#Small thing, but not sure how to solve this. :)
38 messages · Page 1 of 1 (latest)
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$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ad}{bc}$
Romans 12:19
or in this case, let $\frac{\sqrt{2}}{2}=u$ then we have $\frac{u}{-u}=-1$
Romans 12:19
so that comes out to be (sqrt2 * 2)/(2 * -sqrt2), would I cancel out the 2's? 🤔
yes, and cancel out a sqrt(2) as well
Oh. Then what am I left with?
-1
it's simply asking for which angles between $0$ and $2\pi$ for which $\cos x=-\frac{\sqrt{2}}{2}$ and $\sin x=\frac{\sqrt{2}}{2}$
Romans 12:19
do u know when sinx=sqrt(2)/2
I don't think so? I might've been taught but forgot
do u know what sin(pi/4) is
I wrote that down somewhere, hold on
oop np
One sec, let me move to my PC.
it's no biggie if you aren't available, for this specific one I can go to tutoring if needed in a few days :)
Alright. So, are you trying to solve this one?
Well, in any case, here's how you do it:
The points on a unit circle have coordinates (cos(θ), sin(θ)). So:
- Identify the signs of cos(θ) and sin(θ). That will determine the quarter that θ is in.
- Find tan(θ) = sin(θ)/cos(θ), then find θ in the quarter that you determined before.
For example, in (I) we have:
cos(θ) = 1/2 > 0
sin(θ) = √(3)/2 > 0
So, the angle is in the first quarter. Next:
tan(θ) = √(3)
θ = π/3 + πn, n ∈ ℤ
The value lying in the first quarter is θ = π/3. If that's not immediately visible, we can solve a double inequality:
0 ≤ π/3 + πn ≤ π/2
0 ≤ 1/3 + n ≤ 1/2
-1/3 ≤ n ≤ 1/6
The integer solution is n = 0, so θ = π/3.
that's what I was tryna do before, but got -1 D:
