Question 9 a) You are given sin 60^∘ = (√(3))/(2), cos 60^∘...

Question 9 a) You are given $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 60^\circ = \frac{1}{2}$ and $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$. Use these results to find $\sin 15^\circ$, giving your answer in surd form. You must show all of your working. b) Solve $3 \cos 2\theta + \cos \theta = -1$ ($0^\circ \le \theta \le 360^\circ$)

Transcript

00:01 The ultimate goal is to find the sine, cosine, and tangent half angle four values for the given value, for the given information for alpha.

00:08 So the given information we have is that the tangent of alpha is equal to negative four thirds and that pi over two is less than alpha, which is less than pi.

00:17 So the first thing we're going to do is we're going to figure out which quadrant this is in.

00:21 And that's what we're given here.

00:22 This is the restriction.

00:23 Well, remember, pi over two is like 90 degrees and pi is 180 degrees.

00:27 So alpha is somewhere between 90 degrees and 180 degrees, which would be a second quadrant angle.

00:33 So we can go ahead and draw that up.

00:36 So here would be our reference triangle, which means here would be our angle for alpha.

00:41 Now we're going to go to the fact that we know that the tangent of alpha is negative 4 over 3.

00:45 Remember, that tangent is the ratio of the opposite side over the adjacent side.

00:50 So in this case, the opposite side would be positive 4, and the adjacent side will be negative 3 because we're in the second quadrant.

00:56 So now what we need to do is we need to find our hypotenuse, which i'll call x.

01:01 So to find x, we can use pythagorean theorem.

01:04 So we'll have x squared equal to 4 squared plus negative 3 squared.

01:08 Well, 4 squared is equal to 16.

01:10 Negative 3 squared is 9.

01:12 16 plus 9 is 25.

01:14 So now, to solve for x, we take the square of the both sides, and the square of the 25 is 5.

01:20 Perfect.

01:20 So now we know our hypotenuse is 5, so we can go ahead and labeling.

01:24 So now that we have all three sides for the two sides.

01:26 Triangle, now we can go ahead and start solving for the three pieces of information that we want.

01:32 So in part a, we're being asked to find the sign of alpha divided by two.

01:37 So in this case, we have to use our half angle formula.

01:40 Well, the half angle formula for sign is plus or minus the square room of one minus the cosine of alpha, all divided by two.

01:50 So now we just need to know our cosine value.

01:53 So, or let's actually talk about this first.

01:56 How do we know if we're going to use plus or minus.

01:58 Well, notice that this angle is in the second quadrant.

02:00 Like i mentioned before, is between 90 degrees and 180 degrees.

02:04 Well, half of 90 is 45, and half of 180 is equal to 90.

02:09 So that means this angle is going to be in the first quadrant, which means that all of these answers are going to be positive, because all of your trig functions are positive in the first quadrant.

02:18 All right, so now we know it's just a square root of 1 minus, well, the cosine of alpha.

02:22 Remember, the ratio for cosine is it's the adjacent side over the hypotenuse.

02:27 So that's going to be negative 3 divided by 5.

02:30 And this is all still getting divided by 2.

02:32 So now what they want us to do is they just want us to simplify this as much as possible.

02:37 So in the numerator, we have 1 minus negative 3 5ths, which is equal to, excuse me, 8 over 5.

02:45 So we have the square root of 8 5ths getting divided by 2.

02:48 So now what we can do to get rid of our denominator is i can multiply below.

02:52 The numerator and the denominator by five.

02:55 This will at least get rid of the fraction in the numerator because those fives will cancel.

02:58 So now we're going to have the square root of eight divided by two times five, which is 10.

03:04 So now let's go ahead and let's see how much they want to simplify...

Question

Please give Ace some feedback