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Question Number 129063 by benjo_mathlover last updated on 12/Jan/21
If sin x−cos x = (1/5) then { ((sin x=?)),((cos x=?)) :}
Ifsinx−cosx=15then{sinx=?cosx=?
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))) :}
(sinx−cosx)2=125⇒1−2sinxcosx=125let{p=5sinxq=5cosxthenwehave{p−q=1pq=12fromthefirstequationq=p−1sothesecondequationbecomesp2−p−12=0thatthis{p=−3⇒q=−4p=4⇒q=3thus{sinx=−35andcosx=−45sinx=45andcosx=35
Answered by benjo_mathlover last updated on 12/Jan/21
Answered by MJS_new last updated on 12/Jan/21
x=2arctan t ⇒ sin x =((2t)/(t^2 +1))∧cos x =−((t^2 −1)/(t^2 +1)) ((2t)/(t^2 +1))+((t^2 −1)/(t^2 +1))=(1/5) t^2 +(5/2)t−(3/2)=0 t=−3∨t=(1/2) ⇒ sin x =−(3/5)∧cos x =−(4/5) ∨ sin x=(4/5)∧cos x =(3/5)
x=2arctant⇒sinx=2tt2+1∧cosx=−t2−1t2+12tt2+1+t2−1t2+1=15t2+52t−32=0t=−3∨t=12⇒sinx=−35∧cosx=−45∨sinx=45∧cosx=35

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