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

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

1 Answer

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Answer 1 · 2016-11-28T16:36:48

Express both factors in terms of $sin x$ $sinx$

Explanation:

$csc x = \frac{1}{sin x}$ $cscx=1/sinx$

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

and $sin (- x) = - sin x$ $sin(-x) =-sinx$

So:

${csc}^{2} (- x) {cos}^{2} (\frac{π}{2} - x) = \frac{1}{{sin}^{2} (- x)} {sin}^{2} (x) = \frac{1}{{(- sin x)}^{2}} {sin}^{2} (x) = 1$ $csc^2(-x) cos^2(pi/2-x) = 1/sin^2(-x)sin^2(x) =1/(-sinx)^2sin^2(x)=1$

Strictly speaking this is only true for $x \neq k π$ $x!=kpi$ where the original expression is not defined.

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

1 Answer

Explanation:

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How do you simplify csc^2(-x)cos^2x(pi/2-x)csc2(−x)cos2x(π2−x)csc^2(-x)cos^2x(pi/2-x)?

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Explanation:

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