What is expand 2pi(2r^2sin(a)-r^2cos(a)2a) ?

The solution to expand 2pi(2r^2sin(a)-r^2cos(a)2a) is 4pir^2sin(a)-4piar^2cos(a)

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expand 2pi(2r^2sin(a)-r^2cos(a)2a)

expand

2 π (2 r^{2} sin (a) - r^{2} cos (a) 2 a)

4 π r^{2} sin (a) - 4 πa r^{2} cos (a)

Solution steps

2 π (2 r^{2} sin (a) - r^{2} cos (a) \cdot 2 a)

= 2 π (2 r^{2} sin (a) - 2 a r^{2} cos (a))

Apply the distributive law:

a (b - c) = ab - a c

a = 2 π, b = 2 r^{2} sin (a), c = r^{2} cos (a) \cdot 2 a

= 2 π 2 r^{2} sin (a) - 2 π r^{2} cos (a) \cdot 2 a

= 2 \cdot 2 π r^{2} sin (a) - 2 \cdot 2 πa r^{2} cos (a)

Multiply the numbers:

2 \cdot 2 = 4

= 4 π r^{2} sin (a) - 4 πa r^{2} cos (a)

\frac{7 ( 2 + 3 sin ( 2 x ) ) ( 2 x + 3 sin ^{2} ( x ) ) ^{\frac{4}{3}}}{3}

\frac{4 x ^{2}}{( 2 + x ) ( 3 + x )}

\frac{25 ( x ^{2} - 1 )}{( x ^{2} + 1 ) ^{2}}

x^{4} - 2 x - 16 x^{2} + 2 x + 15

\frac{( 4 a + 1 ) ^{2}}{16}

What is expand 2pi(2r^2sin(a)-r^2cos(a)2a) ?
The solution to expand 2pi(2r^2sin(a)-r^2cos(a)2a) is 4pir^2sin(a)-4piar^2cos(a)