anonymous one year ago http://gyazo.com/b8b1ef0f12bc8d0f8667d816336da9b2

1. anonymous

@jim_thompson5910

2. jim_thompson5910

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

3. anonymous

huh lol sorry

4. jim_thompson5910

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

5. jim_thompson5910

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

6. anonymous

so false?

7. jim_thompson5910

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

8. anonymous

yayay thanks

9. anonymous

it wasnt false ):

10. jim_thompson5910

hmm let me think

11. anonymous

12. jim_thompson5910

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

13. anonymous

it's ok lol dont worry :)