At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

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

i think, let me check with a calculator

yeh its correct

ok..m waitin

we must get the angle into the interval 0

and 23pi/20 is a bit more than pi

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

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

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

its a very similar idea

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

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

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

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

no

draw a picture of a sin graph this time

yea i did..

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

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

okk...

isnt it at a distance pi/2 from it?

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

yup..

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

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

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

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

so our answer is -3pi/20

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

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

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

ya..got it