A simplified version of the expression 2sin theta (cos theta)(cot theta) is... I tried to use the reciprocal of cotangent theta and then cancle out cos theta, being left with 2sin theta and sin theta but I was wrong. May someone help me?
Depends what you mean by simplified
Well my options are A.) sin ^2 theta, B.) 2cos^2 theta, C.) cos^2 theta. I want to say it's A.) but I'm not entirely sure. Before I had written 2sin theta(csc theta) but I was marked incorrect. I'm not sure how to explain what it means by simplify.
\[2 \sin \theta \times \cos \theta \cot \theta\] We know \(\large \cot \theta =\frac {\cos \theta}{\sin \theta}\) so we have now \[2 \sin \theta \times \cos \theta \times \frac {\cos \theta}{\sin \theta}\] Now we have \[2 \cancel{\sin \theta} \times \cos \theta \times \frac {\cos \theta}{\cancel{\sin \theta}}\] What do you get now?
Would it be cos^2?
uh-huh, you're missing the 2 !
I'm sorry, what do you mean? (:
Like 2cos^2?
Yeah:] you're correct
Oh, thank you! I really appreciate that you wrote the steps out (: Makes much more sense!
you're welcome:D
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