SugarRainbow
find the value of the expression if sin beta= 0.45, find cos(pi/2-beta)
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SugarRainbow
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do i just do pi/2-0.45?
lgbasallote
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sin beta is 0.45...but beta is not 0.45 does that make sense?
lgbasallote
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\[\Large \cos (\frac \pi 2 - \beta) \implies \cos \frac \pi 2 \cos \beta + \sin \frac \pi 2 \sin \beta\]
are you familiar with this?
SugarRainbow
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no idk what you just did
lgbasallote
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it's a trig formula \[\LARGE \cos (A - B ) \implies \cos A \cos B + \sin A \sin B\]
lgbasallote
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in this case, our A is pi/2 and B is beta
make sense?
SugarRainbow
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yeah but whats beta?
Algebraic!
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|dw:1347254638009:dw|
lgbasallote
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ahh we'll get there
Algebraic!
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already did.
lgbasallote
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so you get how \[\cos(\frac \pi 2 - \beta) = \cos \frac \pi 2 \cos \beta + \sin \frac \pi 2 \sin \beta\]
Algebraic!
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answer is clearly .45
lgbasallote
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i was talking to sugarrainbow
Skaematik
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Is it just not cos(pi/2 - sin^-1(0.45))?
SugarRainbow
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umm kay i don't get algebraic's picture and i not sure how igbasallote's formula works if we don't know what beta is
lgbasallote
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it works because \[\cos \frac \pi 2 = 0\]
\[\sin \frac \pi 2 = 1\]
lgbasallote
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so if you substitute you get \[\implies (0) \cos \beta + (1) \sin \beta\]
SugarRainbow
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what about what Skaematik did? can't i just do that?
SugarRainbow
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and how come we add cos+sin?
Skaematik
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This is a formula that you will need to memorise. Its something that you can use to manipulate trignometric equations to find an answer
cos(A−B)⟹cosAcosB+sinAsinB
Therefore if you use it in this case, like lgbasallote said
cos(π2−β)=cosπ/2cosβ+sinπ/2sinβ
Plug these values into your calculator
cosπ/2
sinπ/2
And you will find they equal to 0 and 1
Therefore cos(π2−β)=0+1*sinβ
Sinβ =0.45
Therefore the answer is 0.45
tanjung
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cos(pi/2-beta) = sin beta
SugarRainbow
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okay that makes more sense but what would the formula be if i had tangent(pi/2-beta)=-5.32 and i have to find cot?
SugarRainbow
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when i did pi/2 i got 0.5
tanjung
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tangent (pi/2-beta) still in the first kuadrant, so impossible has a negative value and tangent (pi/2-beta)=cotangent beta
SugarRainbow
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okay so is that what cosine is or how do we get that
tanjung
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maybe ur problem is if given tangent(pi/2-beta)=5.32, find cot beta?
cot beta = tan(pi/2-beta) = 5.32
Algebraic!
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|dw:1347256883382:dw|
Algebraic!
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cotangent of beta = 1/-5.32
SugarRainbow
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how did you get -5.32
Algebraic!
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see beta?
Algebraic!
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what's the tangent? opp = -5.32 adjacent =1
Algebraic!
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tan(beta) = -5.32
and
cot(beta) = 1/-5.32
SugarRainbow
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but how did you get that if we have sin and cos?
Algebraic!
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what problem are you working on?
SugarRainbow
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oh i'm sorry i was still looking at the first one
SugarRainbow
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can we just go back to the first one cuz i still don't understand it
Algebraic!
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sure, look carefully at the sketch plz
Algebraic!
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|dw:1347257483078:dw|
Algebraic!
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do you agree? sin(beta) = .45/1
Algebraic!
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|dw:1347257532660:dw|
Algebraic!
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opposite over hypotenuse is sine
SugarRainbow
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okay that makes sense
Algebraic!
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cool:)
SugarRainbow
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but now how do i get cos?
Algebraic!
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now:
|dw:1347257619039:dw|
what's that new angle?
Algebraic!
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the whole angle from the x axis to the y axis is 90 or pi/2 ....
Algebraic!
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so that angle must be pi/2 - Beta
SugarRainbow
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does this have anything to do with the unit circle
Algebraic!
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now look at the cosine of that angle:
|dw:1347257708887:dw|
Algebraic!
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yeah, it all is the same thing..
Algebraic!
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see that triangle. one side is .45 the hypotenuse is 1 the other side, no one cares what it is!
Algebraic!
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|dw:1347257807464:dw|
Algebraic!
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the cosine of that new angle is clearly .45 /1 .45 is adjacent, 1 is the hypotenuse.
Algebraic!
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so cos(pi/2 -Beta) is .45
Algebraic!
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it's what @tanjung said :
cos(pi/2-beta) = sin beta
but I showed it graphical so you could believe!
you can't really argue with that picture.
Algebraic!
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basically if two angles are complimentary, the cosine of one angle is the sin of the other.... ie the side that is opposite on one triangle *must be* the side that is adjacent on the other triangle...
Algebraic!
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|dw:1347258175266:dw|
Algebraic!
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|dw:1347258227158:dw|
Algebraic!
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|dw:1347258316959:dw|
SugarRainbow
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umm okay is it also because its sin and cos that its the same?
SugarRainbow
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real quick, since pi/2=1 then would i just divide .45(sine beta) by 1 to get cos?
Algebraic!
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yeah, I guess I get what you're saying... it comes from the definition of sin as opp/hyp and cosine as adj/hyp when you draw those sides for two angles which are complimentary they make a rectangle
Algebraic!
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pi/2 != 1
Algebraic!
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pi/2 = pi/2
SugarRainbow
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umm okay so thinking of it in terms of a rectangle that means cos is the same as sin because the two sides are equal?|dw:1347258695721:dw|
Algebraic!
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ah the hypotenuse =1 you mean?
SugarRainbow
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yeah
Algebraic!
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cosine isn't the same as sine.... that's not what I said, but yeah you're starting to get it, I think...
Algebraic!
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your pic is accurate
Algebraic!
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you see that, in your pic:
|dw:1347258903616:dw|
those two angles aren't the same?
they do add up to give pi/2 (90 degrees) though
so the cosine of one angle = the sine of the other:)
SugarRainbow
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oh so its only the same cuz it the angle adds up to 90?
Algebraic!
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no, it works elsewhere on the circle. but just try to grasp this 1st quadrant sketch for now... then you can move on to more advanced stuff :)
SugarRainbow
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umm okay i do get how you got hyp=1 and opp=.45 but what bout the angles how did you get that they'd = 90?
Algebraic!
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well remember, because if Beta is one of the angles and the other is pi/2 - Beta then Beta + (pi/2 - Beta) = pi/2
SugarRainbow
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but what's pi/2?
Algebraic!
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so we're automatically talking about two angles that add up to pi/2.
Algebraic!
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pi/2 is 90 deg.
Algebraic!
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1/4 the way around a circle.
SugarRainbow
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oh i just put it in my calc and i see how you got 90 but how does that necessarily help us with this because a bunch of different combinations of angles can add up to 90
Algebraic!
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that's good actually. that means it works no matter what Beta is. It's general. For any Beta.
Algebraic!
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|dw:1347259786709:dw|
Algebraic!
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|dw:1347259882928:dw|
SugarRainbow
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okay
Algebraic!
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still talking about first quadrant stuff at the moment..
Algebraic!
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anyway hope that cleared it up. think it over a bit, you'll see why it has to be true.