simple trig problem

[sacred flicker]

its for a unit circle.

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[hollow crest]

What did you get when you solved the equation for θ?

[sacred flicker]

you mean like inverse tan of 12/5?

because i tried to do tan = sin/cos -> sin/cos = 12/5 -> 5sin = 12cos -> 12x = 5y

[rigid flint]

draw a unit circle

and a triangle

[sacred flicker]

i did 💀

[rigid flint]

since tan is positive, theta is in the ..? quadrant

[sacred flicker]

3

idk how that helps me tho

[rigid flint]

yes

so cos is negative and sin is negative

then you have your triangle

opposite over adjacent = 12/5

[sacred flicker]

so 12x = 5y but x and y are both negative

but its still a relation

[rigid flint]

the scale does not matter,

so you can conveniently make the opposite side 12 and the adjacent side 5

[hollow crest]

Well, yeah, but specifically in the given interval.

[sacred flicker]

so sin = -12 and cos = -5?

[rigid flint]

no

sine is opposite over hypotenuse

not just opposite

[sacred flicker]

h is 1

[rigid flint]

h is not 1 if you do this

[sacred flicker]

oh bruh

its not a unit circle then

[rigid flint]

true

[sacred flicker]

[hollow crest]

Ohh, that's how you're trying to solve it...

Well, sure, alright. That works, too.

[sacred flicker]

it didn't work

im lost

[rigid flint]

uh, do you know how to find the hypotenuse

[sacred flicker]

yes but then its not a unit circle

[hollow crest]

You just didn't finish doing it yet.

[rigid flint]

true

doesn't matter, trig functions work for all right triangles

[sacred flicker]

if its not a unit circle then i cant say stuff like cos = x

right

[hollow crest]

Maybe we should just solve it the usual way?

[sacred flicker]

idk what that is

i missed the class on it

[rigid flint]

you have cos = x over the hypotenuse

which
you
can figure out with pythagorean theorem

[hollow crest]

Wait, what? What does cos(x) = x have to do with the problem?

[sacred flicker]

i thought there was a simpler way

[hollow crest]

Yes, there is.

[sacred flicker]

that uses unit circle

[hollow crest]

Oh.

No, my way doesn't use the unit circle, really.

[sacred flicker]

isn't the fact that its unit circle important?

or should I ignore

[hollow crest]

Not sure what you are planning to use it for.

[sacred flicker]

idk our teacher just said we use unit circle

so idk

[hollow crest]

Oh... Well, ok.

Then you can continue with this approach.

[rigid flint]

what?

cos = x/hyp

[hollow crest]

After that I can show the approach I usually use for these problems if you're interested.

[rigid flint]

oh i miswrote

[hollow crest]

Ah, ok.

[rigid flint]

i don't know of any other way to solve this problem

you can theoretically solve sin/cos = (given tan), sin^2 + cos^2 = 1

but triangle is easier i think

[hollow crest]

Well, first you solve the equation, then you use the identities with trig and inverse trig functions.

I can show if you want.

[sacred flicker]

ok sure

[rigid flint]

if kerbaljeb is insistent on using a unit circle then ye

[sacred flicker]

i got the answer but im not entirely sure why

idk why we need to use unit circle in the 1st place

teacher put a lot of emphasis on it so idk

[hollow crest]

Here's how I would do it.
tan(θ) = 12/5
θ = arctan(12/5) + πn, n ∈ ℤ
Let's determine which value belongs to our interval.
π < arctan(12/5) + πn < 2π
1 < arctan(12/5)/π + n < 2
1 - arctan(12/5)/π < n < 2 - arctan(12/5)/π
As 0 < arctan(12/5)/π < 1/2, the only solution is n = 1. So, the root is:
θ = π + arctan(12/5)
Then we use the identities: sin(arctan(x)) = x/√(1 + x^2), cos(arctan(x)) = 1/√(1 + x^2). So:
sin(π + arctan(12/5)) = -sin(arctan(12/5)) = -(12/5)/√(1 + (12/5)^2) = -12/13
cos(π + arctan(12/5)) = -cos(arctan(12/5)) = -1/√(1 + (12/5)^2) = -5/13

[rigid flint]

darpinger please

why dropping a bunch of identities

it's impossible to memorize them all

[rigid flint]

my method concludes with: hypotenuse is 13, sine is -12/13, cos is -5/13
you can check those are actually the same values that x/sqrt(1+x^2) and 1/sqrt(1+x^2) produces

[hollow crest]

Well, these two are really the only ones you should memorize. The others are very easy to derive on the go.

[open ridge]

Isn't the answer sin = -12/13 and cos = -5/13?

[hollow crest]

Yes.

[open ridge]

Wait a second

Why did oyu say that 0 < arctan(12/5)/pi < 1/2?

How do you know this?

[open ridge]

.

[hollow crest]

Well, 0 < arctan(x) < π/2 for positive x, so 0 < arctan(x)/π < 1/2.

[open ridge]

Smart

Thank you