

P≡$$($$−$$sin$$$$($$β−α$$)$$,−$$cos$$β$$)$$≡(x1$$,$$y1)Q$$≡$$($$$$cos$$$$($$β−α$$),$$sin$$β$$)$$≡(x2$$,$$y2)R$$≡$$($$$$cos$$$$($$β−α$$+$$θ$$),$$sin$$$$($$β−θ$$))or$$ R≡(x2cos$$θ$$+x1sin$$θ,$$y2cos$$θ$$+y1sin$$θ$$)If$$ T≡$$(x2cos$$θ$$+x1sin$$θ$$cos$$θ$$+$$$$sin$$θ,$$y2cos$$θ$$+y1sin$$θ$$cos$$θ$$+$$$$sin$$θ$$), for some 'θ' then P,Q and T are collinear. ∴ P,Q,R are collinear if $$cos$$θ$$+$$$$sin$$θ$$=1$$, which is not possible as $$0$$<θ<π$$4$$.Hence, P,Q,R are $$non-collinear$$.

Consider three points
$\begin{array}{l} P \equiv (- \sin (β - α), - cosβ), \\ Q \equiv (\cos (β - α), sinβ), and R \equiv (\cos (β - α + θ), \sin (β - θ)), where 0 < α, β, θ < \frac{π}{4} . Then \end{array}$

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Consider three points P≡(−sin(β−α),−cosβ),Q≡(cos(β−α),sinβ), and R≡(cos(β−α+θ),sin(β−θ)), where 0<α,β,θ<π4. Then

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Consider three points
$\begin{array}{l} P \equiv (- \sin (β - α), - cosβ), \\ Q \equiv (\cos (β - α), sinβ), and R \equiv (\cos (β - α + θ), \sin (β - θ)), where 0 < α, β, θ < \frac{π}{4} . Then \end{array}$