Question

how do i solve cos^2(theta)+sin(theta)=1?

Answer 1

\[1-\sin^2(\theta)+\sin(\theta)-1=0\]
\[-\sin^2(\theta)+\sin(\theta)=0\]
\[-\sin(\theta)(\sin(\theta)-1)=0\]

Answer 2

did you mean the trig identity:
cos^2(theta)+sin^2(theta)=1

Answer 3

probably not ...

Answer 4

the theta isnt part of the exponent and its just sin(theta)

Answer 5

\[-\sin(\theta)=0=> \sin(\theta)=0=> \theta= n \pi, n \in \mathbb{Z}\]
\[\sin(\theta)-1=0=> \sin(\theta)=1=> \theta=\frac{\pi}{2}+2n \pi, \theta=\frac{3\pi}{2}+2n \pi\]

Answer 6

not sin^2

Answer 7

any questions?

Answer 8

yes why did u use a negative identity ?

Answer 9

negative?

Answer 10

whats that mean?

Answer 11

i mean what do you mean by negative identity

Answer 12

i thought you take sin(theta) and put it on the other side

Answer 13

\[\cos^2(\theta)=1-\sin^2(\theta)\] <- this is the only identity i used

Answer 14

oh so u substituted 1-sin^2 where cos^2 is

Answer 15

yes

Answer 16

thanks can u show me the answers i would get?

Answer 17

?
i already wrote answers

Answer 18

ya thanks c them now