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anonymous
 one year ago
find the exact value of cos(sin^1(3/5))
anonymous
 one year ago
find the exact value of cos(sin^1(3/5))

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you have a calculator?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got \( \huge \frac{4}{5} \)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2If \theta is such that: \[\Large \theta = \arcsin \left( {\frac{3}{5}} \right)\] then you have to compute this: \[\Large \cos \theta \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0were not suppose to use calculator :/

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! now we have: \[\Large \sin \theta = \frac{3}{5}\] am I right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it is another way to mean sin^1

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so, 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
 one year ago
Best ResponseYou've already chosen the best response.0let 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
 one year ago
Best ResponseYou've already chosen the best response.2no, we have: cos(A/2) =sqrt(1  sin^2(A/2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait so the way they did it is wrong?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! I think so!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it is not necessary to work with A/2. Please apply my procedure

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I used +/ since we have two square roots

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2for example: \[\Large \sqrt 4 = \pm 2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait so how would your work look like then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because your procedure is confusing me even more lol

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2since 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
 one year ago
Best ResponseYou've already chosen the best response.0is this a half angle formula?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, it is not a bisection formula

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: \[\Large \sqrt {1  {{\left( {\frac{3}{5}} \right)}^2}} = \sqrt {1  \frac{9}{{25}}} = \sqrt {\frac{{25  9}}{{25}}} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there another way to do it? becauase our teacher didn't show us that formula.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Sincerely speaking, it is the unique way that I know

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but whats wrong with the method i used? im still not understanding

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2your method is correct, nevertheless you have to find cos(A/2) not cos(A)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2you have used the subsequent bisection formula: \[\Large {\left\{ {\sin \left( {\frac{\theta }{2}} \right)} \right\}^2} = \frac{{1  \cos \theta }}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This question just got worse so I used something else but idk if its riight..

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the computation of your image is correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So is 7/25 right then??

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we can write this: \[\Large \sin \theta = \frac{3}{5} \Rightarrow \cos \left( {2\theta } \right) = \frac{7}{{25}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino lol what does that mean tho? XD can you explain it?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it 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
 one year ago
Best ResponseYou've already chosen the best response.0ohh alright its the double angle formula too!
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