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Question Number 6762 by 314159 last updated on 24/Jul/16

Commented by Yozzii last updated on 24/Jul/16

$x_{1}^{2} + x_{2}^{2} + 2 x_{1} x_{2} c o s θ = 1,$ $y_{1}^{2} + y_{2}^{2} + 2 y_{1} y_{2} c o s θ = 1,$ $x_{1} y_{1} + x_{2} y_{2} + (x_{1} y_{2} + x_{2} y_{1}) c o s θ = 0 .$ $P r o v e t h a t x_{1}^{2} + y_{1}^{2} = {c o s e c}^{2} θ .$
Commented by Yozzii last updated on 24/Jul/16
$x_1 ^2 +x_2 ^2 +2x_1 x_2 cosθ=1......................(1) y_1 ^2 +y_2 ^2 +2y_1 y_2 cosθ=1.......................(2) x_1 x_2 +y_1 y_2 +(x_1 y_2 +y_1 x_2 )cosθ=0......(3) or x_1 x_2 +y_1 y_2 =−(x_1 y_2 +y_1 x_2 )cosθ for (3). We start by finding x_2 in terms of x_1 in equation (1). (x_2 +x_1 cosθ)^2 +x_1 ^2 =x_2 ^2 +2x_1 x_2 cosθ+x_1 ^2 +x_1 ^2 cos^2 θ (x_2 +x_1 cosθ)^2 +x_1 ^2 =1+x_1 ^2 cos^2 θ (x_2 +x_1 cosθ)^2 =1−x_1 ^2 sin^2 θ x_2 =−x_1 cosθ±(√(1−x_1 ^2 sin^2 θ)) Similarly, y_2 in terms of y_1 for equation (2) is given by y_2 =−y_1 cosθ±(√(1−y_1 ^2 sin^2 θ)). −−−−−−−−−−−−−−−−−−−−−−−−−−− Working at the second form of equation (3) term by term with x_2 and y_2 substituted leads to the following results. (y_1 x_2 +x_1 y_2 )cosθ=−2x_1 y_1 cos^2 θ±(y_1 (√(1−x_1 ^2 sin^2 θ))+x_1 (√(1−y_1 ^2 sin^2 θ)))cosθ −(y_1 x_2 +x_1 y_2 )cosθ=2x_1 y_1 cos^2 θ∓k {Let k=(y_1 (√(1−x_1 ^2 sin^2 θ))+x_1 (√(1−y_1 ^2 sin^2 θ)))cosθ} and x_2 y_2 =x_1 y_1 cos^2 θ∓{x_1 (√(1−y_1 ^2 sin^2 θ))+y_1 (√(1−x_1 ^2 sin^2 θ))}cosθ+(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ))) x_2 y_2 =x_1 y_1 +x_1 y_1 cos^2 θ∓k+(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ))) ∴ second form of (3) becomes, after cancellation of ∓k from both sides, x_1 y_1 +x_1 y_1 cos^2 θ+(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ)))=2x_1 y_1 cos^2 θ x_1 y_1 (1+cos^2 θ−2cos^2 θ)+(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ)))=0 x_1 y_1 (1−cos^2 θ)+(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ)))=0 x_1 y_1 sin^2 θ=−(√((1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ))) ⇒(1−x_1 ^2 sin^2 θ)(1−y_1 ^2 sin^2 θ)=x_1 ^2 y_1 ^2 sin^4 θ 1−x_1 ^2 sin^2 θ−y_1 ^2 sin^2 θ+x_1 ^2 y_1 ^2 sin^4 θ=x_1 ^2 y_1 ^2 sin^4 θ 1−(x_1 ^2 +y_1 ^2 )sin^2 θ=0 sin^2 θ=(1/(x_1 ^2 +y_1 ^2 ))⇒x_1 ^2 +y_1 ^2 =cosec^2 θ This result assumes that in x_2 and y_2 that both ± signs take the same signs. −−−−−−−−−−−−−−−−−−−−−−−−−−−− Suppose that ± takes different signs between x_2 and y_2 ; i.e x_2 =−x_1 cosθ+(√(1−x_1 ^2 sin^2 θ)) y_2 =−y_1 cosθ−(√(1−y_1 ^2 sin^2 θ)). You can show still that cosec^2 θ=x_1 ^2 +y_1 ^2 . ////$
$x_{1}^{2} + x_{2}^{2} + 2 x_{1} x_{2} c o s θ = 1 . . . . . . . . . . . . . . . . . . . . . . (1)$ $y_{1}^{2} + y_{2}^{2} + 2 y_{1} y_{2} c o s θ = 1 . . . . . . . . . . . . . . . . . . . . . . . (2)$ $x_{1} x_{2} + y_{1} y_{2} + (x_{1} y_{2} + y_{1} x_{2}) c o s θ = 0 . . . . . . (3)$ $o r x_{1} x_{2} + y_{1} y_{2} = - (x_{1} y_{2} + y_{1} x_{2}) c o s θ f o r (3) .$ $W e s t a r t b y f i n d i n g x_{2} i n t e r m s o f x_{1} i n e q u a t i o n (1) .$ ${(x_{2} + x_{1} c o s θ)}^{2} + x_{1}^{2} = x_{2}^{2} + 2 x_{1} x_{2} c o s θ + x_{1}^{2} + x_{1}^{2} {c o s}^{2} θ$ ${(x_{2} + x_{1} c o s θ)}^{2} + x_{1}^{2} = 1 + x_{1}^{2} {c o s}^{2} θ$ ${(x_{2} + x_{1} c o s θ)}^{2} = 1 - x_{1}^{2} {s i n}^{2} θ$ $x_{2} = - x_{1} c o s θ \pm \sqrt{1 - x_{1}^{2} {s i n}^{2} θ}$ $S i m i l a r l y, y_{2} i n t e r m s o f y_{1} f o r e q u a t i o n (2) i s g i v e n b y$ $y_{2} = - y_{1} c o s θ \pm \sqrt{1 - y_{1}^{2} {s i n}^{2} θ} .$ $- - - - - - - - - - - - - - - - - - - - - - - - - - -$ $W o r k i n g a t t h e s e c o n d f o r m o f e q u a t i o n (3) t e r m b y t e r m$ $w i t h x_{2} a n d y_{2} s u b s t i t u t e d l e a d s t o t h e f o l l o w i n g r e s u l t s .$ $(y_{1} x_{2} + x_{1} y_{2}) c o s θ = - 2 x_{1} y_{1} {c o s}^{2} θ \pm (y_{1} \sqrt{1 - x_{1}^{2} {s i n}^{2} θ} + x_{1} \sqrt{1 - y_{1}^{2} {s i n}^{2} θ}) c o s θ$ $- (y_{1} x_{2} + x_{1} y_{2}) c o s θ = 2 x_{1} y_{1} {c o s}^{2} θ \mp k {L e t k = (y_{1} \sqrt{1 - x_{1}^{2} {s i n}^{2} θ} + x_{1} \sqrt{1 - y_{1}^{2} {s i n}^{2} θ}) c o s θ}$ $a n d$ $x_{2} y_{2} = x_{1} y_{1} {c o s}^{2} θ \mp {x_{1} \sqrt{1 - y_{1}^{2} {s i n}^{2} θ} + y_{1} \sqrt{1 - x_{1}^{2} {s i n}^{2} θ}} c o s θ + \sqrt{(1 - x_{1}^{2} {s i n}^{2} θ) (1 - y_{1}^{2} {s i n}^{2} θ)}$ $x_{2} y_{2} = x_{1} y_{1} + x_{1} y_{1} {c o s}^{2} θ \mp k + \sqrt{(1 - x_{1}^{2} {s i n}^{2} θ) (1 - y_{1}^{2} {s i n}^{2} θ)}$ $∴ s e c o n d f o r m o f (3) b e c o m e s, a f t e r c a n c e l l a t i o n o f \mp k f r o m b o t h s i d e s,$ $x_{1} y_{1} + x_{1} y_{1} {c o s}^{2} θ + \sqrt{(1 - x_{1}^{2} {s i n}^{2} θ) (1 - y_{1}^{2} {s i n}^{2} θ)} = 2 x_{1} y_{1} {c o s}^{2} θ$ $x_{1} y_{1} (1 + {c o s}^{2} θ - 2 {c o s}^{2} θ) + \sqrt{(1 - x_{1}^{2} {s i n}^{2} θ) (1 - y_{1}^{2} {s i n}^{2} θ)} = 0$ $x_{1} y_{1} (1 - {c o s}^{2} θ) + \sqrt{(1 - x_{1}^{2} {s i n}^{2} θ) (1 - y_{1}^{2} {s i n}^{2} θ)} = 0$ $x_{1} y_{1} {s i n}^{2} θ = - \sqrt{(1 - x_{1}^{2} {s i n}^{2} θ) (1 - y_{1}^{2} {s i n}^{2} θ)}$ $\Rightarrow (1 - x_{1}^{2} {s i n}^{2} θ) (1 - y_{1}^{2} {s i n}^{2} θ) = x_{1}^{2} y_{1}^{2} {s i n}^{4} θ$ $1 - x_{1}^{2} {s i n}^{2} θ - y_{1}^{2} {s i n}^{2} θ + x_{1}^{2} y_{1}^{2} {s i n}^{4} θ = x_{1}^{2} y_{1}^{2} {s i n}^{4} θ$ $1 - (x_{1}^{2} + y_{1}^{2}) {s i n}^{2} θ = 0$ ${s i n}^{2} θ = \frac{1}{x_{1}^{2} + y_{1}^{2}} \Rightarrow x_{1}^{2} + y_{1}^{2} = {c o s e c}^{2} θ$ $T h i s r e s u l t a s s u m e s t h a t i n x_{2} a n d y_{2}$ $t h a t b o t h \pm s i g n s t a k e t h e s a m e s i g n s .$ $- - - - - - - - - - - - - - - - - - - - - - - - - - - -$ $S u p p o s e t h a t \pm t a k e s d i f f e r e n t s i g n s$ $b e t w e e n x_{2} a n d y_{2}; i . e$ $x_{2} = - x_{1} c o s θ + \sqrt{1 - x_{1}^{2} {s i n}^{2} θ}$ $y_{2} = - y_{1} c o s θ - \sqrt{1 - y_{1}^{2} {s i n}^{2} θ} .$ $Y o u c a n s h o w s t i l l t h a t {c o s e c}^{2} θ = x_{1}^{2} + y_{1}^{2} . / / / /$
	Answered by Yozzii last updated on 24/Jul/16

	$C h e c k c o m m e n t s f o r a s o l u t i o n I s h a r e d .$
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