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So let's start with what will you do first?

(by order of ops)

\(\large\color{black}{ \displaystyle \sin(\cos^{-1}x)=\sqrt{1-\left[\cos(\cos^{-1}x)\right]^2~~} }\)

wait i'm confused.. why did you take the sqrt ?

because I had sin²(quantity)
and I want to solve for just sin(quantity)

Support @SolomonZelman \(\pm \sqrt...\)

(Well, it is a single value expression, isn't it ? )

okay, I understand that part, but how am I going to apply it to solve for cos^-1 (-3/5)

\(cos (\alpha) = A \) that is \(cos^{-1} A = \alpha\), right?

Hence \(cos (\alpha) = -3/5= \dfrac{adj}{hypotenuse}\)

owlet your answer, I think that should be positive.

|dw:1443625732989:dw|

now, \(sin(\alpha) = \dfrac{opposite}{hypotenuse}= 4/5\)

However, when alpha is in Quadrant 3, you still have cos alpha = -3/5

What loser is saying is what I was getting at

|dw:1443625863982:dw|

hence sin (alpha) = -4/5 also

Your answer should be both. I think!!

I agree with Loser given no stipulations on quadrant

So it assumes that it is in the 2nd quadrant.

Sure, I am not against -4/5:)

You may have some restriction on the problem, like the angle is in Q2. or (0, pi) for domain.

okay. thanks :)