# Hi guys! Can anyone help me solve this question? Just started learning this chapter so I'm not quite familiar with this : Given sin theta = 1/2 and that theta is an acute angle. Evaluate sin 2 theta & sec 2 theta.

Apr 30, 2018

There are two ways. First the shortcut, the acute angle whose sine is $\frac{1}{2}$ is ${30}^{\circ}$ so $\sin 2 \theta = \sin {60}^{\circ} = \setminus \frac{\sqrt{3}}{2}$ and $\sec {60}^{\circ} = \frac{1}{\cos {60}^{\circ}} = \frac{1}{\frac{1}{2}} = 2.$

#### Explanation:

Sure, we're happy to help. The first way was a trick, an angle we knew, ${30}^{\circ} .$ Of course, 99% of trig as taught in school uses the same trick, 30/60/90 or 45/45/90, so it's good to notice,

Let's solve it in general, say we're given $s = \sin \theta$ and told the sign of the cosine.

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$ which we can solve for $\cos \theta$,

$\cos \theta = \pm \setminus \sqrt{1 - {s}^{2} \theta}$

Let's be definite and say the problem tells us to chose the positive sign, like the acute $\theta$ in the first quadrant would be.

$\cos \theta = + \setminus \sqrt{1 - {s}^{2}}$

Then we're asked for $\sin 2 \theta$ and $\sec 2 \theta = \frac{1}{\cos 2 \theta} .$ These use the double angle formula:

$\sin 2 \theta = 2 \sin \theta \cos \theta = 2 s \setminus \sqrt{1 - {s}^{2}}$

$\cos 2 \theta = 1 - 2 {\sin}^{2} \theta = 1 - 2 {s}^{2}$

$\sec 2 \theta = \frac{1}{\cos 2 \theta} = \frac{1}{1 - 2 {s}^{2}}$

We can check our formulas by trying $s = \frac{1}{2}$,

sin 2 theta = 2 (1/2) \sqrt{1 - (1/2)^2} = sqrt{3}/2 quad sqrt

sec 2 theta = 1/{1 - 2 (1/2)^2} = 2 quad sqrt