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Simplify the trig expression
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\[\frac{\[\tan\] \sin(-\theta) }{ \cos(-\theta) }\] Wouldn't this equal \[\frac{ -\sin \theta }{ -\cos \theta } \] which is just .. \[\tan \theta \] ?
ignore the tan in the first equation
@satellite73
@jim_thompson5910
Rules: cos(-x) = cos(x) for all x sin(-x) = -sin(x) for all x
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\[\Large \frac{\sin(-x)}{\cos(-x)} = \frac{-\sin(x)}{\cos(x)} = -\tan(x)\]
are these the negative identities ?
ohh..i copied the cos identity rule wrong on my book. i see it now, thanks
np
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