Question

http://gyazo.com/b8b1ef0f12bc8d0f8667d816336da9b2

Answer 1

@jim_thompson5910

Answer 2

Hint: The domain and range of \[\Large y = \sqrt{x}\] is [0,infinity)

Answer 3

huh lol sorry

Answer 4

basically you're not allowed to plug in negative x values into a square root function

and

the outputs of the square root function is never negative

Answer 5

if you want negative outputs from a square root function, you have to tack on a negative sign on the outside

Answer 6

so false?

Answer 7

yeah there is no way to get -sin(theta) when it should be +sin(theta)

Answer 8

yayay thanks

Answer 9

it wasnt false ):

Answer 10

hmm let me think

Answer 11

it's ok i already submitted

Answer 12

the only thing I can think of is to test angles in various quadrants

ex: Test Q1. if theta = pi/4, then

sqrt(1 - cos^2(theta) ) = -sin(theta)
sqrt(1 - cos^2(pi/4) ) = -sin(pi/4)
sqrt(1 - 1/2) = -sqrt(2)/2
sqrt(1/2) = -sqrt(2)/2
sqrt(2)/2 = -sqrt(2)/2

that last equation is false, so Q1 doesn't work. Check the other quadrants this way

Answer 13

it's ok lol dont worry :)