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

1. anonymous

Do you have a calculator?

2. anonymous

I got $$\huge \frac{4}{5}$$

3. Michele_Laino

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

4. anonymous

were not suppose to use calculator :/

5. anonymous

whats arc sin?

6. Michele_Laino

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

7. Michele_Laino

it is another way to mean sin^-1

8. Michele_Laino

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

9. 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?

10. Michele_Laino

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

11. anonymous

wait so the way they did it is wrong?

12. Michele_Laino

yes! I think so!

13. Michele_Laino

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

14. Michele_Laino

I used +/- since we have two square roots

15. Michele_Laino

for example: $\Large \sqrt 4 = \pm 2$

16. anonymous

Wait so how would your work look like then?

17. anonymous

Because your procedure is confusing me even more lol

18. Michele_Laino

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

19. anonymous

is this a half angle formula?

20. Michele_Laino

no, it is not a bisection formula

21. Michele_Laino

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

22. anonymous

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

23. Michele_Laino

Sincerely speaking, it is the unique way that I know

24. anonymous

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

25. Michele_Laino

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

26. Michele_Laino

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

27. anonymous

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

28. Michele_Laino

the computation of your image is correct

29. anonymous

So is 7/25 right then??

30. anonymous

@Michele_Laino

31. Michele_Laino

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

32. anonymous

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

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

34. anonymous

ohh alright its the double angle formula too!