JoannaBlackwelder
  • JoannaBlackwelder
Trig question attached
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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JoannaBlackwelder
  • JoannaBlackwelder
1 Attachment
JoannaBlackwelder
  • JoannaBlackwelder
I am getting:|dw:1447274900777:dw| but that isn't an option
Nay_Af
  • Nay_Af
No promise but I am trying, please wait.

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JoannaBlackwelder
  • JoannaBlackwelder
Thanks :-)
JoannaBlackwelder
  • JoannaBlackwelder
@freckles Any ideas?
JoannaBlackwelder
  • JoannaBlackwelder
|dw:1447275252936:dw|
JoannaBlackwelder
  • JoannaBlackwelder
|dw:1447275320995:dw|
freckles
  • freckles
\[\sin(\theta)=\frac{1}{4} \\ \tan(\theta)>0 \text{ so } \cos(\theta)>0 \\ \] \[\sin^2(\theta)+\cos^2(\theta)=1 \\ \frac{1}{16}+\cos^2(\theta)=1 \\ \cos^2(\theta)=1-\frac{1}{16} \\ \cos^2(\theta)=\frac{15}{16} \\ \cos(\theta)=\frac{\sqrt{15}}{4}\]
freckles
  • freckles
we knew cos was positive because sin is positive and tan is positive
JoannaBlackwelder
  • JoannaBlackwelder
Oh! I see my silly mistake. Thanks, @freckles !
freckles
  • freckles
np
freckles
  • freckles
so you do know that they multiplied inside of their fraction by 8/8 to get the answer they have listed right? there is a bit of arithmetic after that but not too bad
freckles
  • freckles
inside of the square root* the fraction inside by 8/8
JoannaBlackwelder
  • JoannaBlackwelder
I thought it was multiplied by 2/2, not 8/8
JoannaBlackwelder
  • JoannaBlackwelder
So, A
JoannaBlackwelder
  • JoannaBlackwelder
But I guess you could do 8/8 and then simplify some
JoannaBlackwelder
  • JoannaBlackwelder
Oh, but I made common denominators first
JoannaBlackwelder
  • JoannaBlackwelder
I see what you mean, yeah, you can just do that instead of the common denominator step
freckles
  • freckles
you can but after doing the 2/2 you are going to have to do 4/4 that is what I did ... anyways... \[\sqrt{\frac{1+\frac{\sqrt{15}}{4}}{2}} \\ \sqrt{\frac{8+8 \cdot \frac{\sqrt{15}}{4}}{8(2)}} \\ \sqrt{\frac{8+2 \sqrt{15}}{16}}\]
freckles
  • freckles
either way yeah you are right the answer is choice a
JoannaBlackwelder
  • JoannaBlackwelder
:-) Thanks so much, again!
freckles
  • freckles
np

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