Trigonometric functions 2

Question:
When $\tan\theta = -2$, find $\cos 2\theta$ and $\sin 2\theta$.

Answer:
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
\[ \sin 2\theta = 2\sin\theta\cos\theta \]
Modify the right hand side.
\[ \sin 2\theta = 2\sin\theta\cos\theta \times \frac{\cos\theta}{\cos\theta} = 2\tan\theta\cos^2\theta \]
Thus, we have
\[ \sin 2\theta = 2\times(-2)\times\frac{1}{5}=-\frac{4}{5}\]

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