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The \(t\) formulas
Introducing the parameter \(t = \tan\dfrac{\theta}{2}\) turns out to be a very useful tool in solving certain types of trigonometric equations and also in finding certain integrals involving trigonometric functions. The basic idea is to relate \(\sin\theta\), \(\cos\theta\) and even \(\tan\theta\) to the tangent of half the angle. This can be done using the double angle formulas.
We let \(t= \tan\dfrac{\theta}{2}\) and so we can draw the following triangle with sides 1, \(t\), \(\sqrt{1+t^2}\).
Detailed description of diagram
Now \(\sin 2\alpha = 2\sin\alpha\cos\alpha\), so replacing \(2\alpha\) with \(\theta\) we have
\[\sin \theta = 2 \sin\dfrac{\theta}{2}\, \cos\dfrac{\theta}{2}.\]
From the triangle, we have
\[\sin \theta = 2 \times \dfrac{t}{\sqrt{1+t^2}} \times \dfrac{1}{\sqrt{1+t^2}} = \dfrac{2t}{1+t^2}.\]
This is referred to as the \(t\) formula for \(\sin \theta\).
Exercise 20
Use the geometric method above to derive the \(t\) formula
\[\cos \theta = \dfrac{1-t^2}{1+t^2}\]and deduce that \(\tan\theta = \dfrac{2t}{1-t^2}\), for \(t \ne \pm 1\).
Exercise 21
The geometric proof of the \(t\) formula for \(\sin\theta\) given above assumes that the angle \(\dfrac{\theta}{2}\) is acute. Give a general algebraic proof of the formula.
One application of the \(t\) formulas is to solving certain types of trigonometric equations.
Example
Solve \(\cos\theta + \sin\theta = \dfrac{1}{2}\), for \(0^\circ \leq \theta \leq 360^\circ\), correct to one decimal place.
Solution
Put \(t = \tan\dfrac{\theta}{2}\). Then
\[\dfrac{1-t^2}{1+t^2} + \dfrac{2t}{1+t^2} = \dfrac{1}{2}.\]
This rearranges to \(3t^2-4t-1=0\), whose solutions are \(t = \dfrac{2\pm\sqrt{7}}{3} \approx 1.549, -0.215\). Taking the inverse tangents (in degrees) and doubling, we obtain the solutions \(\theta \approx 114.3^\circ, 335.7^\circ\) in the given range.
Note. Some care is required when using the \(t\) formulas to find solutions of an equation, as there may be a solution \(\theta\) for which \(\tan {\dfrac\theta2}\) is undefined.
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| This publication is funded by the Australian Government Department of Education, Employment and Workplace Relations | Contributors |
The (t) formulas, arising from the substitution (t = \tan\left(\frac{\theta}{2})), offer a powerful tool in solving trigonometric equations and evaluating integrals involving trigonometric functions. This substitution allows us to establish relationships between (\sin \theta), (\cos \theta), and even (\tan \theta) in terms of (t), enabling a more systematic approach to problem-solving.
The foundational concept lies in the double angle formulas, particularly (\sin 2\alpha = 2\sin \alpha \cos \alpha). Substituting (2\alpha) with (\theta) leads to the equation (\sin \theta = 2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)).
By considering a triangle with sides 1, (t), and (\sqrt{1+t^2}), derived from (t = \tan\left(\frac{\theta}{2}\right)), we can relate (\sin \theta) to (t):
[ \sin \theta = 2 \times \frac{t}{\sqrt{1+t^2}} \times \frac{1}{\sqrt{1+t^2}} = \frac{2t}{1+t^2} ]
This is known as the (t) formula for (\sin \theta). Using similar geometric reasoning and algebraic manipulation, we derive the (t) formulas for (\cos \theta) and (\tan \theta) as:
[ \cos \theta = \frac{1-t^2}{1+t^2} \quad \text{and} \quad \tan \theta = \frac{2t}{1-t^2}, \quad \text{for } t \neq \pm 1 ]
Exercise 21 calls for an algebraic proof of the (t) formula for (\sin \theta) without assuming that (\frac{\theta}{2}) is acute. This proof would provide a more generalized understanding of the formula.
Furthermore, these (t) formulas find application in solving trigonometric equations. For instance, the equation (\cos \theta + \sin \theta = \frac{1}{2}) is solved by substituting (t = \tan\left(\frac{\theta}{2}\right)). After rearranging terms, it leads to a quadratic equation in (t) ((3t^2-4t-1=0)), whose solutions when converted back yield the solutions for (\theta) within the given range.
It's crucial to exercise caution when using the (t) formulas, particularly as certain values of (\theta) might render (\tan\left(\frac{\theta}{2}\right)) undefined, demanding a more nuanced approach in solving equations or evaluating trigonometric functions.
The history and applications of these formulas provide deeper insights into their significance in trigonometry and problem-solving across various mathematical contexts.
