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/sci/ - Science & Math

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File: 32 KB, 488x414, Halplease.jpg [View same] [iqdb] [saucenao] [google]
1200190 No.1200190 [Reply] [Original]

http://www.act.org/compass/sample/trig.html

This is the last and only part I couldn't figure out by myself.

I already know about SOHCAHTOA

sine = opposite/hypotenuse
cosine = adjacent/hypotenuse
tan = opposite/adjacent

That's about it. I are fuckin' retarded.

I'm just trying to pass the introductory exam so I can get into community college so I can join the army since they don't accept GEDs anymore and I dropped out of high school because they used to. I know, i'm retarded lol.

>> No.1200214

1- C
2- E

>> No.1200216

>>1200214
Yeah, I know, the answers are right below. I want to know how to figure them out.

>> No.1200218

>>1200216
Can't help you there. We're all too smart for trig.

>> No.1200231

>>1200218
How can you to be too smart to explain a part of math?

>> No.1200238

>>1200190
<span class="math">adjacent tan27^\circ[/spoiler]

>> No.1200241

First one is just basic rules. Google trig rules/trig identities.

Second one is Tan@= op/ad

So its tan 27 = op/45
(multiply both sides by 45)
45 tan 27 = opposite length.

>> No.1200247

second one:
?/45=tan(27°)
?=45*tan(27°)

>> No.1200265

1. Standard trig identity: sin^2(x) + cos^2(x) = 1
thus sin^2(x) = 1 - cos^2(x)
thus (1-cos^2(x))/cos^2(x) = sin^2(x)/cos^2(x) = tan^2(x)

2. If you know SOHCAHTOA then what's the problem.
tan(angle) = opposite/adjacent
tan(27) = height/45
I'm sure you can figure the rest out.

>> No.1200268

lrn2trig
1 =tan^2theta because 1=sin^2theta*cos^2theta (identities faggot)
2 = 45 tan 27deg because multiplying the tangent of the degree of the angle by teh adjacent leg of the triangle will give you the oppisite leg's legnth
ie: tan27deg= ?/45units
fucking simple shit, what is wrong with you?

>> No.1200269

>>1200231
Because it's pretty damn boring explaining basic rules of trig to somebody who claims to know them

>> No.1200272

>>1200268
oops, wrong operation there on the identities part lol, + not *

>> No.1200277
File: 7 KB, 567x130, trig_ident.png [View same] [iqdb] [saucenao] [google]
1200277

first one

>> No.1200500

Ok, I got the concept behind the second one.

The first one. I can't memorize rules for shit. I need to understand the concept behind the first one. The reasoning for the rules.

Please, last question I promise.

>> No.1200559

>>1200500
Please help, op here.

>> No.1200563 [DELETED] 

Ｗ Ｗ w . ａ ｎ Ｏ ｎ t Ａ ｌ ｋ . s ｅ yxfu icvarqqh t gijwdigvun sd qs cofl oj ca vsvg

>> No.1200616

sin(x)=op/hyp
cos(x)=adj/hyp
sin(x)/cos(x)=(op/hyp)/(adj/hyp)=op/adj=tan(x)

ok?

>> No.1200661

>>1200616
so far so good...

how does this help me solve problem #1 and problems like it though?

>> No.1200718

see trig_ident.png

sin^2(x)+cos^2(x)=1
sin^2(x)=1-cos^2(x)

(1-cos^2(x))/cos^2(x)=sin^2(x)/cos^2(x)=tan^2(x)

>> No.1200743

>>1200661
1 - cos(superscript 2)(theta)
_______________________(over or divided by)
cos(superscript 2)(theta)

by superscript 2 i mean the 2 that we see when numbers are squared by theta i mean the sign for angle which is a vertical oval with a 45 clockwise degree diagonal line relative to a horizontal line)

so basically it's...

1 - (adj)(adj)(hyp)(hyp)
__________________
(adj)(adj)(hyp)(hyp)

adj = adjacent
hyp = hypotenuse

then we get...

1/(adj)(adj)(hyp)(hyp) - 1

right?

am i missing something?

the above is basically...

1/cos(superscript 2 which i guess means squared) - 1

Halp, what does all this mean.

>> No.1200781

>>1200718
yeah i got that...

but i can't remember that

or rather, it'd be easy to understand the reasoning behind it... so i wouldn't have to memorize the rules, but i could figure them out on my own if i forgot, hence i wnat to 'learn' them instead of 'memorizing' them

this stems from really bad memory

basically what i'm asking is


why is...

sin^2(x) = 1 - cos^2(x)?

that would mean that...

(opp)(opp)/(hyp)(hyp) (x) = 1 - (adj)(adj)/(hyp)(hyp)

wouldn't it?

>> No.1200854

>>1200781
com'n i'm so close... just tell me the reasoning behind the rules

>> No.1200882

(o^2/h^2)+(a^2/h^2)=1
simplified:
(o^2+a^2)/h^2=1
now:
(o^2+a^2)=h^2 (pythagoras)
so:
h^2/h^2=1
thus:
(o^2/h^2)+(a^2/h^2)=1
or
sin^2(x)+cos^2(x)=1
called pythagorean identity

>> No.1200938
File: 168 KB, 343x450, mindblown.jpg [View same] [iqdb] [saucenao] [google]
1200938

>>1200882

>> No.1200942

>>1200854
http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity

>> No.1200998

1. /frac{1-\cos^2 \theta}{\cos^2 \theta}=\frac{\sin^2\theta}{\cos^2 \theta}=\tan^2 \theta

2. \tan(27)=\frac{?}{45}
?=45\tan(27)

C
B

>> No.1202436

>>1200718
>>1200882
OK LAST QUESTION I FUCKIN' PROMISE YOU GUYS HAVE BEEN AWESOME SO FAR

Why does it equal 1?

If I can understand that I'm pretty much set on trig

as long as I understand the arbitrary basics I can deduce everything else

>> No.1202489

>>1202436
Try it out in a calculator.

>> No.1202517
File: 33 KB, 481x309, Calculator.jpg [View same] [iqdb] [saucenao] [google]
1202517

>>1202489
lol

that's the thing

how do i do it in a calculator

i was going to ask this

but i figure i should google it instead since it seems so easy

so just tell me how to put it in a calc and your done

preferabbly ms calc

>> No.1202560

>>1202489
>>1202517
Comnnnnnnnnnn!

Pwez

>> No.1202598

>>1202517
how simple is this you guys

just tell me how to enter it in calc

>> No.1202612

I lol'd.

>> No.1202731
File: 13 KB, 500x500, lol trig.png [View same] [iqdb] [saucenao] [google]
1202731

There's a little handwaving at the end of this.

Imagine a unit circle centered at the origin. That is, a circle of radius 1, which has its center at the "zero point" of your graph. Inches, miles, units are irrelevant. 1.

Imagine that a right triangle starts out not as a triangle at all, but as a line segment, from the center of the circle, to the right, or positive, point on the x axis where the circle and the x axis intersect. Shortly, this becomes a right triangle which always has the three properties: one vertex is the center of the circle, one vertex is the point on the circle where the hypotenuse touches, and the third point is still on the x axis.

Consider Theta, the relevant angle. Now, sine and cosine are "opposite/hypotenuse" and "adjacent/hypotenuse". But what is the hypotenuse? In every case, it's 1, the radius of the circle! So the sine of theta is precisly the length of the leg parallel to the y axis, and the cosine of theta is precisley the length of the leg coinciding with the x axis. But what are we dealing with? Right triangles! So Pythagoras's Theorem applies! And since every similar right triangle of whatever size can be corresponded to one of these little right triangles with unit hypotenuse, we see that in every case,

(sin theta)^2, or sin^2 (theta) as it is commonly written, added to cos^2 (theta), is always 1!

A generic EQUATION involving a variable is true when certain values are plugged in. x^2 = 4 is true when x is 2, or -2. An IDENTITY is a special KIND of equation, which is always true, no matter what number you plug in! Thus, the pythagorean identity.

>> No.1202735

1-sin(theta)=cos(theta)

1-cos(theta)=sin(theta)

sin(theta)+cos(theta)=1

in number 1, sin(theta)/cos(theta) = tan(theta)

>> No.1202769

>>1202731

Addendum: my picture fails in that it suggests that the angle theta is always such: 0 degrees < theta < 90 degrees, or, in radians, 0 < theta < pi. We're speaking of the extension of the trig functions to accept any argument. In other words, the angle "spools around" counterclockwise from the positive x-axis. This is admittedly a poor phrasing of what's going on.

>> No.1203095

>>1202731
ok

how do i put it in a calc

like the one here:
>>1202517

and that '1' shit is retarded no offense they should just use x, the symbol for theta is a vertical oval with a diagonal upleft to downright slash in it for a reason, so as to not be confused with x.

the radius of the circle could easily be 2, 3, 4, or any fuckin' of the infinite number of numbers or fractions.

I hate arbitraries in math.

but now that i think about it, it probably isn't arbitrary...

anyways

what i really need help with now

is how to put it in a calc

>> No.1203132

>>1202517
>>1202560
>>1202598
>>1203095
How do I put it in a calculator /sci/entists?

>> No.1203234

>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
>>1203132
/sci/ doesn't think i'm going to give up, does it?

>> No.1203308

>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
>>1203234
How do I do trigonometric functions on a calculator

take two minutes to answer

please

>> No.1203325

OP, read >>1202731 and report back, I beg of you.

>> No.1203414

>>1203325
I read it, I get it.

I just wanna know how to put trigonometric functions in calculators, that's all.

>> No.1203448

>>1203325
>>1203414

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

>> No.1203495

>>1203448
depends on the trigonometric function. tan(whatever the angle is) gives you the tangent. tan^-1(y/x) gives you the angle. harder ones you have to do by hand though

>> No.1203513

>>1203495
can i has pictures?

to explain for each of the three?

please!!!!!!!!!!!!!

i'll love you so much seriously

add me on facebook

facebook.com/paul.obrien.hunter

>> No.1203541

>>1203513

Pictures of what?

>> No.1203552

>>1203541
to explain what this guy meant:
>>1203495

>> No.1203560

Hypotenuse is only equal to 1 when you are referring to the unit circle.

>> No.1203563

>>1203513
Google ffs and figure it out. If you can't even do that you don't have a chance of passing your maths course.

>> No.1203576

>>1203563
this is the only question i don't completely understand

facebook request me

i'm nice to everyone

please help me

i'm dumb

google it for me and link to the site

i'll do the rest

ok>?

>> No.1203612

>>1203576
google it for me and link to the site

i'll do the rest

ok>?

I just wanna know how to put trigonometric functions in calculators, that's all.
I just wanna know how to put trigonometric functions in calculators, that's all.
I just wanna know how to put trigonometric functions in calculators, that's all.

I just wanna know how to put trigonometric functions in calculators, that's all.

SORRY TO SPAM BUT I MUS TKNOW THIS TST IS TOMORROW!!!!!!11111111one

>> No.1203653

>>1203612

I'd actually help if you wanted to learn.

>> No.1203658

>>1203653
I do want to learn.

Teach me. Please. I have aim. paul0brienhunter

i'm online right now

>> No.1203667

>>1203552

How are we supposed to know what kind of calculator you have? Doing trig tends to be different between calculators.

The on you purchased probably had an instruction book or at least a little slip with instructions. Read it.

>> No.1203673

>>1203658
>I just wanna know how to put trigonometric functions in calculators, that's all.

That's not leaning.

>> No.1203688

>>1203673

or learning either.

>> No.1203692

>>1203673
Ok,...

then tell me

about it...

>>1203495


com'n man

tellm e about that, about tan and then about the two others two please ok?

>> No.1203729
File: 2 KB, 252x205, Please explain.jpg [View same] [iqdb] [saucenao] [google]
1203729

>>1203495
Ok please explain

look i even took my time to draw the template for you

>> No.1203864

>>1203729


Do you understand the concept of a function?

Lets define f(x) as 2x
f(x) = 2x
Taking f(2) means
f(2)= 2(2) = 4
The trig functions work in a similar way but they are pre defined.
tan (u) = y/x
Taking tan(pi/4) = 1/1

The inverse of a function "undoes" the function

f^-1(f(2)) = 2
f^-1(4) = 2

tan^-1(tan(pi/4)) = pi/4
tan^-1(1/1) = pi/4

Use the inverse trig functions to find angles.
Use the trig functions to find side lengths.

>> No.1203885

>>1203864

10% of an Internet he will want more.

>> No.1203890

20% of an internet he won't be happy till someone flat out gives him the answer.