## anonymous 5 years ago If sin(x) = 1/3 and sec(y) = 5/4 , where x and y lie between 0 and pie/2, evaluate the expression cos(x+y).

1. myininaya

sinx=1/3 secy=1/cosy=5/4 => cosy=4/5 cos(x+y)=cosxcosy-sinxsiny =cosx(4/5)-(1/3)siny so now we need to know cosx and siny i drew two triangles one where sinx=1/3=20/(3*20) one where cosx=4/5=12*4/(12*5) i got the following conclusion: sin(y)=36/60=3/5 and cosx=40(2)^(1/2)/60=2(2)^(1/2)/3 so we have cos(x+y)=$\frac{2\sqrt{2}}{3}*\frac{4}{5}-\frac{1}{3}*\frac{3}{5}=\frac{8\sqrt{2}-3}{15}$

2. myininaya

i think: i used the Pythagorean thm twice

3. myininaya

im not sure i think i need to think more

4. myininaya

ok right answer but this approach is retarded so here is a better approach

5. anonymous

Hey thanks yo

6. myininaya

np

7. anonymous

could you also answer another one for me?

8. myininaya

im fixing to go to bed :(

9. anonymous

:( oh man

10. myininaya

make a new post im sure someone will help

11. anonymous

kk thnxs a lot

12. anonymous

$\text{Cos}\left[\text{ArcSec}\left[\frac{5}{4}\right]+\text{ArcSin}\left[\frac{1}{3}\right]\right]=-\frac{1}{5}+\frac{8 \sqrt{2}}{15}$

13. myininaya

yay! thanks rob that confirms what i did :) except a different way

14. anonymous

15. anonymous

Thank you for the metals.

16. anonymous

I think I ought to go to bed. Thanks for the "Medals"