Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

I do not understand what comes after 4/5 and -3/5

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question