Is csc^2(θ)(1-cos^2(θ))=sin(6480) ?

The answer to whether csc^2(θ)(1-cos^2(θ))=sin(6480) is False

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prove csc^2(θ)(1-cos^2(θ))=sin(6480)

prove

False

Solution steps

csc^{2} (θ) (1 - cos^{2} (θ)) = sin (648 0^{\circ})

Show that the two sides are not equal

For

csc^{2} (θ) (1 - cos^{2} (θ)) = sin (648 0^{\circ})

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

sin (648 0^{\circ}) = 0

sin (648 0^{\circ})

sin (648 0^{\circ}) = sin (0)

sin (648 0^{\circ})

Rewrite

648 0^{\circ}

36 0^{\circ} \cdot 18 + 0

= sin (36 0^{\circ} 18 + 0)

Apply the periodicity of

sin

sin (x + 36 0^{\circ} \cdot k) = sin (x)

sin (36 0^{\circ} \cdot 18 + 0) = sin (0)

= sin (0)

= sin (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0

The two sides are not equal

\Rightarrow False

Is csc^2(θ)(1-cos^2(θ))=sin(6480) ?
The answer to whether csc^2(θ)(1-cos^2(θ))=sin(6480) is False