Given that $\theta$ is an acute angle and $\sin \theta = m$ , express in terms of $m$ , Another acute angle $\alpha$ is such that $\tan(\theta - \alpha) = -\frac{1}{2}$ .

(i)
$\cos \theta$ ,
(ii)
$\cot \theta$ .
(iii)
What can be deduced about the size of $\theta$ relative to $\alpha$ ?
(iv)
Given further that $\tan \alpha = \frac{3}{2}$ , show $\tan \theta = \frac{4}{7}$ .

Hence, find the value of $m$ .

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Worked solution

2(i)

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

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

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

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

Answer

$\sqrt{1 - m^2}$

2(ii)

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

$\cot \theta = \frac{\cos \theta}{\sin \theta}$

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

From part (i), $\cos \theta = \sqrt{1 - m^2}$ .
$\cot \theta = \frac{\sqrt{1 - m^2}}{m}$

Answer

$\frac{\sqrt{1 - m^2}}{m}$

2(iii)

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

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

2. Deduce the relative sizes of the angles

Therefore, $\theta - \alpha < 0$ , which implies $\theta < \alpha$ .

Answer

$\theta < \alpha$

2(iv)

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

$\tan(\theta - \alpha) = \frac{\tan \theta - \tan \alpha}{1 + \tan \theta \tan \alpha}$
$\frac{\tan \theta - \frac{3}{2}}{1 + \frac{3}{2}\tan \theta} = -\frac{1}{2}$

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

$2\left(\tan \theta - \frac{3}{2}\right) = -\left(1 + \frac{3}{2}\tan \theta\right)$
$2\tan \theta - 3 = -1 - \frac{3}{2}\tan \theta$
$\frac{7}{2}\tan \theta = 2$
$\tan \theta = \frac{4}{7}$ (shown)

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

Since $\tan \theta = \frac{4}{7}$ and $\theta$ is acute, we can consider a right-angled triangle with opposite side $4$ and adjacent side $7$ .
Hypotenuse $= \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$
$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{\sqrt{65}}$
Since $\sin \theta = m$ , $m = \frac{4}{\sqrt{65}}$ (or $\frac{4\sqrt{65}}{65}$ )

Answer

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

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Given that θ\thetaθ is an acute angle and sin⁡θ=m\sin \theta = msinθ=m, express in terms of mmm, Another acute angle α\alphaα is such that tan⁡(θ−α)=−12\tan(\theta - \alpha) = -\frac{1}{2}tan(θ−α)=−21​.

Worked solution

2(i)

2(ii)

2(iii)

2(iv)

More questions on this topic

Given that $\theta$ is an acute angle and $\sin \theta = m$ , express in terms of $m$ , Another acute angle $\alpha$ is such that $\tan(\theta - \alpha) = -\frac{1}{2}$ .