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anonymous
 one year ago
help!
anonymous
 one year ago
help!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436166754136:dw

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1What do you need help solving

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1Do you know the relationship between cosx and sinx?

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1The 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436167048145:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436167321407:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what does this mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sin x =1/3 Find cos x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the question is that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't get what it's asking me??

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1\[cosx = 2\sqrt{2}/3\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Sinx = 1/3 that means the ratio opp/hyp 1/3, so we can make a triangle as followingdw: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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0DOES X MEAN INVERSE?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Huh? No they put x there to represent it's an unkown, and that's what you're solving for

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2dw:1436167875535:dw x is the adjacent side that will help you find cosx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh it shows me the answer is dw:1436167868096:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Well, lets see if its right

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2We'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 \[1^2+b^2 = 3^2 \implies b^2= 3^21^2 \implies b=\sqrt{3^21^2} = \sqrt{91} = \sqrt{8}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2So 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
 one year ago
Best ResponseYou've already chosen the best response.2So yes, it simplifies to \[\frac{ 2\sqrt{2} }{ 3 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @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 \[1^2+b^2 = 3^2 \implies b^2= 3^21^2 \implies b=\sqrt{3^21^2} = \sqrt{91} = \sqrt{8}\] \(\color{blue}{\text{End of Quote}}\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Something not make sense there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait i'm trying to understand.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2No worries, take your time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, the pythagorean identity in terms of trig functions is written: \[ \sin^2x+\cos^2x =1 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0omg so many people viewing my question.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2Haha, don't get overwhelmed @Mindblast3r just see if you understand everything, if not ask questions

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0@Mindblast3r I know how you feel. I had 7 people watching me when I was tutoring sometimes. It was nervewrecking.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You know, you can really only find \(\cos x\). You need more info to know the sign.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436168642801:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0that is from the hypotenuse

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0hypotenuse is the longest side of the triangle which was 3 cos x = adjacent / hypotenuse

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \cos x = \sqrt{1  \sin^2 x} \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so when you were using the Pythagorean theorem, you were solving for one of the sides of the triangle.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2dw:1436168854387:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1@UsukiDoll I am sorry :(

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2What usuki said :), we were just solving for the adjacent side, I should've specified, then we used our cosx ratio adj/hyp

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2where hyp = 3 and adj = sqrt(8)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436168916911:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0@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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436168957305:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1( ͡° ͜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
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0my spelling and grammar is on vacation xD oy!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The formula \[ \sin^2x+\cos^2x=1 \]Is based on the Pythagorean theorem.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2He's not there yet wio lol

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0If 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436168987012:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436169101210:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1problem is over nw party time ( ͡^ ͜ʖ ͡^)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ (\text{adjacent})^2 + (\text{opposite})^2 = (\text{hypotenuse})^2 \\ \iff \frac{(\text{adjacent})^2}{(\text{hypotenuse})^2} + \frac{ (\text{opposite})^2 }{(\text{hypotenuse})^2} = 1 \\ \iff \left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)^2 + \left(\frac{\text{opposite}}{\text{hypotenuse}}\right)^2 = 1 \\ \iff \cos^2x+\sin^2x = 1 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436169251275:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Lol, I think usuki, Lyn, and qwerty should get a big thanks to :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436169321629:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1i met a man who said  "i have two daughters and both are girls."

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436169398594:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok people shows over, do something else.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Haha ok, bye :). And a way I go... dw:1436169509148:dw
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