Is csc^2(θ)(1-cos^2(θ))=-cos(pi/2) ?

The answer to whether csc^2(θ)(1-cos^2(θ))=-cos(pi/2) is False

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prove csc^2(θ)(1-cos^2(θ))=-cos(pi/2)

prove

csc^{2} (θ) (1 - cos^{2} (θ)) = - cos (\frac{π}{2})

False

Solution steps

csc^{2} (θ) (1 - cos^{2} (θ)) = - cos (\frac{π}{2})

Show that the two sides are not equal

For

csc^{2} (θ) (1 - cos^{2} (θ)) = - cos (\frac{π}{2})

plug in

θ = 1

csc^{2} (1) (1 - cos^{2} (1)) = 1

csc^{2} (1) (1 - cos^{2} (1))

Refine to a decimal form

= 1

- cos (\frac{π}{2}) = 0

- cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

= - 0

Simplify

= 0

The two sides are not equal

\Rightarrow False

Is csc^2(θ)(1-cos^2(θ))=-cos(pi/2) ?
The answer to whether csc^2(θ)(1-cos^2(θ))=-cos(pi/2) is False