Question

simplify the expression:
2sin^2x-1/sinx-cosx
Can someone please explain all the steps...i'm really confused...

Answer 1

\[2\sin ^{2}x-1\div sinx-cosx\] here it is in equation form

Answer 2

Use the fact that:
\[\cos^2x-\sin^2x=1-2\sin^2x\]

Answer 3

http://puu.sh/qjVR

Answer 4

http://puu.sh/qjXJ

Hope you can understand better with this ;D

Answer 5

@zepp ty so much ^^ i understand now

Answer 6

\[\cos(2*x)=(\cos(x))^{2}-(\sin(x))^{2}\rightarrow 1-2*(\sin(x))^{2}=\cos(2*x)\]
your equation will be:
\[-\cos(2*x)/(\sin(x)-\cos(x))\]
multiplying all by \[\sin(x)+\cos(x)\]
we have:
\[-[\cos(2*x)/(\sin(x)^{2}-\cos(x)^{2})]*(\sin(x)+\cos(x))=\sin(x)+\cos(x)\]

Answer 7

@RaphaelFilgueiras ty ^^

Answer 8

Haha, you're welcome ;D
Good luck on simplifying other fractions! ^_^