help!

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|dw:1436166754136:dw|
What do you need help solving
find cosx

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Other answers:

Do you know the relationship between cosx and sinx?
no
|dw:1436174086784:dw|
cosx = (root8)/3
The relationship between the cosine and sine graphs is that the cosine is the same as the sine — only it’s shifted to the left by 90 degrees, or π/2
|dw:1436167048145:dw|
x=2?
|dw:1436174284089:dw|
i don't even understand what i'm asking to be honest but i have a test on this soon. so pls explain to me what i'm asking.
when i say,
|dw:1436167321407:dw|
what does this mean?
sin x =1/3 Find cos x
the question is that.
i don't get what it's asking me??
\[cosx = 2\sqrt{2}/3\]
Sinx = 1/3 that means the ratio opp/hyp -1/3, so we can make a triangle as following|dw:1436167626215:dw| now we need to figure out cosx, and remember the ratio of cosx is adj/hyp, so solve for the missing length using pythagorean theorem, and you will get your ratio, that make sense?
DOES X MEAN INVERSE?
Huh? No they put x there to represent it's an unkown, and that's what you're solving for
|dw:1436167875535:dw| x is the adjacent side that will help you find cosx
ohh it shows me the answer is |dw:1436167868096:dw|
Well, lets see if its right
We'll use \[a^2+b^2 = c^2\] where c is the hypotenuse. We'll let a = 1, b the side we are solving for, and c = 3 \[1^2+b^2 = 3^2 \implies b^2= 3^2-1^2 \implies b=\sqrt{3^2-1^2} = \sqrt{9-1} = \sqrt{8}\]
So that's our x on the triangle |dw:1436168040178:dw| now we're looking for cosx right, \[cosx = \frac{ adj }{ hyp } = \frac{ \sqrt{8} }{ 3 }\]
So yes, it simplifies to \[\frac{ 2\sqrt{2} }{ 3 }\]
\(\color{blue}{\text{Originally Posted by}}\) @Astrophysics We'll use \[a^2+b^2 = c^2\] where c is the hypotenuse. We'll let a = 1, b the side we are solving for, and c = 3 \[1^2+b^2 = 3^2 \implies b^2= 3^2-1^2 \implies b=\sqrt{3^2-1^2} = \sqrt{9-1} = \sqrt{8}\] \(\color{blue}{\text{End of Quote}}\)
Something not make sense there?
wait i'm trying to understand.
No worries, take your time
Yeah, the pythagorean identity in terms of trig functions is written: \[ \sin^2x+\cos^2x =1 \]
omg so many people viewing my question.
@imqwerty please refrain from giving direct answers because it's against OpenStudy's Code of Conduct. "Give Help, Not Answers I will encourage and guide those needing help, and not just give them an answer" "Don’t post only answers - guide the asker to a solution"
Haha, don't get overwhelmed @Mindblast3r just see if you understand everything, if not ask questions
@Mindblast3r I know how you feel. I had 7 people watching me when I was tutoring sometimes. It was nervewrecking.
You know, you can really only find \(|\cos x|\). You need more info to know the sign.
|dw:1436168642801:dw|
come*
that is from the hypotenuse
hypotenuse is the longest side of the triangle which was 3 cos x = adjacent / hypotenuse
\[ |\cos x| = \sqrt{1 - \sin^2 x} \]
so when you were using the Pythagorean theorem, you were solving for one of the sides of the triangle.
|dw:1436168854387:dw|
@UsukiDoll I am sorry :(
What usuki said :), we were just solving for the adjacent side, I should've specified, then we used our cosx ratio adj/hyp
where hyp = 3 and adj = sqrt(8)
|dw:1436168916911:dw|
@wio 's method is another way of solving this problem, but you should only use it if you know the trig identities...otherwise it's best to use the Pythagorean theorem and drawing a triangle
|dw:1436168957305:dw|
Haha
( ͡° ͜ʖ ͡°) u need cosx which is adjacent side/hypotenuse u knw that the hypotenuse is 3 nd nw u need adj. side u solve for adj. side by pythagorus theo. and plug it in the equation nd get answer=(root8)/3
@imqwerty that's ok...just remember next time that the main reason why OpenStudy exists is to provide guidance towards the question, not a massive online forum to trade answers.
my spelling and grammar is on vacation xD oy!
The formula \[ \sin^2x+\cos^2x=1 \]Is based on the Pythagorean theorem.
He's not there yet wio lol
If OP doesn't know those identities or if the problem specifies to use the Pythagorean theorem and draw the triangle method only, so be it.
|dw:1436168987012:dw|
Yes
so very true
|dw:1436169101210:dw|
problem is over nw party time ( ͡^ ͜ʖ ͡^)
\[ (\text{adjacent})^2 + (\text{opposite})^2 = (\text{hypotenuse})^2 \\ \iff \frac{(\text{adjacent})^2}{(\text{hypotenuse})^2} + \frac{ (\text{opposite})^2 }{(\text{hypotenuse})^2} = 1 \\ \iff \left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)^2 + \left(\frac{\text{opposite}}{\text{hypotenuse}}\right)^2 = 1 \\ \iff \cos^2x+\sin^2x = 1 \]
|dw:1436169251275:dw|
Lol, I think usuki, Lyn, and qwerty should get a big thanks to :)
|dw:1436169321629:dw|
( ͡° ͜ʖ ͡°)
i met a man who said - "i have two daughters and both are girls."
|dw:1436169398594:dw|
ok people shows over, do something else.
Haha ok, bye :). And a way I go... |dw:1436169509148:dw|
lOL!
Now that's funny!

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