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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.