## anonymous one year ago Find the exact value of the inverse trigonometric function cos(sin^-1(root 2/3))

1. anonymous

@jim_thompson5910

2. jim_thompson5910

Let $\Large \theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)$ So that means $\Large \sin(\theta) = \frac{\sqrt{2}}{3}$

3. jim_thompson5910

since $\Large \sin(\theta) = \frac{\sqrt{2}}{3}$ this means we can draw out this triangle |dw:1433388810567:dw|

4. jim_thompson5910

what is the missing side?

5. anonymous

wait how did you know what value corresponds with what angle/side?

6. jim_thompson5910

because sine = opposite/hypotenuse |dw:1433388989918:dw|

7. anonymous

oh ok so you wouldn't adjacent/hypotenuse cause there's a cos?

8. jim_thompson5910

well that comes later into play, but when setting up sine, you only deal with opposite/hypotenuse

9. anonymous

okay, sorry I might ask a lot of question as a forewarning

10. jim_thompson5910

that's fine

11. jim_thompson5910

what's the missing side?

12. anonymous

root 7?

13. jim_thompson5910

yep

14. jim_thompson5910

|dw:1433389408003:dw|

15. jim_thompson5910

so saying $\Large \cos\left(\sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\right)$ is the same exact thing as saying $\Large \cos\left(\theta\right)$

16. jim_thompson5910

Because remember I said "Let $\Large \theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)$" back up at the top

17. anonymous

yeah

18. jim_thompson5910

so whats the final answer

19. anonymous

so is it just root 7? or am I misinterpreting?

20. jim_thompson5910

close

21. jim_thompson5910

|dw:1433389714910:dw|

22. anonymous

OHHHH ok I get it now~ so it would be root 7/3?

23. jim_thompson5910

yep $\Large \frac{\sqrt{7}}{3}$

24. anonymous

cool! Thankyou!

25. jim_thompson5910

yw