展开 (-2cos(2x)-sin(3x))^2

解答

展开

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

4 cos^{2} (2 x) + 4 cos (2 x) sin (3 x) + sin^{2} (3 x)

求解步骤

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

使用完全平方公式:

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

(- 2 cos (2 x) - sin (3 x))^{2} = (- 2 cos (2 x))^{2} - 2 (- 2 cos (2 x)) sin (3 x) + sin^{2} (3 x)

= (- 2 cos (2 x))^{2} - 2 (- 2 cos (2 x)) sin (3 x) + sin^{2} (3 x)

化简

(- 2 cos (2 x))^{2} - 2 (- 2 cos (2 x)) sin (3 x) + sin^{2} (3 x) : 4 cos^{2} (2 x) + 4 cos (2 x) sin (3 x) + sin^{2} (3 x)

= 4 cos^{2} (2 x) + 4 cos (2 x) sin (3 x) + sin^{2} (3 x)

x \to \frac{1}{2} lim (arcsin (x))

s l o p e f (x) = x^{2} + x - 3

3 t^{4} y^{^{''}} + 7 t^{3} y^{^{'}} = 0

n = 0 \sum \infty \frac{1}{( n + 10 )}

\int 14 tan^{3} (x) d x