展开 (1+6sin(θ))^2

解答

展开

(1 + 6 sin (θ))^{2}

1 + 12 sin (θ) + 36 sin^{2} (θ)

求解步骤

(1 + 6 sin (θ))^{2}

使用完全平方公式:

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

(1 + 6 sin (θ))^{2} = 1^{2} + 2 \cdot 1 \cdot 6 sin (θ) + (6 sin (θ))^{2}

= 1^{2} + 2 \cdot 1 \cdot 6 sin (θ) + (6 sin (θ))^{2}

化简

1^{2} + 2 \cdot 1 \cdot 6 sin (θ) + (6 sin (θ))^{2} : 1 + 12 sin (θ) + 36 sin^{2} (θ)

= 1 + 12 sin (θ) + 36 sin^{2} (θ)

e x p an d (2 n + 3)!

e x p an d P (- 3, - 5) y Q (- 8, - 3)

e x p an d (1 + \frac{2 x}{n})^{3}

e x p an d \frac{s}{( s ^{2} + 1 ) ( s ^{2} - 2 s + 2 )}

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