If angle α is reflex, and cos α = -9/41, without using a calculator, evaluate
(a) sin α (b) tan α (c) cos (α - 180 (degrees) .
I asked this question before but was unsure how to ask it again or renew it without creating another thread.
I really don't understand the process of working this out.
Would I bet right in saying: Cos a = -9/41. -9/41 = -0.225. cos-(cos inverse) -0.225 = 103.0028781629139. So Cos A = 103.0028781629139? (But what am I even trying to say in that instance. lol).
A reflex angle is one that is between $180^\circ$ and $360^\circ$. Meaning, it's either in quadrant III or IV.
Recall that $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. Hypotenuse is never negative, so the adjacent must be the negative one. If the adjacent is the negative one, it must be in quadrant III. This also means that the opposite must be negative as well.
I'm not the best artist in the world (although I'm definitely close to being that), but this is an image to show you what I mean by this^. When you have a side that extends to the negative $x$, it will be negative. When you have a side that's to the negative $y$, it will be negative. Otherwise, it's positive.
In this case, because $\cos(\theta)$ is negative, it means that the adjacent side is negative (because hypotenuse is always positive). The adjacent side is the side right next to the angle that's not the hypotenuse (and the angle is always from the x-axis to the hypotenuse). When we look at the diagram, there are only two places where the adjacent side can be negative: In quadrants II and III. However, this is a reflex angle so it can't be in quadrant II. Hence, it must be in quadrant III:
We are not looking for specific lengths, because these are trigonometric ratios. We know the ratio of the lengths of the adjacent to the hypotenuse is $9:41$. Maybe the lengths are $18$ and $82$, but it won't matter because it will reduce to the ultimate ratio.
With that in mind, let's just assume the length of the adjacent is $9$ and that of the hypotenuse is $41$. Pythagorean theorem gives us the length of the opposite side: $\sqrt{41^2-9^2}=40$. Back to the ratio thing, just another confirmation of my previous assertion, if we had $82$ and $18$, we would get the last side to be $80$.
Cool, now we know everything we need to know. The ratio of the lengths is $41:9:40$ (for hypotenuse:adjacent:opposite).
HOWEVER, because it's in quadrant III, the lengths (bit really lengths because length is only positive) will be: $41:-9:-40$.
a) $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}=-\frac{40}{41}$
b) $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{40}{9}$
c) $\cos(\theta-1800)$ - Well, what does this mean. You're just moving the angle back $1800/360=5$ revolutions. You're at the same exact place. I mean, $\cos(0)=\cos(360)=\cos(720)=\dots$, and same thing here. So it will be the same value, $-\frac{9}{41}$.