Bisection and Duplication with Negative Cosine

Text

Given that $\cos\alpha=-\dfrac15$ and $180°<\alpha<270°$, use trigonometric formulas to find $\sin\alpha$, $\cos 2\alpha$ and $\tan\dfrac\alpha2$.

Solution

$\alpha$ lies in the third quadrant, so $\sin\alpha<0$: $\sin\alpha = -\sqrt{1-\frac{1}{25}} = -\frac{2\sqrt6}{5} \approx -0.980.$ From the duplication formula: $\cos 2\alpha = 2\cos^2\alpha - 1 = -\frac{23}{25} = -0.92.$ Since $\dfrac\alpha2\in(90°,135°)$, we have $\tan\dfrac\alpha2<0$; from the bisection formula: $\tan\frac\alpha2 = \frac{1-\cos\alpha}{\sin\alpha} = \frac{1.2}{-0.980} \approx -1.225.$

$\boxed{\sin\alpha=-\tfrac{2\sqrt6}5,\ \cos2\alpha=-\tfrac{23}{25},\ \tan\tfrac\alpha2\approx-1.225}$