Question

Helped by: Michele_Laino , amistre64

Find the equivalent algebraic expression for the composition below.
\(\sf cos(arctan(x))=?\)
(Simplify your answer.)

Answer 1

@amistre64 @Michele_Laino

Answer 2

Find an equivalent algebraic expression for each composition.?

Answer 3

here this mite help @TheSmartOne
For part A: Draw a right triangle. Inverse cos(x) is really telling you that the adjacent side and the hypotenuse are x and 1 respectively. Label these on the triangle. Using the pythagorean theorem, u find that the opposite side is the sqrt(1-x^2). So sin(arccos(x)) is sqrt(1-x^2).

Using the same logic you should get 1/(sqrt(x^2+1)) for cos(arctan(

Answer 4

The answer is in terms of x. And that is what the question said exactly.

Answer 5

im with love on it ....

Answer 6

The example says to make \(\sf arctan(x)=\theta\)

Answer 7

so then \(\sf tan\theta =x\)

Answer 8

sorry is sopost to say
Using the same logic you should get 1/(sqrt(x^2+1)) for cos(arctan(x0

Answer 9

And then using that information to draw the right angle triangle.

Answer 10

right

Answer 11

thanks for the medal

Answer 12

And then \(\sf\Large cos\theta = \frac{adj}{hyp}=\frac{1}{\sqrt{x^2+1}}\)

Answer 13

Now it all makes sense. Thanks a lot! @Loveheart @amistre64 :)

Answer 14

@TheSmartOne do you need help with any thing els

Answer 15

Not at the moment. :)

Answer 16

kk just tag me when you do

Answer 17

let's suppose that
\[\theta = \arctan x\]
then we have:
\[\tan \theta = x\]

Answer 18

so we can write:
\[\sin \theta = x\cos \theta \]

Answer 19

now I use the fundamental identity, and I get:
\[{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1\]

Answer 20

after a substitution, I can write:
\[{\left( {x\cos \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1\]

Answer 21

and finally:
\[\cos \theta = \pm \frac{1}{{\sqrt {1 + {x^2}} }}\]

Answer 22

Interesting way to arrive at the same answer. :)

Answer 23

i stretching ropes is quicker lol

Answer 24

thats just the ancient egyptian in me tho .... all these new fangled maths are just hogwash ...

Answer 25

Well, I like Michele's method. Saves time because you don't have to graph. While the book said to graph it so that maybe you can have a visual aspect to it.

Thank you all! :D

Answer 26

thank you! @TheSmartOne @amistre64

Answer 27

@loveheart don't forget to cite your sources, add that you got that from yahoo.

Answer 28

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