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

- anonymous

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

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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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