证明恒等式 sin⁡2A+cos⁡A1−cos⁡2A+sin⁡A=cot⁡A\frac{\sin 2A + \cos A}{1 - \cos 2A + \sin A} = \cot A 由此,针对 −π≤x≤π-\pi \le x \le \pi,解方程 1−cos⁡4x+sin⁡2xsin⁡4x+cos⁡2x=5−2sec⁡22x\frac{1 - \cos 4x + \sin 2x}{\sin 4x + \cos 2x} = 5 - 2\sec^2 2x 结果保留两位小数。

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完整解法

9(a)

1. Apply double angle formulae to the numerator and denominator

LHS=sin⁡2A+cos⁡A1−cos⁡2A+sin⁡A\text{LHS} = \frac{\sin 2A + \cos A}{1 - \cos 2A + \sin A}
Using sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A \cos A and cos⁡2A=1−2sin⁡2A\cos 2A = 1 - 2\sin^2 A:
LHS=2sin⁡Acos⁡A+cos⁡A1−(1−2sin⁡2A)+sin⁡A\text{LHS} = \frac{2\sin A \cos A + \cos A}{1 - (1 - 2\sin^2 A) + \sin A}

2. Simplify the denominator

LHS=2sin⁡Acos⁡A+cos⁡A2sin⁡2A+sin⁡A\text{LHS} = \frac{2\sin A \cos A + \cos A}{2\sin^2 A + \sin A}

3. Factorise numerator and denominator, then cancel common factors to prove the identity

LHS=cos⁡A(2sin⁡A+1)sin⁡A(2sin⁡A+1)\text{LHS} = \frac{\cos A (2\sin A + 1)}{\sin A (2\sin A + 1)}
LHS=cos⁡Asin⁡A\text{LHS} = \frac{\cos A}{\sin A}
LHS=cot⁡A=RHS\text{LHS} = \cot A = \text{RHS}

答案

LHS=cot⁡A=RHS\text{LHS} = \cot A = \text{RHS} (shown)

9(b)

1. Use the identity from part (a) to simplify the left-hand side of the equation.

From part (a), sin⁡2A+cos⁡A1−cos⁡2A+sin⁡A=cot⁡A\frac{\sin 2A + \cos A}{1 - \cos 2A + \sin A} = \cot A.
Let A=2xA = 2x. The left-hand side of the given equation is the reciprocal of the left-hand side of the identity in (a).
Therefore, 1−cos⁡4x+sin⁡2xsin⁡4x+cos⁡2x=1cot⁡2x=tan⁡2x\frac{1 - \cos 4x + \sin 2x}{\sin 4x + \cos 2x} = \frac{1}{\cot 2x} = \tan 2x.

2. Substitute into the equation and use the Pythagorean identity to form a quadratic equation in $\tan 2x$.

The equation becomes:
tan⁡2x=5−2sec⁡22x\tan 2x = 5 - 2\sec^2 2x
Using the identity sec⁡22x=1+tan⁡22x\sec^2 2x = 1 + \tan^2 2x:
tan⁡2x=5−2(1+tan⁡22x)\tan 2x = 5 - 2(1 + \tan^2 2x)
tan⁡2x=5−2−2tan⁡22x\tan 2x = 5 - 2 - 2\tan^2 2x
2tan⁡22x+tan⁡2x−3=02\tan^2 2x + \tan 2x - 3 = 0

3. Solve the quadratic equation for $\tan 2x$.

Factorising the quadratic equation:
(2tan⁡2x+3)(tan⁡2x−1)=0(2\tan 2x + 3)(\tan 2x - 1) = 0
tan⁡2x=−1.5\tan 2x = -1.5 or tan⁡2x=1\tan 2x = 1

4. Find the basic angles and solve for $x$ in the given range.

For −π≤x≤π-\pi \le x \le \pi, the range for 2x2x is −2π≤2x≤2π-2\pi \le 2x \le 2\pi.

Case 1: tan⁡2x=1\tan 2x = 1
Basic angle α=π4\alpha = \frac{\pi}{4}
2x=−7π4,−3π4,π4,5π42x = -\frac{7\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4}
x=−7π8,−3π8,π8,5π8x = -\frac{7\pi}{8}, -\frac{3\pi}{8}, \frac{\pi}{8}, \frac{5\pi}{8}
x≈−2.75,−1.18,0.39,1.96x \approx -2.75, -1.18, 0.39, 1.96

Case 2: tan⁡2x=−1.5\tan 2x = -1.5
Basic angle α=tan⁡−1(1.5)≈0.98279\alpha = \tan^{-1}(1.5) \approx 0.98279
2x=−π−0.98279,−0.98279,π−0.98279,2π−0.982792x = -\pi - 0.98279, -0.98279, \pi - 0.98279, 2\pi - 0.98279
2x≈−4.1244,−0.9828,2.1588,5.30042x \approx -4.1244, -0.9828, 2.1588, 5.3004
x≈−2.06,−0.49,1.08,2.65x \approx -2.06, -0.49, 1.08, 2.65

答案

x=−2.75,−2.06,−1.18,−0.49,0.39,1.08,1.96,2.65x = -2.75, -2.06, -1.18, -0.49, 0.39, 1.08, 1.96, 2.65

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