Question:

When \(\tan\theta = -2\), find \(\cos 2\theta\) and \(\sin 2\theta\).

Answer:

Maybe, you don’t believe that you can derive \(\sin 2\theta\) from \(\tan \theta\). This might be a little tricky, but if you come up with a relationship between \(\tan^2 \theta\) and \(\cos^2 \theta\) first. Then, \(\cos^2 \theta\) can be converted into \(\cos 2\theta\).

Remember

\[ 1+\tan^2\theta = \frac{1}{\cos^2\theta} \]

This can be derived from \(\sin^2\theta + \cos^2\theta = 1\) and \(\tan\theta = \frac{\sin\theta}{\cos\theta}\).

Since \(\tan\theta = -2\), we have

\[ \cos^2\theta = \frac{1}{1+\tan^2\theta} = \frac{1}{1+(-2)^2} = \frac{1}{5} \]

Now, recall the double-angle formula as for cosine:

\[ \cos 2\theta = 2\cos^2\theta -1 \]

Plug above result into this.

\[ \cos 2\theta = 2\times\frac{1}{5} -1 = -\frac{3}{5}\]

Also remember

\[ \sin 2\theta = 2\sin\theta\cos\theta \]

Mmm, well, we don’t know \(\sin\theta\)… What should we do? Wait! Modify the right hand side as follows:

\[ \sin 2\theta = 2\sin\theta\cos\theta \times \frac{\cos\theta}{\cos\theta} = 2\tan\theta\cos^2\theta \]

Thus, we have

\[ \sin 2\theta = 2\times(-2)\times\frac{1}{5}=-\frac{4}{5}\]

Isn’t this cool? Let me know your opinion!