Question

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).

Answer 1

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)=\[\frac{2\sqrt{2}}{3}*\frac{4}{5}-\frac{1}{3}*\frac{3}{5}=\frac{8\sqrt{2}-3}{15}\]

Answer 2

i think:
i used the Pythagorean thm twice

Answer 3

im not sure i think i need to think more

Answer 4

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

Answer 5

Hey thanks yo

Answer 6

np

Answer 7

could you also answer another one for me?

Answer 8

im fixing to go to bed :(

Answer 9

:( oh man

Answer 10

make a new post im sure someone will help

Answer 11

kk thnxs a lot

Answer 12

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

Answer 13

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

Answer 14

Glad you were there first. Adios.

Answer 15

Thank you for the metals.

Answer 16

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