give me the proof of this question :
Sin(-theta) = - Sin theta
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Answer:
The terminal side of \theta and -\theta intersect the unit circle at points P and P’. The coordinates of
P are (\cos \theta, \sin \theta)
P' are (\cos(- \theta), \sin (- \theta)).
Since P and P' are symmetric with respect to the x-axis,
\sin (\theta) = - \sin \theta
\cos (- \theta) = \cos \theta.
Now, we know that
\tan (- \theta) = \displaystyle \frac{\sin (- \theta)}{\cos(-\theta)}
= \displaystyle \frac{- \sin \theta}{\cos \theta}
= - \tan \theta.
And we are done.
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