What is expand-(lk(lsin((2pix)/l)+2Cpi))/(4pi^2) ?

The solution to expand-(lk(lsin((2pix)/l)+2Cpi))/(4pi^2) is -(l^2ksin((2pix)/l))/(4pi^2)-(lCk)/(2pi)

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expand-(lk(lsin((2pix)/l)+2Cpi))/(4pi^2)

expand

- \frac{l k ( l sin ( \frac{2 π x}{l} ) + 2 C π )}{4 π ^{2}}

- \frac{l ^{2} k sin ( \frac{2 π x}{l} )}{4 π ^{2}} - \frac{lC k}{2 π}

Solution steps

- \frac{l k ( l sin ( \frac{2 π x}{l} ) + 2 C π )}{4 π ^{2}}

Expand

l k (l sin (\frac{2 π x}{l}) + 2 C π) : l^{2} k sin (\frac{2 π x}{l}) + 2 π lC k

= - \frac{l ^{2} k sin ( \frac{2 π x}{l} ) + 2 π lC k}{4 π ^{2}}

Apply the fraction rule:

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

\frac{l ^{2} k sin ( \frac{2 π x}{l} ) + 2 π lC k}{4 π ^{2}} = - (\frac{l ^{2} k sin ( \frac{2 π x}{l} )}{4 π ^{2}}) - (\frac{2 π lC k}{4 π ^{2}})

= - (\frac{l ^{2} k sin ( \frac{2 π x}{l} )}{4 π ^{2}}) - (\frac{2 π lC k}{4 π ^{2}})

Remove parentheses:

(a) = a

= - \frac{l ^{2} k sin ( \frac{2 π x}{l} )}{4 π ^{2}} - \frac{2 π lC k}{4 π ^{2}}

Cancel

\frac{2 π lC k}{4 π ^{2}} : \frac{lC k}{2 π}

= - \frac{l ^{2} k sin ( \frac{2 π x}{l} )}{4 π ^{2}} - \frac{lC k}{2 π}

3 (y - 40 = - 38)

f (∣ x ∣ - 1) e^{∣ x ∣}

1 - x ((- 2 - x) (4 - x))

\frac{[ x ( x - 1 ) - 2 ]}{2}

2 x 2 x^{1}

What is expand-(lk(lsin((2pix)/l)+2Cpi))/(4pi^2) ?
The solution to expand-(lk(lsin((2pix)/l)+2Cpi))/(4pi^2) is -(l^2ksin((2pix)/l))/(4pi^2)-(lCk)/(2pi)