Question

simplify the expression:
2sin^2x-1/sinx-cosx

Answer 1

i always get confused with the 2 in front

Answer 2

Is this the right way to rewrite your question?
\[2\sin^2\theta-1/\sin(\theta)-\cos(\theta)\]
Where the "1" divides both sine and cosine?

Answer 3

yes

Answer 4

i'm not sure what to do

Answer 5

Same here. Do you have any useful/relevant identities?

Answer 6

Hint: 1 = sin^2 (x) + cos^2 (x)

Answer 7

i think its one of the pythagorean identities

Answer 8

so what does the 2 in front do?

Answer 9

Something really interesting :)
\[2\sin^2 x - 1\]
If 1 = sin^2 (x) + cos^2 (x)
\[2\sin^2(x) - (\sin^2x + \cos^2x)\]

Then RELEASE the parentheses!!1! :D

Answer 10

hint: it's gonna be like 2x - x where x = sin^2 x

Answer 11

still confused...

Answer 12

\[\large \frac{2sin^{2}x - sin^{2}x - cos^{2}x}{sinx - cosx}\]

like i said 2sin^2 x - sin^ x is like 2x - x where x = sin^2 x

when given 2x - x..what do you do??

Answer 13

idk >.<' i'm still lost

Answer 14

hmm maybe i shouldnt have used x...lemme try again..you have 2sin^2 x - sin^2 x right?

let a = sin^2 x so...

2a - a = a right?

and a = sin^2 x so...2 sin^2 x - sin^2 x = sin^2 x

getting it now? :D

Answer 15

not really... _ _||| i'm not really good at this

Answer 16

hmmm..then i'll do this the traditional way

2sin^2 x - sin^2 x...

imagine sin^2 x is an apple

2 apples - 1 apple...what do you have?

Answer 17

1 apple

Answer 18

and apple = sin^2 x

so 1 apple = 1 (sin^2 x)..anything multiplied to 1 is the number itself so it's...sin^2 x

do you get it now?

Answer 19

yes i think so

Answer 20

is this what it looks like factored out?