Question

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

Answer 1

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

Answer 2

(by order of ops)

Answer 3

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

Answer 4

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)

Answer 5

And also, you should know that:

\(\large\color{black}{ \displaystyle \left[\cos(\cos^{-1}\theta)\right]=\theta }\)

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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 ±....

Answer 11

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

Answer 12

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

Answer 13

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 +/-

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

owlet your answer, I think that should be positive.

Answer 18

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

Answer 19

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

Answer 20

What loser is saying is what I was getting at