Help me solve this please: \(\sf sin(cos^{-1} (-3/5))\)

- owlet

Help me solve this please: \(\sf sin(cos^{-1} (-3/5))\)

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- schrodinger

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- FibonacciChick666

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

- FibonacciChick666

(by order of ops)

- SolomonZelman

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

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## More answers

- SolomonZelman

using, the fact that:
sin²(w) + cos²(w) = 1
sin²(w) = 1 - cos²(w)
sin(w) = \(\sqrt{1 - \cos^2(w)}\)
in this case w is: \(\cos^{-1}(-3/5)\)
(That makes it even easier for you)

- SolomonZelman

And also, you should know that:
\(\large\color{black}{ \displaystyle \left[\cos(\cos^{-1}\theta)\right]=\theta }\)

- owlet

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

- SolomonZelman

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

- Loser66

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

- FibonacciChick666

@SolomonZelman nice use of identity, but it's a special triangle. If one recognizes that it can be done in two steps

- SolomonZelman

Well, it is a number in our case, and I was assuming that just like by taking a square root of any ordinary number such as √35 you don't have a ±, so it would be here without a ±....

- SolomonZelman

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

- owlet

@FibonacciChick666 my first attempt, I also used special triangles and the answer that I got is -4/5 but it's not right

- Loser66

cos (theta) = negative number, then theta can be in Quadrant 2 or Quadrant 3. That is why when taking off the square root, you need +/-

- owlet

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

- Loser66

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

- Loser66

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

- SolomonZelman

owlet your answer, I think that should be positive.

- Loser66

|dw:1443625732989:dw|

- Loser66

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

- Loser66

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

- FibonacciChick666

What loser is saying is what I was getting at

- Loser66

|dw:1443625863982:dw|

- Loser66

hence sin (alpha) = -4/5 also

- Loser66

Your answer should be both. I think!!

- FibonacciChick666

I agree with Loser given no stipulations on quadrant

- SolomonZelman

i checked wolfram gives only 4/5 positive.
http://www.wolframalpha.com/input/?i=sin%28cos%5E%28-1%29+of+%28-3%2F5%29%29

- SolomonZelman

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

- Loser66

@SolomonZelman It's nice but to us (@FibonacciChick666 and me) we need the firm logic to make sure that we don't miss any cases on our argument. :)

- SolomonZelman

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

- owlet

i was doing the same thing as loser.. but yeah i think that the right answer is positive 4/5 but idk why the right is only positive

- Loser66

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

- owlet

okay. thanks :)

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