Veeky Forums have any hacks for Trigonometry?

I found trig super easy because I had already taken an elementary physics course where my teacher went in depth about the concept of vectors. So read up on vectors and their components.

Also, obligatory unit circle is obligatory. It really does make everything so very easy. The key is to remember
Sin=y/r
Cos=x/r
Tan=y/x

So, at 0°, the y component of the line is obviously 0. At 90°, the y component of the line is the entire line, so call it 1. Now realize, when you go around the circle, from 0° to 90°, sin starts at zero, and then increases to 1 as the angle increases. You can use this as a self-check. If the angle is larger than 45°, that means the y component is longer than the x component, so sin must be larger than cosin. This can help you decide whether or not to use 1/2 or √3/2 for the commonly used angles 30° and 60°. At 60°, you know sin is bigger that cosin, so sin(60°)=√3/2, since √3/2 is larger than 1/2.

Also, it helps to know the identies, and how to derive them.

The most common one is sin^2(x)+cos^2(x)=1. That comes from the equation of a circle, which is f(x)=y^2+x^2. You can then play around with this using simple algebra, for instance, sin^2(x)=1-cos^2(x), or since tan(x)=sin(x)/cos(x), tan^2(x)+1=1/cos^2(x).

So to recap:
Learn how vector components work.
Understand how the unit circle works, making sure to get an intuition for how the values of x and y change with regards to how the angle changes.
Learn the identities by learning how to derive them instead of memorizing them.

For some video resources, intro trig is some pretty light math, so KhanAcademy and PatrickJMT offer good videos on them. Also, sites like mathnotes and stack exchange are incredibly useful tools.

Use complex numbers to get your angle sum formulas.

Thank you very much you all!

End yourself you faggot

>no rational trigonometry in this thread
guys...

i'm sorry i upset you guys, take care anons.

i meant

sin = y/r
cos = x/r
tan = y/x

cot = 1/tan
sec = 1/cos
csc = 1/sin

x^2 + y^2 = r^2

/thread

Could I get some quick help with something here?

How the fuck do I find a side length of a triangle when I'm only given the area and two angles?

The sine and cosine rules in my textbook don't show any examples of how to do this when only given the area.

45-45-90 rule

AN = NM (because the angle is 45º, so half of a square)
(AN)2 = 3200/2
AN = √1600 = 40 = NM
AM = 40√2

Area is half absintheta. Both sides same length therefore area is a squared sun theta

[math]
\overline{AN} = \overline{NM} = x \\
x^{2} = \frac{3600}{2} \\
x = \sqrt{1600} \\
x = 40 \\
\overline{AM} = 40\sqrt{2}
[/math]

Thanks for the help guys.

Apparently, the solution is:

Area = 0.5 x base x height = 0.5 x AN x MN = 0.5 x MN^2
0.5 x MN^2 = 3200m2, giving MN = √6400 = 80 m

my reply is wrong, sorry, I took the time to read on latex though

>unit circle
>good advice
>not learning how to sketch radian angles and know what a reference angle is
Holy kek, you're a memorization monkey

Some old hippie
Caught another hippie
Tripping on acid

That's what my teacher from high school used to say.