Question

Rewrite with only sin x and cos x.

sin 2x - cos 2x

2 sin2x - 2 sin x cos x + 1

2 sin x

2 sin2x + 2 sin x cos x - 1

2 sin2x - 2 sin x cos x - 1

Answer 1

Im confused @mustafa2014

Answer 2

@RadEn

Answer 3

\[2 \sin2x + 2 \sin x \cos x - 1\]
or
\[sin^2x + 2 sin x \cos x - 1\]

Answer 4

\[2\sin^2x + 2 \sin x \cos x - 1 \]

Answer 5

@mondona

Answer 6

ohh i think i understand

Answer 7

\[\sin 2x=2 sin(x) cos(x)\]
\[\cos 2x=cos^2x-sin^2x\]
\[\sin 2x-cos 2x=2 sin(x) cos(x)-(cos^2x-sin^2x)\]
\[\sin 2x-cos 2x=2 sin(x) cos(x)-cos^2x+sin^2x\]
*
\[\sin^2x+cos^2x=1\]
\[\cos^2x=1-sin^2x\]*
\[\sin 2x-cos 2x=2 sin(x) cos(x)-(1-sin^2x)+sin^2x\]
\[\sin 2x-cos 2x=2 sin(x) cos(x)-1+sin^2x+sin^2x\]
\[\sin 2x-cos 2x=2 sin(x) cos(x)-1+2sin^2x\]

Answer 8

so it could be C

Answer 9

@mustafa2014

Answer 10

yes :)