What is expand (4v^2-vsin(θ))^2 ?

The solution to expand (4v^2-vsin(θ))^2 is 16v^4-8v^3sin(θ)+v^2sin^2(θ)

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expand (4v^2-vsin(θ))^2

expand

(4 v^{2} - v sin (θ))^{2}

16 v^{4} - 8 v^{3} sin (θ) + v^{2} sin^{2} (θ)

Solution steps

(4 v^{2} - v sin (θ))^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

(4 v^{2} - v sin (θ))^{2} = (4 v^{2})^{2} - 2 \cdot 4 v^{2} v sin (θ) + (v sin (θ))^{2}

= (4 v^{2})^{2} - 2 \cdot 4 v^{2} v sin (θ) + (v sin (θ))^{2}

Simplify

(4 v^{2})^{2} - 2 \cdot 4 v^{2} v sin (θ) + (v sin (θ))^{2} : 16 v^{4} - 8 v^{3} sin (θ) + v^{2} sin^{2} (θ)

= 16 v^{4} - 8 v^{3} sin (θ) + v^{2} sin^{2} (θ)

(4 x y + 2 x^{2} y) + (2 x^{2} + 3 y^{2}) y^{'} (x)

x^{3} \cdot e^{3 x^{2}} (2 x + 3)

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

6 (- e^{3 x} (2 sin (2 x) - 3 cos (2 x)))

(sec (x) + tan (x))^{3} (sec (x) - tan (x))^{4}

What is expand (4v^2-vsin(θ))^2 ?
The solution to expand (4v^2-vsin(θ))^2 is 16v^4-8v^3sin(θ)+v^2sin^2(θ)