Question

Verify the trig identity

Answer 1

\[\frac{ \sin ^2(-x) }{ \tan^2x }=\cos ^2\]

Answer 2

@dan815

Answer 3

@mathmath333 @jim_thompson5910

Answer 4

@bibby

Answer 5

use the idea that tan(x) = sin(x)/cos(x) and try to simplify the left side

Answer 6

also, sin(-x) = -sin(x) for all values of x

Answer 7

so

\[\Large \sin^2(-x) = [\sin(-x)]^2\]
\[\Large \sin^2(-x) = [-\sin(x)]^2\]
\[\Large \sin^2(-x) = [\sin(x)]^2\]
\[\Large \sin^2(-x) = \sin^2(x)\]

Answer 8

\[\frac{ \frac{ -\sin^2(x) }{ 1 } }{ \frac{ \sin^2 }{ \cos^2 } }=\frac{ -\sin^2(x) }{ 1 }\times \frac{ \cos^2 }{ \sin^2 }\]

Answer 9

im getting a negative cos^2x

Answer 10

look back up at my steps, you'll see that \[\Large \sin^2(-x) = \sin^2(x)\]

Answer 11

oo missed that. got it though. thanks

Answer 12

np