What is (-2cos(npi))/n sin(nx) ?

The solution to (-2cos(npi))/n sin(nx) is -(2(-1)^nsin(nx))/n

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simplify (-2cos(npi))/n sin(nx)

simplify

\frac{- 2 cos ( nπ )}{n} sin (n x)

- \frac{2 ( - 1 ) ^{n} sin ( n x )}{n}

Solution steps

\frac{- 2 cos ( nπ )}{n} sin (n x)

Use the basic trigonometric identity:

cos (nπ) = (- 1)^{n}

= sin (x n) \frac{- 2 ( - 1 ) ^{n}}{n}

Apply the fraction rule:

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

= (- \frac{2 ( - 1 ) ^{n}}{n}) sin (n x)

Remove parentheses:

(- a) = - a

= - sin (x n) \frac{2 ( - 1 ) ^{n}}{n}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{2 ( - 1 ) ^{n} sin ( x n )}{n}

sin (θ) - sin (\frac{π}{2} - θ)

cos (9 π + a)

csc (\frac{π}{2} + a)

csc (- x + \frac{π}{2})

csc (arccos (\frac{2 x}{3}))

What is (-2cos(npi))/n sin(nx) ?
The solution to (-2cos(npi))/n sin(nx) is -(2(-1)^nsin(nx))/n