/sci/ have any hacks for Trigonometry?

>>8333535
Yes.
Understand the unit circle and you can quickly find back any trig formula you need.

Say it loud

SOH CAH TOA

Sintheta =Opposite /Hypotenuse

Costheta = adjacent /hypo

Tantheta = opposite /adjacent

>>8333547
>>8333551
Great shit anon's, thank you!

>>8333547
good advice
>>8333551
brainlet advice

>>8333601
>SOH CAH TOA
>Useful mnemonics is degenerate.
Please, end yourself.

>>8333616

He didnt say its degenerate.
Go back to /pol/, containment board for idiots like you.

>>8333547
Yeah, do this.

It might be useful to memorize the unit circle for quick computation but you need to understand how the values of the trig functions are derived. Think of Cartesian coordinates when thinking about side lengths.

Once you understand one quadrant of the unit circle, the others are just a reflection across x/y axis with a different principal angle but a similar reference angle.

File: unit-circle.gif (59KB, 1024x1024px)

59KB, 1024x1024px

>>8333535
[math] \displaystyle
\begin{matrix}
angle & sin & cos \\
0 & \sqrt{0}/2 & \sqrt{4}/2 \\
\pi/6 & \sqrt{1}/2 & \sqrt{3}/2 \\
\pi/4 & \sqrt{2}/2 & \sqrt{2}/2 \\
\pi/3 & \sqrt{3}/2 & \sqrt{1}/2 \\
\pi/2 & \sqrt{4}/2 & \sqrt{0}/2
\end{matrix}
[/math]

>>8333535

Make sure you learn this great, easy system called "rational trigonometry" ;)

>>8333535
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.

>>8333535
Use complex numbers to get your angle sum formulas.

File: motivation.jpg (745KB, 1324x1101px) Image search: [Google]

745KB, 1324x1101px

>>8333641
>>8333651
>>8333703
>>8333707
Thank you very much you all!

>>8333601
End yourself you faggot

>no rational trigonometry in this thread
guys...

>>8333601
>>8333944
i'm sorry i upset you guys, take care anons.

>>8334100
i meant >>8333616

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

File: triangle.png (23KB, 228x310px) Image search: [Google]

23KB, 228x310px

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.

>>8334169
45-45-90 rule

>>8334169
Area is half absintheta. Both sides same length therefore area is a squared sun theta

>>8334169
[math]
\overline{AN} = \overline{NM} = x \\
x^{2} = \frac{3600}{2} \\
x = \sqrt{1600} \\
x = 40 \\
\overline{AM} = 40\sqrt{2}
[/math]

>>8334175
>>8334202
>>8334214
>>8334226
>>8334228
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

>>8334233
my reply is wrong, sorry, I took the time to read on latex though

>>8333601
>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

>>8333551
Some old hippie
Caught another hippie
Tripping on acid

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