- anonymous

Find the exact value of the following expression (multiple choice)

- jamiebookeater

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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.

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- anonymous

http://tigger.uic.edu/~calkafka/spring2011math121practiceexam3.pdf - number 17

- anonymous

method or answers?

- anonymous

method. i'll figure out the answer from the method..that way, i learn the method and figure the answer

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## More answers

- anonymous

or calculator?

- anonymous

k

- anonymous

number 1. think of a number (angle) between 0 and pi whose cosine is \[\frac{\sqrt{3}}{2}\]

- anonymous

look at trig cheat sheet here
http://tutorial.math.lamar.edu/cheat_table.aspx if you do not remember

- mathteacher1729

Drawing pictures is SUPER helpful. The trig sheet Satellite mentioned is also very nice. :)

- anonymous

oh i only nueed help on number 17. thanks though

- anonymous

aaahhhhhhhh

- mathteacher1729

Still,l a pic would help.
Lemmie sketch out one... brb.

- anonymous

addition angle formula:
\[sin(a+b)=sin(a)cos(b)+cos(a)sin(b)\]

- mathteacher1729

You won't even have to do that, I think, cuz you have to do the inverse sin & cos stuff inside the main sin argument.

- anonymous

thats what i thought..at math teacher

- anonymous

i just dont know how to go about solving it

- anonymous

typeset is goofy but i assume it means
\[sin(sin^{-1}(\frac{2}{3}) + cos^{-1}(\frac{1}{3}))\]

- anonymous

yuppers. thats what it means

- anonymous

put \[a=sin^{-1}(\frac{2}{3}), b=cos^{-1}(\frac{1}{3})\]

- anonymous

use formula above. you aready know
\[sin(sin^{-1}(\frac{2}{3}))=\frac{2}{3}\]

- anonymous

you need \[cos^{-1}(\frac{2}{3})=\frac{\sqrt{5}}{3}\] by pythagoras

- anonymous

oops typo

- anonymous

you need \[cos(sin^{-1}(\frac{2}{3}))=\frac{\sqrt{5}}{3}\]

- anonymous

by pythagoras.

- mathteacher1729

Here is the first part.

##### 1 Attachment

- anonymous

ahh the pythagorean thm

- anonymous

what math teacher said. you only need to find
\[sin(cos^{-1}(\frac{1}{3}))=\frac{\sqrt{2}}{3}\]

- anonymous

now you have everything you need to plug in to the "addition angle" formula

- anonymous

but i think you must use the formula now. you have all 4 numbers that you need.
\[sin(a)=\frac{2}{3}\]
\[cos(a)=\frac{\sqrt{5}}{3}\]

- anonymous

\[sin(b)=\frac{\sqrt{2}}{3}\]
\[cos(b)=\frac{1}{3}\]

- anonymous

write out the formula, substitute the numbers, and be done.

- anonymous

right. gimme a sec.

- anonymous

whoooooooooooops

- anonymous

\[sin(b)=\frac{\sqrt{8}}{3}\]

- anonymous

fraid both math teacher and i made a mistake.

- anonymous

its okay. i'm still working on it. thanks

- anonymous

if you look at the picture math teachers sent unfortunately the second triangle is wrong. adjacent side is 1, hypotenuse is 3, opposite side should be \[{\sqrt{8}}=2\sqrt{2}\]

- anonymous

get
\[\frac{2}{3}\times \frac{\sqrt{5}}{3}+\frac{1}{3}\times \frac{2\sqrt{2}}{3}\]

- anonymous

is the answer for number 17 "A"? thats what i'm getting. http://tigger.uic.edu/~calkafka/spring2011math121practiceexam3.pdf

- mathteacher1729

Oh my gosh, I can't believe I did that. :( here is the corrected version.

##### 1 Attachment

- anonymous

dont worry mathteacher. satellite clarified. i'm debating on whether its A or B for number 17 in the link i posted

- anonymous

A yes!

- anonymous

SWEET!
thanky you so much!

- anonymous

welcome

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