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?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- Christos

- jamiebookeater

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

for one thing, the square root of 9 is 3

- anonymous

so if you were typing in an answer, it probably wanted it to look like
\[\cos(\theta)=-\frac{\sqrt{5}}{3}\]

- anonymous

other than that, it is all correct

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Christos

@satellite73 So I didnt make any mistakes at all?

- geerky42

No.

- Christos

@satellite73 confirm to me please if all of my steps above were correct, and even if they had a mistake which line?

- anonymous

just draw a triangle and with 0 = 2 and h = 3.. then find a then do a/h to get cos theta

- anonymous

|dw:1362625931832:dw|

- anonymous

there is no mistake, although you should have
\[\pm\sqrt{1-\left(\frac{2}{3}\right)^2}=\cos(\theta)\]

- Christos

I should have used another fundamental identity?

- anonymous

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

- Christos

got it its the same just modified, meeh

- anonymous

it is right, everything you did is right

- Christos

I know
why its negative yea

- Christos

problem is that My solution is different from the one on the book

- anonymous

so there is no mistake here, although you should have a 3 in the denominator, rather than the square root of 9

- anonymous

what did the book get?

- Christos

They the solutions might be equal but, I sill found a different solution

- Christos

5/9

- anonymous

a different solution, or a different method

- Christos

yea right..

- anonymous

nothing wrong with your method
answer is \(-\frac{\sqrt{5}}{3}\) for sure

- Christos

Books solution: 5/9

- anonymous

no that is \(\cos^2(\theta)\)

- anonymous

or else the book made a mistake
it happens
in any case you are right for sure, so don't fret about it

- Christos

what do you mean "no that is cos2(θ)" ?

- anonymous

\[\cos(\theta)=-\frac{\sqrt{5}}{3}\]
\[\cos^2(\theta)=\frac{5}{9}\]

- Christos

hmmmmm

- Christos

Is there any other method to approche the situation, like another identity to use for the specidif prob

- Christos

I AM SORRY the answer in my book was: -sqrt(5)/3 !!!!!

- anonymous

|dw:1362626751880:dw|

- Christos

So my final outcome is - sqrt(5)/sqrt3 HOW DO I PROCEED?

- Christos

- Christos

- Christos

- Christos

SOLVED THANK YOU ALL, SPECIAL THANKS TO SATELITE

Looking for something else?

Not the answer you are looking for? Search for more explanations.