Question

find the exact value of cos(sin^-1(3/5))

Answer 1

Do you have a calculator?

Answer 2

I got \( \huge \frac{4}{5} \)

Answer 3

If \theta is such that:
\[\Large \theta = \arcsin \left( {\frac{3}{5}} \right)\]
then you have to compute this:
\[\Large \cos \theta \]

Answer 4

were not suppose to use calculator :/

Answer 5

whats arc sin?

Answer 6

ok!
now we have:
\[\Large \sin \theta = \frac{3}{5}\]
am I right?

Answer 7

it is another way to mean sin^-1

Answer 8

so, if we apply the fundamental identity:
\[\Large {\left( {\cos \theta } \right)^2} + {\left( {\sin \theta } \right)^2} = 1\]
we can write:
\[\Large \cos \theta = \pm \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} = \pm \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = ...?\]

Answer 9

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?

Answer 10

no, we have:
cos(A/2) =sqrt(1 - sin^2(A/2)

Answer 11

wait so the way they did it is wrong?

Answer 12

yes! I think so!

Answer 13

it is not necessary to work with A/2. Please apply my procedure

Answer 14

I used +/- since we have two square roots

Answer 15

for example:
\[\Large \sqrt 4 = \pm 2\]

Answer 16

Wait so how would your work look like then?

Answer 17

Because your procedure is confusing me even more lol

Answer 18

since we have:
\[\Large \sin \theta = \frac{3}{5}\]
then we can write:
\[\Large \cos \theta = \pm \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} = \pm \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = ...?\]

Answer 19

is this a half angle formula?

Answer 20

no, it is not a bisection formula

Answer 21

hint:
\[\Large \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = \sqrt {1 - \frac{9}{{25}}} = \sqrt {\frac{{25 - 9}}{{25}}} = ...?\]

Answer 22

Is there another way to do it? becauase our teacher didn't show us that formula.

Answer 23

Sincerely speaking, it is the unique way that I know

Answer 24

but whats wrong with the method i used? im still not understanding

Answer 25

your method is correct, nevertheless you have to find cos(A/2) not cos(A)

Answer 26

you have used the subsequent bisection formula:
\[\Large {\left\{ {\sin \left( {\frac{\theta }{2}} \right)} \right\}^2} = \frac{{1 - \cos \theta }}{2}\]

Answer 27

This question just got worse so I used something else but idk if its riight..

Answer 28

the computation of your image is correct

Answer 29

So is 7/25 right then??

Answer 30

@Michele_Laino

Answer 31

we can write this:
\[\Large \sin \theta = \frac{3}{5} \Rightarrow \cos \left( {2\theta } \right) = \frac{7}{{25}}\]

Answer 32

@Michele_Laino lol what does that mean tho? XD can you explain it?

Answer 33

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:
\[\Large \begin{gathered}
\cos \left( {2\theta } \right) = {\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2} = \hfill \\
\hfill \\
= {\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2} = \frac{{16}}{{25}} - \frac{9}{{25}} = \frac{{16 - 9}}{{25}} = \frac{7}{{25}} \hfill \\
\end{gathered} \]

Answer 34

ohh alright its the double angle formula too!