Question

helpp

Answer 1

ok did you do problems earlier where you were told if
\[\tan(x)=\frac{1}{2}\] find
\[\sin(x)\]?

Answer 2

do i find tan first and then solve for sin?

Answer 3

because that is what this means, it is just phrased differently. make a triangle, label "opposite" 1, "adjacent" 2 and find the hypotenuse by pythag

Answer 4

no you know what the tangent is. it is
\[\frac{1}{2}\] that is what
\[\tan^{-1}(x)=\frac{1}{2}\] means, the angle whose tangent is one half

Answer 5

i will draw the picture

Answer 6

ooh i see

Answer 7

there is the angle whose tangent is
\[\frac{1}{2}\] and you want the sine of that angle, so all that is missing is the hypotenuse

Answer 8

what are you trying to find then?

Answer 9

the sine of that angle

Answer 10

so you need the hypotenuse, which you find via pythagoras, more or less in your head
it is
\[h=\sqrt{2^2+1^2}=\sqrt{5}\]

Answer 11

would the answer be 2/2squareroot of 5?

Answer 12

so now you know the sine of that angle, because it is "opposite over hypotenuse" get
\[\sin(\tan^{-1}(\frac{1}{2}))=\frac{1}{\sqrt{5}}\]

Answer 13

sine is opposite over hypotenuse. opposite is 1, hypotenuse is
\[\sqrt{5}\] so sine is
\[\frac{1}{\sqrt{5}}\]

Answer 14

you also from this picture know
\[\cos(\tan^{-1}(\frac{1}{2}))=\frac{2}{\sqrt{5}}\]

Answer 15

yeah one of my answers is 2/2 square root of 5 which is basically what you said that is simplified

Answer 16

i guess so right. but why in the earth would anyone write
\[\frac{2}{2\sqrt{5}}\]??

Answer 17

i have no idea but that is one of the options that makes more sense from all the choices

Answer 18

k got it

Answer 19

thanks for your help!