The t-Formulae Revisited

Let $t = \tan\dfrac{\theta}{2}$. We derive $\sin\theta$ using the double-angle formula $\sin\theta = 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}$.

From a right triangle with angle $\dfrac{\theta}{2}$, opposite $= t$, adjacent $= 1$, hypotenuse $= \sqrt{1+t^2}$:

For $\cos\theta$, use the double-angle formula $\cos\theta = \cos^2\dfrac{\theta}{2} - \sin^2\dfrac{\theta}{2}$:

For $\tan\theta$, divide: $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{2t/(1+t^2)}{(1-t^2)/(1+t^2)} = \dfrac{2t}{1-t^2}$.

Answer the hook: $t = \tan 15°$, so $\theta = 30°$. Thus $\dfrac{2t}{1+t^2} = \sin 30° = \dfrac{1}{2}$.

t = {2}; triangle: opp = t, adj = 1, hyp = 1+t^2; = 2t{1+t^2}; = 1-t^2{1+t^2}; = 2t{1-t^2}

Pause — copy the three $t$-formulas into your book along with the derivation triangle: opposite = $t$, adjacent = $1$, hypotenuse = $\sqrt{1+t^2}$, then use double-angle formulas to get $\sin\theta$ and $\cos\theta$.

We just saw that $t = \tan\frac{\theta}{2}$ gives $\sin\theta = \frac{2t}{1+t^2}$ and $\cos\theta = \frac{1-t^2}{1+t^2}$. That raises a question: identity proofs often contain $1+\cos\theta$ or $1-\cos\theta$ — can we simplify these directly using $t$, without going through the full $\cos\theta$ formula? This card answers it → yes: $1+\cos\theta = \frac{2}{1+t^2}$ and $1-\cos\theta = \frac{2t^2}{1+t^2}$.

The t-formulas are particularly useful for proving identities that would be harder with direct trig manipulation. The strategy is: convert both sides to $t$, simplify each, show they are equal.

Example: prove that $\dfrac{\sin\theta}{1+\cos\theta} = \tan\dfrac{\theta}{2}$.

1+ = 1 + 1-t^2{1+t^2} = 2{1+t^2}; 1- = 1 - 1-t^2{1+t^2} = 2t^2{1+t^2}

Pause — copy the derived $t$-forms into your book: $1+\cos\theta = \frac{2}{1+t^2}$ and $1-\cos\theta = \frac{2t^2}{1+t^2}$, where $t = \tan\frac{\theta}{2}$ — use these to simplify identity proofs.