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

The solution to expand (sin(pi(n-1)))/(n-1) is 0

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expand (sin(pi(n-1)))/(n-1)

expand

\frac{sin ( π ( n - 1 ) )}{n - 1}

0

Solution steps

\frac{sin ( π ( n - 1 ) )}{n - 1}

Expand

π (n - 1) : πn - π

= \frac{sin ( πn - π )}{n - 1}

Rewrite using trig identities

= \frac{- sin ( πn )}{n - 1}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{sin ( πn )}{n - 1}

Use the basic trigonometric identity:

sin (nπ) = 0

= - \frac{0}{- 1 + n}

Simplify

= 0

\frac{( 3 x + 2 y ) ( 3 x - 2 y )}{36}

(x + 9 x) e^{x}

\frac{( 2 x + 1 )}{( 4 + x ^{2} ) d x}

(e^{2 x} + 1) (e^{2 x} - 1)

\frac{1}{3 ( x - 5 )}

What is expand (sin(pi(n-1)))/(n-1) ?
The solution to expand (sin(pi(n-1)))/(n-1) is 0