How do you simplify $cos (\frac{π}{2} - x) \cdot csc (- x)$ $cos(pi/2 - x) * csc(-x)$ ?

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How do you simplify $cos (\frac{π}{2} - x) \cdot csc (- x)$ $cos(pi/2 - x) * csc(-x)$ ?

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Answer 1 · 2018-04-05T20:12:39

$cos (\frac{π}{2} - x) \cdot csc (- x) = - 1$ $cos(pi/2-x)*csc(-x)=-1$ for $x$ $x$ not a multiple of $π$ $pi$ .

Explanation:

We know that the angles of a right triangle must sum to $π$ $pi$ radians. Since the right angle is $\frac{π}{2}$ $pi/2$ radians, the other two angles must sum to $\frac{π}{2}$ $pi/2$ radians. Therefore $\frac{π}{2}$ $pi/2$ minus the adjacent angle must be the opposite angle. Another way of saying this is $cos (\frac{π}{2} - x) = sin x$ $cos(pi/2-x)=sinx$ , so

$cos (\frac{π}{2} - x) \cdot csc (- x) = sin x \cdot csc (- x)$ $cos(pi/2-x)*csc(-x)=sinx*csc(-x)$ .

The definition of $csc x$ $cscx$ says that $csc x = \frac{1}{sin x}$ $cscx=1/sinx$ , so

$sin x \cdot csc (- x) = \frac{sin x}{sin (- x)}$ $sinx*csc(-x)=sinx/sin(-x)$

We know that $sin x$ $sinx$ is an odd function so $sin (- x) = - sin x$ $sin(-x)=-sinx$ and we can write

$\frac{sin x}{sin (- x)} = - \frac{sin x}{sin x} = - 1$ $sinx/sin(-x)=-sinx/sinx=-1$ .

However, this breaks down when $x$ $x$ is a multiple of $π$ $pi$ because $csc (n π)$ $csc(npi)$ is not defined.

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How do you simplify $cos (\frac{π}{2} - x) \cdot csc (- x)$ $cos(pi/2 - x) * csc(-x)$ ?

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Explanation:

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