owlet
  • owlet
Help me solve this please: \(\sf sin(cos^{-1} (-3/5))\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
FibonacciChick666
  • FibonacciChick666
So let's start with what will you do first?
FibonacciChick666
  • FibonacciChick666
(by order of ops)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sin(\cos^{-1}x)=\sqrt{1-\left[\cos(\cos^{-1}x)\right]^2~~} }\)

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SolomonZelman
  • 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
  • SolomonZelman
And also, you should know that: \(\large\color{black}{ \displaystyle \left[\cos(\cos^{-1}\theta)\right]=\theta }\)
owlet
  • owlet
wait i'm confused.. why did you take the sqrt ?
SolomonZelman
  • SolomonZelman
because I had sin²(quantity) and I want to solve for just sin(quantity)
Loser66
  • Loser66
Support @SolomonZelman \(\pm \sqrt...\)
FibonacciChick666
  • FibonacciChick666
@SolomonZelman nice use of identity, but it's a special triangle. If one recognizes that it can be done in two steps
SolomonZelman
  • 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
  • SolomonZelman
(Well, it is a single value expression, isn't it ? )
owlet
  • 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
  • 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
  • owlet
okay, I understand that part, but how am I going to apply it to solve for cos^-1 (-3/5)
Loser66
  • Loser66
\(cos (\alpha) = A \) that is \(cos^{-1} A = \alpha\), right?
Loser66
  • Loser66
Hence \(cos (\alpha) = -3/5= \dfrac{adj}{hypotenuse}\)
SolomonZelman
  • SolomonZelman
owlet your answer, I think that should be positive.
Loser66
  • Loser66
|dw:1443625732989:dw|
Loser66
  • Loser66
now, \(sin(\alpha) = \dfrac{opposite}{hypotenuse}= 4/5\)
Loser66
  • Loser66
However, when alpha is in Quadrant 3, you still have cos alpha = -3/5
FibonacciChick666
  • FibonacciChick666
What loser is saying is what I was getting at
Loser66
  • Loser66
|dw:1443625863982:dw|
Loser66
  • Loser66
hence sin (alpha) = -4/5 also
Loser66
  • Loser66
Your answer should be both. I think!!
FibonacciChick666
  • FibonacciChick666
I agree with Loser given no stipulations on quadrant
SolomonZelman
  • 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
  • SolomonZelman
So it assumes that it is in the 2nd quadrant.
Loser66
  • 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
  • SolomonZelman
Sure, I am not against -4/5:)
owlet
  • 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
  • Loser66
You may have some restriction on the problem, like the angle is in Q2. or (0, pi) for domain.
owlet
  • owlet
okay. thanks :)

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