anonymous
  • anonymous
what is the principal value of cos^-1{cos 23pi/20}
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{17\pi}{20}\]
anonymous
  • anonymous
i think, let me check with a calculator
anonymous
  • anonymous
yeh its correct

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anonymous
  • anonymous
its an interesting method to solve , Ill do a quick picture now on paint and hopefully try to explain
anonymous
  • anonymous
ok..m waitin
anonymous
  • anonymous
Well, not the best picture , but good enough . Now, first thing to notice is that the answer is not 23pi / 20 , because 23pi/20 is not in the range of cos^-1
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anonymous
  • anonymous
we must get the angle into the interval 0
anonymous
  • anonymous
and 23pi/20 is a bit more than pi
anonymous
  • anonymous
hey can you tell me what is meant by principal value?? i dont know if i get the idea u knw
anonymous
  • anonymous
however, the cos graph is symmetrical about the minimum points , that is cos(23pi /20) does equal the same value as cos (17pi /20) because they are both the same distance from the minimums
anonymous
  • anonymous
principal value just means , "your answer must be in the range 0
anonymous
  • anonymous
if you really want to check your understanding then try \[\sin^{-1} (\sin(\frac{23\pi}{20}))\]
anonymous
  • anonymous
its a very similar idea
anonymous
  • anonymous
using translations of the graph , you get the angle into the range -pi/2 to pi/2 ( because the range of sin^-1 is -pi/2 to pi/2 )
anonymous
  • anonymous
then apply sin^-1(sin(x)) = x
anonymous
  • anonymous
hmm.. i understand what you are saying. i would like to know how you got the value 17pi/20? i know its at a distance pi/4 from 23pi/4. i wanna know if there is some other way to work it out.
anonymous
  • anonymous
there is other ways of working it our but I dont remember them at this point in time
anonymous
  • anonymous
hey is the prin. value for \sin^{-1} (\sin(\frac{23\pi}{20})) 13 pi/20?
anonymous
  • anonymous
\[\sin^{-1} (\sin(\frac{23\pi}{20})) \]
anonymous
  • anonymous
no
anonymous
  • anonymous
draw a picture of a sin graph this time
anonymous
  • anonymous
yea i did..
anonymous
  • anonymous
my picture is not good, but you get the idea
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anonymous
  • anonymous
23pi / 20 , is a bit more than pi , yes?
anonymous
  • anonymous
now, we need to get the angle 23pi/20 into the range -pi/2 <= <= pi/2
anonymous
  • anonymous
okk...
anonymous
  • anonymous
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anonymous
  • anonymous
isnt it at a distance pi/2 from it?
anonymous
  • anonymous
now, I know the graph of sin is symmetric about pi/2
anonymous
  • anonymous
so if I find the distance between 23pi/20 and pi/2 , then this is the distance I must go to the left and I will still have the equivalent value of the function
anonymous
  • anonymous
yup..
anonymous
  • anonymous
now, the distance between pi/2 and 23pi/ 20 = (23/20 - 1/2) pi = 13pi/20
anonymous
  • anonymous
this means that if I go 13pi/20 units left from pi/2 , then I would be at a point that is a reflection in the line x=pi/2
anonymous
  • anonymous
ie sin(23pi / 20 ) = sin ( pi/2 - 13pi/20 ) = sin ( -3pi/20 )
anonymous
  • anonymous
now, since the angle of the sin function is in the range of the invese sin function
anonymous
  • anonymous
I can apply sin^-1(sin(x) ) = x
anonymous
  • anonymous
so our answer is -3pi/20
anonymous
  • anonymous
oh..i understand now. thankyou so much=)
anonymous
  • anonymous
which can be checked on the calculator , put cal in radian mode , type in " sin^-1(sin(23pi /20) )"
anonymous
  • anonymous
with the brackets, and you will get something like -0.47 .... which is equal to -3pi/20
anonymous
  • anonymous
ya..got it

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