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.

Question

Nolan Alexander · Accepted Answer

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.# 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 #

Victoria Adebayo · Answer

undefined Sure! Given $ \sin \theta = \frac{1}{2} $, and knowing that $ \theta $ is an acute angle, we can use the double-angle identities to find $ \sin(2\theta) $ and $ \sec(2\theta) $.

1. $ \sin(2\theta) = 2\sin \theta \cos \theta $
2. $ \sec(2\theta) = \frac{1}{\cos(2\theta)} $

Given that $ \sin \theta = \frac{1}{2} $, and $ \cos \theta $ can be found using the Pythagorean identity $ \sin^2 \theta + \cos^2 \theta = 1 $, we can solve for $ \cos \theta $.

1. $ \sin \theta = \frac{1}{2} $
2. $ \cos \theta = \sqrt{1 - \sin^2 \theta} = \frac{\sqrt{3}}{2} $

Now, we can substitute $ \sin \theta $ and $ \cos \theta $ into the double-angle formulas:

1. $ \sin(2\theta) = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \sqrt{3} $
2. $ \sec(2\theta) = \frac{1}{\cos(2\theta)} = \frac{1}{\cos^2 \theta - \sin^2 \theta} = \frac{1}{(\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2} = \frac{1}{\frac{3}{4} - \frac{1}{4}} = \frac{4}{3} $

So, $ \sin(2\theta) = \sqrt{3} $ and $ \sec(2\theta) = \frac{4}{3} $.

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 &amp; sec 2 theta.

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.