anonymous
  • anonymous
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).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • 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)=\[\frac{2\sqrt{2}}{3}*\frac{4}{5}-\frac{1}{3}*\frac{3}{5}=\frac{8\sqrt{2}-3}{15}\]
myininaya
  • myininaya
i think: i used the Pythagorean thm twice
myininaya
  • myininaya
im not sure i think i need to think more

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myininaya
  • myininaya
ok right answer but this approach is retarded so here is a better approach
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anonymous
  • anonymous
Hey thanks yo
myininaya
  • myininaya
np
anonymous
  • anonymous
could you also answer another one for me?
myininaya
  • myininaya
im fixing to go to bed :(
anonymous
  • anonymous
:( oh man
myininaya
  • myininaya
make a new post im sure someone will help
anonymous
  • anonymous
kk thnxs a lot
anonymous
  • anonymous
\[\text{Cos}\left[\text{ArcSec}\left[\frac{5}{4}\right]+\text{ArcSin}\left[\frac{1}{3}\right]\right]=-\frac{1}{5}+\frac{8 \sqrt{2}}{15} \]
myininaya
  • myininaya
yay! thanks rob that confirms what i did :) except a different way
anonymous
  • anonymous
Glad you were there first. Adios.
anonymous
  • anonymous
Thank you for the metals.
anonymous
  • anonymous
I think I ought to go to bed. Thanks for the "Medals"

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