Duplication and Bisection Formulas from a Known Value

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Use trigonometric formulas (without evaluating $\alpha$ with a calculator) to answer the following:

(a) $\sin\alpha=\dfrac14$ and $90°<\alpha<180°$: find $\cos\alpha$, then $\sin 2\alpha$ and $\cos\dfrac\alpha2$.

(b) $\tan\alpha=\dfrac12$ and $0°<\alpha<180°$: find $\sin\alpha$, $\cos\alpha$, $\cos\left(\alpha+\dfrac\pi2\right)$, $\sin\left(\alpha-\dfrac\pi3\right)$.

(c) $\cos\alpha=\dfrac13$ and $270°<\alpha<360°$: find $\sin\dfrac\alpha2$ and $\cos\dfrac\alpha2$.

Solution

(a) $\alpha$ lies in the second quadrant, so $\cos\alpha<0$: $\cos\alpha = -\sqrt{1-\frac{1}{16}} = -\frac{\sqrt{15}}{4} \approx -0.968.$ From the duplication formula: $\sin 2\alpha = 2\sin\alpha\cos\alpha = -\frac{\sqrt{15}}{8} \approx -0.484.$ Since $\dfrac\alpha2\in(45°,90°)$, we have $\cos\dfrac\alpha2>0$; from the bisection formula: $\cos\frac\alpha2 = \sqrt{\frac{1+\cos\alpha}{2}} \approx 0.126.$

(b) $\tan\alpha=\dfrac12>0$ with $0°<\alpha<180°$ means $\alpha$ lies in the first quadrant: $\sin\alpha = \frac{1}{\sqrt5} \approx 0.447, \qquad \cos\alpha = \frac{2}{\sqrt5} \approx 0.894.$ With the associated-angles formula: $\cos\left(\alpha+\frac\pi2\right) = -\sin\alpha \approx -0.447.$ With the subtraction formula: $\sin\left(\alpha-\frac\pi3\right) = \sin\alpha\cos\frac\pi3 - \cos\alpha\sin\frac\pi3 \approx 0.447\cdot0.5 - 0.894\cdot0.866 \approx -0.551.$

(c) $\alpha$ lies in the fourth quadrant; since $\dfrac\alpha2\in(135°,180°)$, we have $\sin\dfrac\alpha2>0$ and $\cos\dfrac\alpha2<0$: $\sin\frac\alpha2 = \sqrt{\frac{1-\cos\alpha}{2}} = \sqrt{\frac13} \approx 0.577,$ $\cos\frac\alpha2 = -\sqrt{\frac{1+\cos\alpha}{2}} = -\sqrt{\frac23} \approx -0.816.$

$\boxed{\cos\alpha=-\tfrac{\sqrt{15}}4,\ \sin2\alpha=-\tfrac{\sqrt{15}}8,\ \cos\tfrac\alpha2\approx0.126;\ \ \sin\alpha=\tfrac1{\sqrt5},\ \cos\alpha=\tfrac2{\sqrt5};\ \ \sin\tfrac\alpha2=\tfrac1{\sqrt3},\ \cos\tfrac\alpha2=-\sqrt{\tfrac23}}$