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