anonymous
  • anonymous
find the exact value of cos(sin^-1(3/5))
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Do you have a calculator?
anonymous
  • anonymous
I got \( \huge \frac{4}{5} \)
Michele_Laino
  • Michele_Laino
If \theta is such that: \[\Large \theta = \arcsin \left( {\frac{3}{5}} \right)\] then you have to compute this: \[\Large \cos \theta \]

Looking for something else?

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

More answers

anonymous
  • anonymous
were not suppose to use calculator :/
anonymous
  • anonymous
whats arc sin?
Michele_Laino
  • Michele_Laino
ok! now we have: \[\Large \sin \theta = \frac{3}{5}\] am I right?
Michele_Laino
  • Michele_Laino
it is another way to mean sin^-1
Michele_Laino
  • Michele_Laino
so, if we apply the fundamental identity: \[\Large {\left( {\cos \theta } \right)^2} + {\left( {\sin \theta } \right)^2} = 1\] we can write: \[\Large \cos \theta = \pm \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} = \pm \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = ...?\]
anonymous
  • anonymous
let 2 sin^-1(3/5) = A sin^-1(3/5) = A/2 sin(A/2) = 3/5 cos A = 1 - 2sin^2(A/2) = 1 - 2(9/25) = 7/25 Thus cos [ 2sin^-1(3/5)] = cos A = 7/25 I found this answer but where could they have gotten 9 from?
Michele_Laino
  • Michele_Laino
no, we have: cos(A/2) =sqrt(1 - sin^2(A/2)
anonymous
  • anonymous
wait so the way they did it is wrong?
Michele_Laino
  • Michele_Laino
yes! I think so!
Michele_Laino
  • Michele_Laino
it is not necessary to work with A/2. Please apply my procedure
Michele_Laino
  • Michele_Laino
I used +/- since we have two square roots
Michele_Laino
  • Michele_Laino
for example: \[\Large \sqrt 4 = \pm 2\]
anonymous
  • anonymous
Wait so how would your work look like then?
anonymous
  • anonymous
Because your procedure is confusing me even more lol
Michele_Laino
  • Michele_Laino
since we have: \[\Large \sin \theta = \frac{3}{5}\] then we can write: \[\Large \cos \theta = \pm \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} = \pm \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = ...?\]
anonymous
  • anonymous
is this a half angle formula?
Michele_Laino
  • Michele_Laino
no, it is not a bisection formula
Michele_Laino
  • Michele_Laino
hint: \[\Large \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = \sqrt {1 - \frac{9}{{25}}} = \sqrt {\frac{{25 - 9}}{{25}}} = ...?\]
anonymous
  • anonymous
Is there another way to do it? becauase our teacher didn't show us that formula.
Michele_Laino
  • Michele_Laino
Sincerely speaking, it is the unique way that I know
anonymous
  • anonymous
but whats wrong with the method i used? im still not understanding
Michele_Laino
  • Michele_Laino
your method is correct, nevertheless you have to find cos(A/2) not cos(A)
Michele_Laino
  • Michele_Laino
you have used the subsequent bisection formula: \[\Large {\left\{ {\sin \left( {\frac{\theta }{2}} \right)} \right\}^2} = \frac{{1 - \cos \theta }}{2}\]
anonymous
  • anonymous
This question just got worse so I used something else but idk if its riight..
Michele_Laino
  • Michele_Laino
the computation of your image is correct
anonymous
  • anonymous
So is 7/25 right then??
anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
we can write this: \[\Large \sin \theta = \frac{3}{5} \Rightarrow \cos \left( {2\theta } \right) = \frac{7}{{25}}\]
anonymous
  • anonymous
@Michele_Laino lol what does that mean tho? XD can you explain it?
Michele_Laino
  • Michele_Laino
it is simple: since \[\Large \sin \theta = \frac{3}{5}\] then: \[\Large \cos \theta = \pm \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} = \pm \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = \pm \frac{4}{5}\] Now we use the duplication formula: \[\Large \begin{gathered} \cos \left( {2\theta } \right) = {\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2} = \hfill \\ \hfill \\ = {\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2} = \frac{{16}}{{25}} - \frac{9}{{25}} = \frac{{16 - 9}}{{25}} = \frac{7}{{25}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
ohh alright its the double angle formula too!

Looking for something else?

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