Question

Trig identity simplification check

Answer 1

\[\frac{ \sin(-x) }{ cscx}+\frac{ \cos(-x) }{ secx }\]

Answer 2

Each one is -1, so I believe this adds up to -2.

Answer 3

@SithsAndGiggles

Answer 4

That's not true. \(\dfrac{\sin x}{\csc x}=\dfrac{\sin x}{\frac{1}{\sin x}}=\sin^2x\neq1\).

Answer 5

\(\cdots\neq-1\) for that matter.

Answer 6

Oh, divided this incorrectly.

Answer 7

well in that case, would this be..

sin^2x+cos^2x=1?

Answer 8

Not quite.
\[\sin(-x)=-\sin x\]
which means
\[\frac{\sin(-x)}{\csc x}=-\sin^2x\]
while
\[\cos(-x)=\cos x\]
so
\[\frac{\cos x}{\sec x}=\cos^2x\]
but \(-\sin^2x+\cos^2x\neq1\).

Answer 9

uhh..

Answer 10

what rule should i use?

Answer 11

The last line is another identity. Do you remember your double angle identities?

Answer 12

dont even think we got up to that lol