How do you find the derivative of $sin θ csc θ$ $sinthetacsctheta$ ?

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Answer 1 · 2016-12-09T14:51:38

This is a question partly to see whether you remember your trigonometry and partly to teach you that it's OK to rewrite before differentiating..

$csc θ = \frac{1}{sin θ}$ $csc theta = 1/sin theta$ , so

$sin θ csc θ = 1$ $sin theta csc theta = 1$ ,

and the derivative is $0$ $0$

If you want to do more trigonometry , then use the product rule to get

$(cos θ) (csc θ) + (sin θ) (- csc θ cot θ)$ $(cos theta) (csc theta) + (sin theta) (-csc theta cot theta)$ .

Now simplify to get

$cot θ - cot θ = 0$ $cot theta - cot theta = 0$ .

How do you find the derivative of sinθcscθsinthetacsctheta?