Question

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.

Answer 1

There are two ways. First the shortcut, the acute angle whose sine is `1/2` is `30^circ` so `sin 2 theta = sin 60^circ = \sqrt{3}/2` and `sec 60^circ = 1/{cos 60^circ} = 1/(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 \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 = + \sqrt{ 1- s^2 }`

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

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

`cos 2 theta = 1 - 2 sin ^2 theta = 1 - 2s^2`

`sec 2 theta = 1/{cos 2 theta} = 1/{1-2s^2}`

We can check our formulas by trying `s=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`