Mindblast3r
  • Mindblast3r
help!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mindblast3r
  • Mindblast3r
|dw:1436166754136:dw|
Compassionate
  • Compassionate
What do you need help solving
Mindblast3r
  • Mindblast3r
find cosx

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Compassionate
  • Compassionate
Do you know the relationship between cosx and sinx?
Mindblast3r
  • Mindblast3r
no
LynFran
  • LynFran
|dw:1436174086784:dw|
imqwerty
  • imqwerty
cosx = (root8)/3
Compassionate
  • Compassionate
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
Mindblast3r
  • Mindblast3r
|dw:1436167048145:dw|
Mindblast3r
  • Mindblast3r
x=2?
LynFran
  • LynFran
|dw:1436174284089:dw|
Mindblast3r
  • Mindblast3r
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.
Mindblast3r
  • Mindblast3r
when i say,
Mindblast3r
  • Mindblast3r
|dw:1436167321407:dw|
Mindblast3r
  • Mindblast3r
what does this mean?
Mindblast3r
  • Mindblast3r
sin x =1/3 Find cos x
Mindblast3r
  • Mindblast3r
the question is that.
Mindblast3r
  • Mindblast3r
i don't get what it's asking me??
imqwerty
  • imqwerty
\[cosx = 2\sqrt{2}/3\]
Astrophysics
  • Astrophysics
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?
Mindblast3r
  • Mindblast3r
DOES X MEAN INVERSE?
Astrophysics
  • Astrophysics
Huh? No they put x there to represent it's an unkown, and that's what you're solving for
Astrophysics
  • Astrophysics
|dw:1436167875535:dw| x is the adjacent side that will help you find cosx
Mindblast3r
  • Mindblast3r
ohh it shows me the answer is |dw:1436167868096:dw|
Astrophysics
  • Astrophysics
Well, lets see if its right
Astrophysics
  • 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 \[1^2+b^2 = 3^2 \implies b^2= 3^2-1^2 \implies b=\sqrt{3^2-1^2} = \sqrt{9-1} = \sqrt{8}\]
Astrophysics
  • Astrophysics
So that's our x on the triangle |dw:1436168040178:dw| now we're looking for cosx right, \[cosx = \frac{ adj }{ hyp } = \frac{ \sqrt{8} }{ 3 }\]
Astrophysics
  • Astrophysics
So yes, it simplifies to \[\frac{ 2\sqrt{2} }{ 3 }\]
Mindblast3r
  • Mindblast3r
\(\color{blue}{\text{Originally Posted by}}\) @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 \[1^2+b^2 = 3^2 \implies b^2= 3^2-1^2 \implies b=\sqrt{3^2-1^2} = \sqrt{9-1} = \sqrt{8}\] \(\color{blue}{\text{End of Quote}}\)
Astrophysics
  • Astrophysics
Something not make sense there?
Mindblast3r
  • Mindblast3r
wait i'm trying to understand.
Astrophysics
  • Astrophysics
No worries, take your time
anonymous
  • anonymous
Yeah, the pythagorean identity in terms of trig functions is written: \[ \sin^2x+\cos^2x =1 \]
Mindblast3r
  • Mindblast3r
omg so many people viewing my question.
UsukiDoll
  • UsukiDoll
@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"
Astrophysics
  • Astrophysics
Haha, don't get overwhelmed @Mindblast3r just see if you understand everything, if not ask questions
UsukiDoll
  • UsukiDoll
@Mindblast3r I know how you feel. I had 7 people watching me when I was tutoring sometimes. It was nervewrecking.
anonymous
  • anonymous
You know, you can really only find \(|\cos x|\). You need more info to know the sign.
Mindblast3r
  • Mindblast3r
|dw:1436168642801:dw|
Mindblast3r
  • Mindblast3r
come*
UsukiDoll
  • UsukiDoll
that is from the hypotenuse
UsukiDoll
  • UsukiDoll
hypotenuse is the longest side of the triangle which was 3 cos x = adjacent / hypotenuse
anonymous
  • anonymous
\[ |\cos x| = \sqrt{1 - \sin^2 x} \]
UsukiDoll
  • UsukiDoll
so when you were using the Pythagorean theorem, you were solving for one of the sides of the triangle.
Astrophysics
  • Astrophysics
|dw:1436168854387:dw|
imqwerty
  • imqwerty
@UsukiDoll I am sorry :(
Astrophysics
  • Astrophysics
What usuki said :), we were just solving for the adjacent side, I should've specified, then we used our cosx ratio adj/hyp
Astrophysics
  • Astrophysics
where hyp = 3 and adj = sqrt(8)
Mindblast3r
  • Mindblast3r
|dw:1436168916911:dw|
UsukiDoll
  • UsukiDoll
@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
Mindblast3r
  • Mindblast3r
|dw:1436168957305:dw|
Astrophysics
  • Astrophysics
Haha
imqwerty
  • imqwerty
( ͡° ͜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
UsukiDoll
  • UsukiDoll
@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.
UsukiDoll
  • UsukiDoll
my spelling and grammar is on vacation xD oy!
anonymous
  • anonymous
The formula \[ \sin^2x+\cos^2x=1 \]Is based on the Pythagorean theorem.
Astrophysics
  • Astrophysics
He's not there yet wio lol
UsukiDoll
  • UsukiDoll
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.
Mindblast3r
  • Mindblast3r
|dw:1436168987012:dw|
Astrophysics
  • Astrophysics
Yes
UsukiDoll
  • UsukiDoll
yes @Mindblast3r :D
imqwerty
  • imqwerty
so very true
Mindblast3r
  • Mindblast3r
|dw:1436169101210:dw|
imqwerty
  • imqwerty
problem is over nw party time ( ͡^ ͜ʖ ͡^)
anonymous
  • anonymous
\[ (\text{adjacent})^2 + (\text{opposite})^2 = (\text{hypotenuse})^2 \\ \iff \frac{(\text{adjacent})^2}{(\text{hypotenuse})^2} + \frac{ (\text{opposite})^2 }{(\text{hypotenuse})^2} = 1 \\ \iff \left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)^2 + \left(\frac{\text{opposite}}{\text{hypotenuse}}\right)^2 = 1 \\ \iff \cos^2x+\sin^2x = 1 \]
Mindblast3r
  • Mindblast3r
|dw:1436169251275:dw|
Astrophysics
  • Astrophysics
Lol, I think usuki, Lyn, and qwerty should get a big thanks to :)
Mindblast3r
  • Mindblast3r
|dw:1436169321629:dw|
imqwerty
  • imqwerty
( ͡° ͜ʖ ͡°)
imqwerty
  • imqwerty
i met a man who said - "i have two daughters and both are girls."
Mindblast3r
  • Mindblast3r
|dw:1436169398594:dw|
Mindblast3r
  • Mindblast3r
ok people shows over, do something else.
Astrophysics
  • Astrophysics
Haha ok, bye :). And a way I go... |dw:1436169509148:dw|
Mindblast3r
  • Mindblast3r
lOL!
Mindblast3r
  • Mindblast3r
Now that's funny!

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