What is expand ((3sin(θ)+3)^2)/2 ?

The solution to expand ((3sin(θ)+3)^2)/2 is (9sin^2(θ))/2+9sin(θ)+9/2

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expand ((3sin(θ)+3)^2)/2

expand

\frac{( 3 sin ( θ ) + 3 ) ^{2}}{2}

\frac{9 sin ^{2} ( θ )}{2} + 9 sin (θ) + \frac{9}{2}

Solution steps

\frac{( 3 sin ( θ ) + 3 ) ^{2}}{2}

(3 sin (θ) + 3)^{2} = 9 sin^{2} (θ) + 18 sin (θ) + 9

= \frac{9 sin ^{2} ( θ ) + 18 sin ( θ ) + 9}{2}

Apply the fraction rule:

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

\frac{9 sin ^{2} ( θ ) + 18 sin ( θ ) + 9}{2} = \frac{9 sin ^{2} ( θ )}{2} + \frac{18 sin ( θ )}{2} + \frac{9}{2}

= \frac{9 sin ^{2} ( θ )}{2} + \frac{18 sin ( θ )}{2} + \frac{9}{2}

Divide the numbers:

\frac{18}{2} = 9

= \frac{9 sin ^{2} ( θ )}{2} + 9 sin (θ) + \frac{9}{2}

\frac{1}{2 u ^{3}} (u \cdot \frac{1}{2} x - v + u^{3} = 0)

(sinh (x)) (coth^{2} (x) + csch^{3} (x)) d x

\frac{3 x ^{2} - 2 x}{( 2 x - 1 ) ^{\frac{3}{2}}}

t^{2} (a t + b) e^{t}

\frac{2 ( 3Δ x - 2 x - 1 )}{Δ x ( 1 - 2 x - 2Δ x ) ( 1 - 2 x )}

What is expand ((3sin(θ)+3)^2)/2 ?
The solution to expand ((3sin(θ)+3)^2)/2 is (9sin^2(θ))/2+9sin(θ)+9/2