kaylala
  • kaylala
TOPIC: Trigonometry GIVEN THE FOLLOWING CONDITIONS: sin A = 4/5, A Σ I sin B = -3/5, B Σ III Find the value of: I. sin (A +- B) II. cos (A +- B) III. tan (A +- B)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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kaylala
  • kaylala
@Callisto
kaylala
  • kaylala
@eliassaab
anonymous
  • anonymous
I do not understand what comes after 4/5 and -3/5

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kaylala
  • kaylala
let me fix the question
anonymous
  • anonymous
Also look at this summary of trig formulas, your problem follows easily http://www.math.missouri.edu/escgi/mucgi-bin/trigidentities.cgi
anonymous
  • anonymous
It is still not fixed
kaylala
  • kaylala
it is. see this pic that i will post. it's where i got the question. wait
anonymous
  • anonymous
You need these formulae sin(a + b ) = sin(a) cos(b) + sin(b) cos(a) sin(a - b ) = sin(a) cos(b) - sin(b) cos(a) cos(a + b ) = cos(a) cos(b) - sin(a) sin(b) cos(a - b ) = cos(a) cos(b) + sin(a) sin(b)
kaylala
  • kaylala
oh it's (+_) the plus on top of the minus for I - III
anonymous
  • anonymous
You do first + than -
anonymous
  • anonymous
You need six answers
anonymous
  • anonymous
Where is the picture?
kaylala
  • kaylala
wait. i just took a picture of it
kaylala
  • kaylala
1 Attachment
Yttrium
  • Yttrium
@eliassaab , I think the information beside 4/5 and -3/5 are the position of angle in the cartesian plane. (specifically Quadrant I for
kaylala
  • kaylala
@eliassaab here's the picture
kaylala
  • kaylala
and yes those are values for the carthesian plane @Yttrium
kaylala
  • kaylala
i dont think i have to find 6 answers @eliassaab
kaylala
  • kaylala
somebody. anyone. please help
Yttrium
  • Yttrium
You really have to find 6 ansers @kaylala. Let me tell you this: \[\sin (A \pm B)\] is the notation for\[\sin (A+B)\] and \[\sin(A-B)\] I hope you understand now.
kaylala
  • kaylala
oh okay i see now @Yttrium
kaylala
  • kaylala
i just dont know how to start this
Yttrium
  • Yttrium
let's start with angle A
Yttrium
  • Yttrium
the condition says sinA = 4/5, where it is in quadrant 1. we know that sinx = opp/hyp Hence, the figure much look like this. |dw:1391508636187:dw|
kaylala
  • kaylala
okay
Yttrium
  • Yttrium
|dw:1391508787128:dw| let's say that adjacent side of A is x. Can you solve x now?
kaylala
  • kaylala
okay x=sqrt-9 ??? @Yttrium
Yttrium
  • Yttrium
isn't is sqrt(+9)? which is 3?
kaylala
  • kaylala
oh my bad
kaylala
  • kaylala
it do is 3
Yttrium
  • Yttrium
Yea it is. And, with that you can now actually solve for cosA, tanA.
kaylala
  • kaylala
how?
kaylala
  • kaylala
i really dont know how to do this please guide me @Yttrium
Yttrium
  • Yttrium
cosA = adj/hyp tanA = opp/adj did you forget that?
kaylala
  • kaylala
sorta
kaylala
  • kaylala
cos A = 3/5 ???
kaylala
  • kaylala
tanA=4/3
kaylala
  • kaylala
did i get it?
kaylala
  • kaylala
what do i do next?
Yttrium
  • Yttrium
You got this. I think it is better if you write sinA, cosA, and tanA in a piece of paper. They are really important later. Next thing we are going to do is to solve for cosB and tanB
kaylala
  • kaylala
yeah. just wrote it down already how do i solve for that?
kaylala
  • kaylala
y=4
kaylala
  • kaylala
? @Yttrium
Yttrium
  • Yttrium
you sure? isn't it -4? take note it is in the quadrant III.
kaylala
  • kaylala
okay so the quadrant determines the sign? right?
Yttrium
  • Yttrium
Yeah. for (x,y) quadrant I (+,+) II (-,+) III (-,-) IV (+,-) Now. Find cosB and tanB. Be careful with the signs.
kaylala
  • kaylala
okay then
Yttrium
  • Yttrium
Solve for cosB and tanB
kaylala
  • kaylala
cos B= -4/5
kaylala
  • kaylala
tanB= 3/-4
Yttrium
  • Yttrium
Ohhh wait!!!
kaylala
  • kaylala
did i get it?
kaylala
  • kaylala
ookay
Yttrium
  • Yttrium
the graph will be like this |dw:1391510194742:dw|
Yttrium
  • Yttrium
x then is -4 solve again for cosB and tanB
kaylala
  • kaylala
i see now but the answer for cos b and tan B still remains?
kaylala
  • kaylala
so are those the answers already?
kaylala
  • kaylala
what to do next?
kaylala
  • kaylala
@Yttrium
Yttrium
  • Yttrium
ooops tanB = -3/-4 = 3/4
kaylala
  • kaylala
oh you're right
Yttrium
  • Yttrium
Now is the time to evaluate them. I. sin (A +- B) = sinAcosB +- cosAsinB II. cos (A +- B) = cosAcos -+ sinAsinB
Yttrium
  • Yttrium
III. \[\tan(A \pm B) = \frac{ tanA \pm tanB }{ 1\mp tanAtanB }\]
kaylala
  • kaylala
how's that? do i just substitute what i got? to the given formulas?
Yttrium
  • Yttrium
Yeah. All the operation found upside are the only operations used when applying upside identities. Look at this. \[\tan(A +B) = \frac{ tanA + tanB }{ 1 - tanAtanB }\]
kaylala
  • kaylala
so sin(a+b) = 1 sin(a-b) = -7/25 ??? @Yttrium
kaylala
  • kaylala
cos (a+b)=0 cos (a-b)=-24/25 did i get it correctly @Yttrium ?
Yttrium
  • Yttrium
i'll just check the sin function, you take care of the others :D sin(a+b) = -1
kaylala
  • kaylala
okay :) thanks
kaylala
  • kaylala
tan (A+B)= undefined??? is it? please check this @Yttrium
Yttrium
  • Yttrium
what is tanA and tanbB?
kaylala
  • kaylala
tanA=4/3 tanB=3/4
Yttrium
  • Yttrium
Yeah. That's undefined. And that is normal. :))
kaylala
  • kaylala
i see. okay thanks again :)
Yttrium
  • Yttrium
No problem :))

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