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sin(arc cos(-0.2))=?

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Use a calculator?

[EDIT: sorry, a more helpful answer might be:

On a calculator you should have a cos-1 button. Put in -0.2, and press it (or the other way round, depending on what your calculator likes). Then press sin and you have your answer. I think.

Miles]

[edited by - Miles Clapham on September 10, 2002 5:51:22 PM]

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Whipping out my handy Ti-83+, I see that the answer is 0.9797958971...

Of course, physically finding the answer would be more complicated...

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Okay follow the steps

Sin (arc cos (-0.2) = x

Apply arc sin to both sides of equation , this becomes

Arc cos (-0.2) = arc sin (x) …………………… because arc sin (sin) = 1

Arc cos (-0.2) means “angle whose cosine is - 0.2” ,

from calculator arc cos (-0.2) = 101.5 degrees

therefore arc sin(x) = 101.5

apply sin to both sides

x = sin (101.5) ………………..again because sin (arc sin) = 1

from calculator again :

x=0.979 !!

cheers
moshe

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quote:
Original post by Fella
I know that But how do you solve this without a calculator?


Use maths tables , or Taylor series and lots of paper . I dunno. I don''t think it''s really possible without a calculator (or tables). Why don''t you want to use one?

Miles

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quote:
Original post by Beer Hunter
Remember that sin2 ¦¨ + cos2 ¦¨ = 1
From this, you can find that sin(arccos(x)) = sqrt(1 - x2)

Could you write the intermediary steps? I'm missing something, here.

Oh, and be quick, this thread is going to be closed soon

Cédric

[edited by - cedricl on September 10, 2002 7:06:39 PM]

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quote:
Original post by Beer Hunter
Remember that sin2 È + cos2 È = 1
From this, you can find that sin(arccos(x)) = sqrt(1 - x2)

And then, sqrt(1 - 0.22) = 0.979795897

/me adds a medal to his collection

[edited by - Beer Hunter on September 10, 2002 6:34:56 PM]


Hey, that''s really cool! Never seen it before.

Miles

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quote:
Original post by cedricl
Could you write the intermediary steps? I''m missing something, here.
Hmmm... I forgot the ± last time.. but anyway:

sin2 Θ + cos2 Θ = 1
sin2 Θ = 1 - cos2 Θ
sqrt(sin2 Θ) = ± sqrt(1 - cos2 Θ)
sin Θ = ± sqrt(1 - cos2 Θ)

let Θ = arccos(x)

sin(arccos(x)) = ± sqrt(1 - cos2(arccos(x)))
sin(arccos(x)) = ± sqrt(1 - x2)

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quote:
Original post by cedricl
Could you write the intermediary steps? I''m missing something, here.



You can do it geometrically. Draw a unit circle centre the origin. Draw the line x = -0.2. It crosses the circle at two points. The one we are interested in is in the positive y quadrant. It has coordinates (-0.2, y).

Draw the line from this point to the centre. The angle between this line and the x axis, A, has the property

cos (A) = -0.2

i.e.

A = arccos (-0.2)

so sin (arccos(-0.2)) is just sin (A), and from the diagram this is just y.

You can just draw the diagram as described in the first paragraph and measure y. Or use Pythagoras''s theorem to calculate it: the rihgt angled triangle is the one with the radius of the circle as it''s hypotenuse, giving

y2 + (0.2)2 = 1

and so y = sqrt (1 - (0.2)2) = sqrt (0.96)

(and I would say this IS game related: using such non-trig methods to solve problems involving trig is a big win in games where trig is too slow and innacurate to be widely used. And this is perhaps the single most useful method to know).

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Guest Anonymous Poster
Cosine is Adjacent over Hypotenues, right?

arccos(-0.2/1)

So the angle you are looking at has the hypotenues of 1, and the adjacent leg of -.02.

The opposite leg will be
H^2 - adjacent^2 = opposite^2
or

1 - .04 = .96

oppostie leg = sqrt(.96)

so the sine of the triangle is sqrt(.96)/1

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I would be interested to know what this specific problem has to do with game development. If its homework, a very quick question to the teacher or fellow student would have yielded the answer. But since trig is used in game development, and because the question was trivial, I don''t see any major harm in leaving it open.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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