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WHERE IS MY MISTAKE?!?!? Problem: Given that sinƟ = 2/3 and Π/2 < Ɵ < Π find the exact value of cosƟ My solution: cos^2(Ɵ) + sin^2(Ɵ) = 1 sin^2(Ɵ) = 1 - cos^2(Ɵ) 4/9=1- cos^2(Ɵ) -5/9 = -cos^2(Ɵ) 5/9 = cos^2(Ɵ) cos(Ɵ) = -sqrt(5/9) Where is my mistake till now?

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for one thing, the square root of 9 is 3
so if you were typing in an answer, it probably wanted it to look like \[\cos(\theta)=-\frac{\sqrt{5}}{3}\]
other than that, it is all correct

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Other answers:

@satellite73 So I didnt make any mistakes at all?
@satellite73 confirm to me please if all of my steps above were correct, and even if they had a mistake which line?
just draw a triangle and with 0 = 2 and h = 3.. then find a then do a/h to get cos theta
there is no mistake, although you should have \[\pm\sqrt{1-\left(\frac{2}{3}\right)^2}=\cos(\theta)\]
I should have used another fundamental identity?
or if you write \[\frac{5}{9}=\cos^2(\theta)\] then \[\pm\frac{\sqrt{5}}{3}=\cos(\theta)\] the reason you know it is negative is because you are in quadrant 2
got it its the same just modified, meeh
it is right, everything you did is right
I know why its negative yea
problem is that My solution is different from the one on the book
so there is no mistake here, although you should have a 3 in the denominator, rather than the square root of 9
what did the book get?
They the solutions might be equal but, I sill found a different solution
a different solution, or a different method
yea right..
nothing wrong with your method answer is \(-\frac{\sqrt{5}}{3}\) for sure
Books solution: 5/9
no that is \(\cos^2(\theta)\)
or else the book made a mistake it happens in any case you are right for sure, so don't fret about it
what do you mean "no that is cos2(θ)" ?
\[\cos(\theta)=-\frac{\sqrt{5}}{3}\] \[\cos^2(\theta)=\frac{5}{9}\]
Is there any other method to approche the situation, like another identity to use for the specidif prob
I AM SORRY the answer in my book was: -sqrt(5)/3 !!!!!
So my final outcome is - sqrt(5)/sqrt3 HOW DO I PROCEED?

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