已知 θ\theta 是锐角且 sin⁡θ=m\sin \theta = m，用 mm 表示： 另一个锐角 α\alpha 满足 tan⁡(θ−α)=−12\tan(\theta - \alpha) = -\frac{1}{2}。

  1. (i)

    cos⁡θ\cos \theta，

  2. (ii)

    cot⁡θ\cot \theta。

  3. (iii)

    关于 θ\theta 相对于 α\alpha 的大小，可以得出什么结论？

  4. (iv)

    进一步已知 tan⁡α=32\tan \alpha = \frac{3}{2}，证明 tan⁡θ=47\tan \theta = \frac{4}{7}。

    由此，求 mm 的值。

三角恒等式三角函数解三角方程三角式化简English
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完整解法

2(i)

1. Use the Pythagorean identity to relate $\sin \theta$ and $\cos \theta$

We know that sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.
Given sin⁡θ=m\sin \theta = m, we have:
m2+cos⁡2θ=1m^2 + \cos^2 \theta = 1
cos⁡2θ=1−m2\cos^2 \theta = 1 - m^2

2. Solve for $\cos \theta$, taking the positive root since $\theta$ is acute

Since θ\theta is an acute angle, cos⁡θ>0\cos \theta > 0.
Therefore, cos⁡θ=1−m2\cos \theta = \sqrt{1 - m^2}.

答案

1−m2\sqrt{1 - m^2}

2(ii)

1. Express $\cot \theta$ in terms of $\sin \theta$ and $\cos \theta$

cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}

2. Substitute the expressions for $\sin \theta$ and $\cos \theta$ to find $\cot \theta$

From part (i), cos⁡θ=1−m2\cos \theta = \sqrt{1 - m^2}.
cot⁡θ=1−m2m\cot \theta = \frac{\sqrt{1 - m^2}}{m}

答案

1−m2m\frac{\sqrt{1 - m^2}}{m}

2(iii)

1. Determine the sign of the angle difference based on the tangent value

Since θ\theta and α\alpha are acute angles, −90∘<θ−α<90∘-90^\circ < \theta - \alpha < 90^\circ.
Given tan⁡(θ−α)=−12<0\tan(\theta - \alpha) = -\frac{1}{2} < 0, the angle (θ−α)(\theta - \alpha) must be in the fourth quadrant (i.e., negative).

2. Deduce the relative sizes of the angles

Therefore, θ−α<0\theta - \alpha < 0, which implies θ<α\theta < \alpha.

答案

θ<α\theta < \alpha

2(iv)

1. Apply the compound angle formula for tangent to set up an equation for $\tan \theta$.

tan⁡(θ−α)=tan⁡θ−tan⁡α1+tan⁡θtan⁡α\tan(\theta - \alpha) = \frac{\tan \theta - \tan \alpha}{1 + \tan \theta \tan \alpha}
tan⁡θ−321+32tan⁡θ=−12\frac{\tan \theta - \frac{3}{2}}{1 + \frac{3}{2}\tan \theta} = -\frac{1}{2}

2. Solve the equation to show $\tan \theta = \frac{4}{7}$.

2(tan⁡θ−32)=−(1+32tan⁡θ)2\left(\tan \theta - \frac{3}{2}\right) = -\left(1 + \frac{3}{2}\tan \theta\right)
2tan⁡θ−3=−1−32tan⁡θ2\tan \theta - 3 = -1 - \frac{3}{2}\tan \theta
72tan⁡θ=2\frac{7}{2}\tan \theta = 2
tan⁡θ=47\tan \theta = \frac{4}{7} (shown)

3. Use the value of $\tan \theta$ to find $\sin \theta$ and hence $m$.

Since tan⁡θ=47\tan \theta = \frac{4}{7} and θ\theta is acute, we can consider a right-angled triangle with opposite side 44 and adjacent side 77.
Hypotenuse =42+72=16+49=65= \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}
sin⁡θ=oppositehypotenuse=465\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{\sqrt{65}}
Since sin⁡θ=m\sin \theta = m, m=465m = \frac{4}{\sqrt{65}} (or 46565\frac{4\sqrt{65}}{65})

答案

m=465m = \frac{4}{\sqrt{65}}

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