Use the formula tan 2θ =
2 tan θ
1
− tan2 θ
to mark two sides of a right-angled triangle whose included
angle is 2θ. Use Pythagoras theorem to find the length of the third side in terms of tan
θ. Hence,
write down a formula for sin 2θ in terms of tan θ and use it to find the smallest positive value of
θ for which 2 tan θ = cosec θ.
- Use the formula \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\) to express \(\tan 2\theta\) in terms of \(\tan \theta\).
- \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\)
- Consider a right-angled triangle where one of the angles is \(2\theta\). Let the opposite side be \(a\) and the adjacent side be \(b\). Use the expression for \(\tan 2\theta\) to relate \(a\) and \(b\).
- \(\frac{a}{b} = \frac{2 \tan \theta}{1 - \tan^2 \theta}\)
- Assume \(b = 1\) for simplicity, and express \(a\) in terms of \(\tan \theta\).
- \(a = \frac{2 \tan \theta}{1 - \tan^2 \theta}\)
- Use the Pythagorean theorem to find the hypotenuse \(c\) of the triangle in terms of \(\tan \theta\).
- \(c = \sqrt{1 + \left(\frac{2 \tan \theta}{1 - \tan^2 \theta}\right)^2}\)
- Write down the formula for \(\sin 2\theta\) in terms of \(\tan \theta\) using the sides of the triangle.
- \(\sin 2\theta = \frac{\frac{2 \tan \theta}{1 - \tan^2 \theta}}{\sqrt{1 + \left(\frac{2 \tan \theta}{1 - \tan^2 \theta}\right)^2}}\)
- Simplify the expression for \(\sin 2\theta\) in terms of \(\tan \theta\).
- \(\sin 2\theta = \frac{2 \tan \theta}{\sqrt{(1 - \tan^2 \theta)^2 + 4 \tan^2 \theta}}\)
- Use the equation \(2 \tan \theta = \csc \theta\) to find the smallest positive value of \(\theta\).
- \(\theta = \frac{\pi}{6}\) (or 30 degrees)
