If we write \(t = \tan \theta\), then the following equations are true.
\[\begin{align*} \tan 2\theta &= \frac{2t}{1-t^2}, \\ \sin 2\theta &= \frac{2t}{1+t^2}, \\ \cos 2\theta &= \frac{1-t^2}{1+t^2}. \end{align*}\]Can you use this diagram to obtain these formulae?
We have various interesting triangles to consider in this diagram. Let’s study them to see which lengths and angles we can find. In order to talk about the triangle, it will be convenient to label the important points.
In triangle \(ABC\), since \(B\) is the right angle, \(\tan \theta = \tfrac{AB}{BC}\). Since we are told that \(\tan \theta = t\) and \(BC=2\) we get length \(AB=2t\).
Let’s say that \(AD\) has length \(x\).
We see that we have a triangle with two angles that are both \(\theta\), namely triangle \(ADC\), so this is an isosceles triangle, and so \(DC\) also has length \(x\).
Now \(BD\) has length \(2 - x\).
Triangle \(ABD\) is right-angled, and so we can apply Pythagoras’s theorem: we have \(x^2 = (2-x)^2 + 4t^2\). Rearranging this gives \(0 = 4 - 4x + 4t^2\), and so \(x = 1 + t^2\).
We now know that \(AD = DC = 1 + t^2\), and \(BD = 2 - (1 + t^2) = 1 - t^2\).
Are there any angles in the diagram that might be \(2\theta\)?
\(\angle ADC = 180 - 2\theta\), so \(\angle BDA = 2\theta\).
We are now well placed to consider the trig functions applied to \(2\theta\), because not only do we have a triangle that contains this angle, but we have a right-angled triangle containing the angle (namely triangle \(ABD\)). So let’s focus on that triangle.
Considering the right-angled triangle \(ABD\), we see that \[\sin 2\theta = \frac{2t}{1+t^2},\] \[\cos 2\theta = \frac{1-t^2}{1+t^2},\] and \[\tan 2\theta = \frac{2t}{1-t^2}.\]
For what range of values of \(\theta\) does this argument work?
If \(0^{\circ} < \theta < 45^{\circ}\), then we can draw the diagram given in the question and the argument goes through.
If \(\theta = 45^{\circ}\), then the argument degenerates: point \(D\) is the same as point \(B\), we have \(t = 1\), and \(\tan 2\theta\) is not defined anyway.
If \(45^{\circ} < \theta < 90^{\circ}\), then point \(D\) lies outside the triangle. Does the argument still work?
We can sketch triangle \(ACB\) so that \(\angle C=\theta\) where \(45^{\circ} < \theta < 90^{\circ}\) and \(\angle B = 90^{\circ}\). If \(\angle CAD=\theta\) and \(C, B\) and \(D\) are on a straight line, it puts \(D\) outside triangle \(ACB\). Can we now use a similar process to above?
Do any lengths keep the same algebraic values?
The angle in triangle \(BDA\) is now \(180^\circ -2\theta\). How can we use our knowledge of the functions to relate the trigonometric ratios of these angles to \(\sin {2\theta}\), \(\cos {2\theta}\), \(\tan {2\theta}?\)
