Question

Is sqrt(1-sin^2Θ)=cosΘ true?

Answer 1

\[\huge \sqrt{(1-\sin^2Θ)} =\sqrt{\cos^2Θ} = \cos \theta\]
thus it is true.

Answer 2

Ok thank you. Can you help me figure out which quadrant it lies in?

Answer 3

if you provide some boundaries then i may help you..

Answer 4

This is all I have to go off of. Since cosine is positive, I think it is Quadrant II or Quadrant III. Am I correct?

Answer 5

@ganeshie8 can you tell me if I'm right please?

Answer 6

hint : Cosine is positive in I and IV quadrants

Answer 7

So, it must lie in I and IV quadrants right ?

Answer 8

Oh ok. Is cosine x or y?

Answer 9

cosine is x

Answer 10

sine is y

Answer 11

Ok I see where I went wrong now. Thank you!

Answer 12

good :) u wlc !

Answer 13

it is not true if cosine is negative

Answer 14

\[\large \sqrt{1-\sin^2(\frac{2\pi}{3})}\neq \cos(\frac{2\pi}{3})\] for example

Answer 15

I'm confused now. So that means sqrt(1-sin^2Θ)=cosΘ is false then?

Answer 16

\[\sqrt{1-\sin^2(x)}\] is always positive

Answer 17

but cosine is not always positive

Answer 18

it is true always cuz sqrt is always postive

Answer 19

however the converse is not true always, but that shouldnt matter here

Answer 20

what is true is that
\[\cos(\theta)=\pm\sqrt{1-\sin(\theta)}\]

Answer 21

also, if u live in I and IV quadrants, you dont have to bother cuz cos(x) is always positive

Answer 22

Ok because sqrt(1-sin^2Θ)=sqrt(sin^2Θ+cos2Θ-sin2Θ=cosΘ) right?

Answer 23

Ok wait, I get what @satellite73 is saying now

Answer 24

that looks bit convincing,
but satellite is telling an important concept... oh good :)

Answer 25

Thank you both :). You explained it a lot better than my teacher

Answer 26

glad to hear :)