what is the principal value of cos^-1{cos 23pi/20}

- anonymous

what is the principal value of cos^-1{cos 23pi/20}

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- anonymous

\[\frac{17\pi}{20}\]

- anonymous

i think, let me check with a calculator

- anonymous

yeh its correct

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- anonymous

its an interesting method to solve , Ill do a quick picture now on paint and hopefully try to explain

- anonymous

ok..m waitin

- 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

we must get the angle into the interval 0

- anonymous

and 23pi/20 is a bit more than pi

- anonymous

hey can you tell me what is meant by principal value?? i dont know if i get the idea u knw

- 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

principal value just means , "your answer must be in the range 0

- anonymous

if you really want to check your understanding then try \[\sin^{-1} (\sin(\frac{23\pi}{20}))\]

- anonymous

its a very similar idea

- 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

then apply sin^-1(sin(x)) = x

- 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

there is other ways of working it our
but I dont remember them at this point in time

- anonymous

hey is the prin. value for \sin^{-1} (\sin(\frac{23\pi}{20})) 13 pi/20?

- anonymous

\[\sin^{-1} (\sin(\frac{23\pi}{20})) \]

- anonymous

no

- anonymous

draw a picture of a sin graph this time

- anonymous

yea i did..

- anonymous

my picture is not good, but you get the idea

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- anonymous

23pi / 20 , is a bit more than pi , yes?

- anonymous

now, we need to get the angle 23pi/20 into the range -pi/2 <= <= pi/2

- anonymous

okk...

- anonymous

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- anonymous

isnt it at a distance pi/2 from it?

- anonymous

now, I know the graph of sin is symmetric about pi/2

- 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

yup..

- anonymous

now, the distance between pi/2 and 23pi/ 20 = (23/20 - 1/2) pi = 13pi/20

- 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

ie sin(23pi / 20 ) = sin ( pi/2 - 13pi/20 ) = sin ( -3pi/20 )

- anonymous

now, since the angle of the sin function is in the range of the invese sin function

- anonymous

I can apply sin^-1(sin(x) ) = x

- anonymous

so our answer is -3pi/20

- anonymous

oh..i understand now. thankyou so much=)

- anonymous

which can be checked on the calculator , put cal in radian mode , type in " sin^-1(sin(23pi /20) )"

- anonymous

with the brackets, and you will get something like -0.47 ....
which is equal to -3pi/20

- anonymous

ya..got it

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