[Help] Exact Value - Trig Question

[jamonathin]

Hey guys, whats goin on. I have a big ass calc. test tomorrow and I have no idea how to figure out exact values. And yesh I know I prolyl have a lot of other things I need to work on, but i'm incredibly retarded.

Here's a chart I found on the internet. If anyone can fill that badboy in or at least tell me how to find exact values of cosine, sine and tangent, that'd be great.

Thanks for your help

[wtd]

How do you find exact values? Keep things in terms of fractions. Fractions are exact.

[jamonathin]

that's what I dont know, you're supposed to "remember them".

All i know is that sin(pi)/6 = 1/2

[wtd]

Ah. Well, remember cosine and sine and you have tangent.

For cosine and sine, visualize the waves. That won't give you the exact value, but it'll let you see the basic pattern, which will make it easier to retain the specifics in your mind.

[jamonathin]

Ok, so basically the waves will tell me the size of the fraction, like if it should be big or not.

[brenn]

Eh.. I'm probably not the best person to be teaching you math. But here goes.

Have you ever learnt about the "special triangles" that can help you "calculate" the cosines, sines, and tangents of select angles?

Since you're learning trigonometry, I'll assume you know what a right angle triangle is.

First: Draw an isoceles right angle triangle and label the angle as such: 90, 45 and 45 (in degrees). Now you label the sides:

1) Two of the side lengths are equal. Label these as "1".
2) Using Pythagorean Theorem, you know that the third side is root(2).

Since

Sine = Opposite / Hypoteneuse

sin45 = 1/root(2) <-EXACT VALUE.

Similarly, You can "calculate" the exact values of cos45 and tan45 using this triangle.

=====

If you would like to know the exact values of sine, cosine, and tangent of 30 or 60 degrees, you can construct a different triangle to do so.

First, you create an equilateral triangle, and label the length of each side as "2". /\

If you split this triangle into two equal halves, you have just made two right-angle triangles. /| |\ (I really hope you understood that @___@) Take one of the triangles and ignore the other one--they're the same anyway.

Knowing that each of the angles in the equilateral triangle were 60 degrees, you now know that the angles in the new triangle are 90, 60 and 30. You can also calculate the lengths of each side.

So... like above... just use sine = opp/hyp, cos = adjacent/hyp, and tan = opposite/adjacent to produce the exact values.


===


Phew @___@ It's also really useful to know how to sketch the sine and cosine waves so you can figure out angles that weren't included above. But I'll let someone else explain that ^^;;;

[wtd]

jamonathin wrote:

Ok, so basically the waves will tell me the size of the fraction, like if it should be big or not.

Essentially, yes.

[jamonathin]

Thanks for the crash course there brenn it all makes sense, and I even undertand your /\'s .

Thanks again, both of yous

[Bacchus]

special triangles are used because for them, the ratio of sides is quite easy to remember.

..............|\...<-- 30 degrees
..............|..\
...root(3).|....\.2
..............|......\
..............|........\
..90........|______\.<-- 60 degrees
degrees.......1

those are the ratios of the sides. they can be that small because with angles if you increase one side, to keep the same angle you must increase the other sides. this is the 30-60 special triangle. so to find the sine/cosine/tangent of 30 degrees, 60 degrees, or even 90 degrees lol you can use these ratios
ex) to find the sine of 60 degrees
= sin(opposite/hypotenues)
=sin(root(3)/2)

thats how you use a special triangle. the theres the 45 triangle like brenn said
.............45 degrees
...........|
...........|................*slashes wont work, to steep. pretend points connect
........1.|..............root(2)
...........|
.......90|_________..45 degrees
..degrees....1

and you use that the same way as the 30-60 triangle
ex) sine of 45 degrees:
=sin(opposite/hypoteneus)
=sin(1/root(2))

hope that helps a bit

[jamonathin]

It does, it'll work out for my test in peroids