Question

write algebraic expression for sec(arcsin(x/sqrtx^2+4))

Answer 1

\(\large\color{black}{ \displaystyle \sec\left(\arcsin\frac{x}{\sqrt{x^2+4}}\right) }\)

like this?

Answer 2

yes

Answer 3

oh ok

Answer 4

\(\large\color{black}{ \displaystyle \sec\left(A\right)= \frac{1}{\cos(A)}=\frac{1}{\sqrt{1-\sin^2(A)}}}\)

Answer 5

And we can use that here

Answer 6

\(\large\color{black}{ \displaystyle \sec\left(\arcsin\frac{x}{\sqrt{x^2+4}}\right)=\frac{1}{\sqrt{1-\sin^2\left(\arcsin\dfrac{x}{\sqrt{x^2+4}}\right)}} }\)

Answer 7

The thing is, that:

\(\color{red}{\sin\left(\arcsin B\right)=B}\)

Answer 8

And so,

\(\color{red}{\sin~^m~\left(\arcsin B\right)=B~^m}\)

Answer 9

sorry computer is super slow right now

Answer 10

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?

Answer 11

ummm in a way. but the directions say that i must use the right triangle method.

Answer 12

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

Answer 13

My bad. i thought there was only one method to solve this.

Answer 14

(I worked this out the answer is pretty simple)

Answer 15

all right so then it is\[\frac{ 1 }{\sqrt{1-\frac{ x }{\sqrt{x^2+4} }} }\]

Answer 16

no not quite

Answer 17

\(\Large\color{black}{ \displaystyle \sin^{\rm P}\left(\arcsin {\bf Q}\right)={\bf Q}^{\rm P} }\)

Answer 18

and that follows logically, just knowing the rules of exponents and knowing the fact that

\(\Large\color{black}{ \displaystyle \sin\left(\arcsin {\bf Q}\right)={\bf Q} }\)

Answer 19

So,

\(\large\color{black}{ \displaystyle \frac{1}{\sqrt{1-\sin^{\color{red}{\LARGE 2}}\left(\arcsin\dfrac{x}{\sqrt{x^2+4}}\right)}}=?}\)

Answer 20

umm \[\frac{ 1 }{ \sqrt{1-\frac{ x^2 }{ (x+4)^2 }} }\]

Answer 21

right?

Answer 22

\(\large\color{black}{ \displaystyle \frac{1}{\sqrt{1-\sin^2\left(\arcsin\dfrac{A}{B}\right)}}=\frac{1}{\sqrt{1-\frac{A^2}{B^2}}} }\)

Answer 23

\(\large\color{black}{ \displaystyle \frac{1}{\sqrt{1-\sin^2\left(\arcsin\dfrac{x}{\sqrt{x^2+4}}\right)}}=\frac{1}{\sqrt{1-\dfrac{(x)^2}{(\sqrt{x^2+4})^2}}}}\)

Answer 24

is that the final

Answer 25

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)

Answer 26

\(\large\color{black}{ \displaystyle \frac{1}{\sqrt{1-\dfrac{x^2}{x^2+4}}}}\)

Answer 27

oh ok.

Answer 28

Hint to simplify:

\(\large\color{black}{ \displaystyle \frac{1}{\sqrt{\dfrac{x^2+4}{x^2+4}-\dfrac{x^2}{x^2+4}}}}\)

Answer 29

Can we do the right triangle method?

Answer 30

this is the right triangle method. i have to solve for A. so i will use the pythagoras theory

Answer 31

So.... A^2+x^2=(sqrtx^2+4)^2

Answer 32

yep and solve for A

Answer 33

then you find sec(u)

Answer 34

yeah but i dont know how to set it up

Answer 35

@freckles

Answer 36

you already did
just solve for A

Answer 37

\[A^2+x^2=(\sqrt{x^2+4})^2 \\ A^2+x^2=x^2+4 \\ \\ \text{ subtract } x^2 \text{ on both sides } \\ A^2=4\]