anonymous
  • anonymous
write algebraic expression for sec(arcsin(x/sqrtx^2+4))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sec\left(\arcsin\frac{x}{\sqrt{x^2+4}}\right) }\) like this?
anonymous
  • anonymous
yes
anonymous
  • anonymous
oh ok

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SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sec\left(A\right)= \frac{1}{\cos(A)}=\frac{1}{\sqrt{1-\sin^2(A)}}}\)
SolomonZelman
  • SolomonZelman
And we can use that here
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sec\left(\arcsin\frac{x}{\sqrt{x^2+4}}\right)=\frac{1}{\sqrt{1-\sin^2\left(\arcsin\dfrac{x}{\sqrt{x^2+4}}\right)}} }\)
SolomonZelman
  • SolomonZelman
The thing is, that: \(\color{red}{\sin\left(\arcsin B\right)=B}\)
SolomonZelman
  • SolomonZelman
And so, \(\color{red}{\sin~^m~\left(\arcsin B\right)=B~^m}\)
anonymous
  • anonymous
sorry computer is super slow right now
SolomonZelman
  • SolomonZelman
All that matters is whether you are able to follow what I am trying to get you to do or not.... so are you with me?
anonymous
  • anonymous
ummm in a way. but the directions say that i must use the right triangle method.
SolomonZelman
  • SolomonZelman
Ok, you didn't mention that... sadly. But, still can you give me the answer based on what I have suggested, and then we will do this your way
anonymous
  • anonymous
My bad. i thought there was only one method to solve this.
SolomonZelman
  • SolomonZelman
(I worked this out the answer is pretty simple)
anonymous
  • anonymous
all right so then it is\[\frac{ 1 }{\sqrt{1-\frac{ x }{\sqrt{x^2+4} }} }\]
SolomonZelman
  • SolomonZelman
no not quite
SolomonZelman
  • SolomonZelman
\(\Large\color{black}{ \displaystyle \sin^{\rm P}\left(\arcsin {\bf Q}\right)={\bf Q}^{\rm P} }\)
SolomonZelman
  • SolomonZelman
and that follows logically, just knowing the rules of exponents and knowing the fact that \(\Large\color{black}{ \displaystyle \sin\left(\arcsin {\bf Q}\right)={\bf Q} }\)
SolomonZelman
  • SolomonZelman
So, \(\large\color{black}{ \displaystyle \frac{1}{\sqrt{1-\sin^{\color{red}{\LARGE 2}}\left(\arcsin\dfrac{x}{\sqrt{x^2+4}}\right)}}=?}\)
anonymous
  • anonymous
umm \[\frac{ 1 }{ \sqrt{1-\frac{ x^2 }{ (x+4)^2 }} }\]
anonymous
  • anonymous
right?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{1}{\sqrt{1-\sin^2\left(\arcsin\dfrac{A}{B}\right)}}=\frac{1}{\sqrt{1-\frac{A^2}{B^2}}} }\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{1}{\sqrt{1-\sin^2\left(\arcsin\dfrac{x}{\sqrt{x^2+4}}\right)}}=\frac{1}{\sqrt{1-\dfrac{(x)^2}{(\sqrt{x^2+4})^2}}}}\)
anonymous
  • anonymous
is that the final
SolomonZelman
  • SolomonZelman
so you will get \(\large\color{black}{ \displaystyle \frac{1}{\sqrt{1-\dfrac{x^2}{|x^2+4|}}}}\) absolute value is purely technical, but we assume the x is real so it is anyways positive. (and no you can simplify)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{1}{\sqrt{1-\dfrac{x^2}{x^2+4}}}}\)
anonymous
  • anonymous
oh ok.
SolomonZelman
  • SolomonZelman
Hint to simplify: \(\large\color{black}{ \displaystyle \frac{1}{\sqrt{\dfrac{x^2+4}{x^2+4}-\dfrac{x^2}{x^2+4}}}}\)
anonymous
  • anonymous
Can we do the right triangle method?
anonymous
  • anonymous
|dw:1446417271563:dw|
anonymous
  • anonymous
this is the right triangle method. i have to solve for A. so i will use the pythagoras theory
anonymous
  • anonymous
So.... A^2+x^2=(sqrtx^2+4)^2
freckles
  • freckles
yep and solve for A
freckles
  • freckles
|dw:1446418463857:dw| then you find sec(u)
anonymous
  • anonymous
yeah but i dont know how to set it up
anonymous
  • anonymous
@freckles
freckles
  • freckles
you already did just solve for A
freckles
  • freckles
\[A^2+x^2=(\sqrt{x^2+4})^2 \\ A^2+x^2=x^2+4 \\ \\ \text{ subtract } x^2 \text{ on both sides } \\ A^2=4\]

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