If sin x−cos x = (1/5) then { ((sin x=?)),((cos x=?)) :}


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Question Number 129063 by benjo_mathlover last updated on 12/Jan/21

$If \sin x - \cos x = \frac{1}{5}$ $then {\begin{cases} \sin x = ? \\ \cos x = ? \end{cases}$
	Answered by liberty last updated on 12/Jan/21
	$(sin x−cos x)^2 =(1/(25)) ⇒1−2sin x cos x = (1/(25)) let { ((p=5sin x)),((q=5cos x)) :} then we have { ((p−q=1)),((pq=12)) :} from the first equation q=p−1 so the second equation becomes p^2 −p−12=0 that this { ((p=−3⇒q=−4)),((p=4⇒q=3)) :} thus { ((sin x=−(3/5) and cos x=−(4/5))),((sin x=(4/5) and cos x=(3/5))) :}$
	${(\sin x - \cos x)}^{2} = \frac{1}{25} \Rightarrow 1 - 2 \sin x \cos x = \frac{1}{25}$ $let {\begin{cases} p = 5 \sin x \\ q = 5 \cos x \end{cases} t h e n we have {\begin{cases} p - q = 1 \\ p q = 12 \end{cases}$ $f r o m t h e f i r s t equation q = p - 1 so the$ $second equation becomes p^{2} - p - 12 = 0$ $t h a t t h i s {\begin{cases} p = - 3 \Rightarrow q = - 4 \\ p = 4 \Rightarrow q = 3 \end{cases}$ $t h u s {\begin{cases} \sin x = - \frac{3}{5} a n d \cos x = - \frac{4}{5} \\ \sin x = \frac{4}{5} a n d \cos x = \frac{3}{5} \end{cases}$
	Answered by benjo_mathlover last updated on 12/Jan/21

	Answered by MJS_new last updated on 12/Jan/21

	$x = 2 \arctan t \Rightarrow \sin x = \frac{2 t}{t^{2} + 1} \land \cos x = - \frac{t^{2} - 1}{t^{2} + 1}$ $\frac{2 t}{t^{2} + 1} + \frac{t^{2} - 1}{t^{2} + 1} = \frac{1}{5}$ $t^{2} + \frac{5}{2} t - \frac{3}{2} = 0$ $t = - 3 \lor t = \frac{1}{2}$ $\Rightarrow \sin x = - \frac{3}{5} \land \cos x = - \frac{4}{5}$ $\lor$ $\sin x = \frac{4}{5} \land \cos x = \frac{3}{5}$
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