What is 1/(npi)sin(n-pi) ?

The solution to 1/(npi)sin(n-pi) is -(sin(n))/(pin)

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simplify 1/(npi)sin(n-pi)

simplify

\frac{1}{nπ} sin (n - π)

- \frac{sin ( n )}{πn}

Solution steps

\frac{1}{nπ} sin (n - π)

Rewrite using trig identities

= \frac{1}{nπ} (- sin (n))

Simplify

\frac{1}{nπ} (- sin (n)) : - \frac{sin ( n )}{πn}

= - \frac{sin ( n )}{πn}

cos (sin (- 1 x))

1 - (sin (θ))^{2}

\frac{11.9}{63.89 cos ( θ )}

sin (- y) cos (- y)

(4 cos (x + 18 0^{\circ}) - 4) - 8