
How do you convert from radiants to degrees?
for example pi/4 to degrees?
You would multiply by 180/pi to go from radiants to degrees.

How would you convert degrees to radiants?
for ex) 90 degrees is how many radiants?
You would multiply by pi/180 so that the degrees cancel out.

what is angular momentum?
 w.
 the units are radiants per second
 so if you are given a circle turns pi/4 radiants in 30 seconds. then you must divide (pi/4)/30 to get radiants per second, which is your angular momentum (w).

If you are given two angles and one side, how would you find the other side's length?
You would use the sin rule. sinA/a = sinB/b for example.

How would you find the arc length given the angle measurement and the radius?
s=r(theta) where s is the arc length, r is the radius of the cirlce, and theta is the angle measurement. REMEMBER theta MUST BE IN RADIANTS. it will not work in degrees.
if the problem gives you theta as 90 degrees then plug in pi/2 for theta in the formula.

If you want to convert a negative angle to the correspondin smallest positive angle what would you do?
if given in degrees add 360. for example 60 degrees is the same as (60+360) degrees which is 300 degrees.

If you are given 135 degrees (or some other random large number) how would you find the six trig functions of that angle?
since it is a large number you know to use the sum or difference formula for sin and cos. then those values can be used to find the rest of the trig functions tan,cot,csc,sec.
in this case if 135 is our angle then you do sin(90+45) = sin90cos45+cos90sin45 and cos(90+45) = cos90cos45sin90sin45. (you must remember the sum/diff formulas)

If you are given two sides of a right angle triangle, how do you find the theird side's length?
You must use the pythagorean theorem. a^2+b^2=c^2 the carrots indicate that the number 2 is an exponent.

if you have to consider a function y=36sin(2x+pi/2) how do you define the period, range, phase shift, amplitude and yintercept?
(1) put the given function in standard form, so that you can identify the a,b,c, and d. standard form is y = c + a*sin[b(xd)] if given sin function or y = c + a*cos[b(xd)] if given cos function.
 for this example it would be y = 3 + (6)sin2(xpi/4) notice that you factored out the 2, which is the coefficent of x. you do this in every case.
 so now identify c = 3, a = 6, b = 2, d = pi/4
 so that the period is 2pi/b = 2pi/2 = pi
 the range is all possible y values it be [amplitude, amplitude]
 The amplitude is a = 6 = 6. so the range is from [6,6]
 phase shift, c tells you vertical (up and down) phase shift. this graph moves up 3. (because c is positive.)
 d tells you horizontal (left/right) phase shift. This graph moves left pi/4 (because d is positive)
 the yintercept occurs when x=0. so plug x=0 into the equation and you get the yintercept is at y = 36sin(0+pi/2) which is just 3.

how do you find theta if you are given that cos(theta) = .985 and you are told to give and APPROXIMATE answer?
Approximate means you want to use a calculator. so in this case solve for theta by theta = acos(.985) where acos is inverse cosine. it can also be written arc cos or cos^1 where 1

How do you find what theta is if given cos(theta) = 1/2 and you want an exact answer?
you us the left hand rule. put down your pointing finger becasue then only one finger will be held up to the left of your now folded finger. which indicates that the cos value is 1/2. this tells you that the theta is the angle that corresponds to your pointing finger, 60 degrees or pi/3.

How do you go from decimal degrees to degrees, minutes, seconds?
 (1) take decimal portion of degrees and multiply by 60.
 (2) the numbers left of the decimal point is your minutes. (indicated by ')
 (3) take the decimal portion of the minutes and multiply by 60 again.
 (4) round, and there is your seconds. (indicated by '')

how do you go from given degrees minutes seconds to just decimal degrees?
divide the seconds by sixty, attach that decimal number to your minutes, divide by sixty again, and attach those numbers to your degrees to get degrees in decimal form

what is csc(theta) = ?
1/sin(theta) or the reciprocal of the sin value

what is sec(theta) =?
1/cos(theta) or the reciprocal of the cos value

what is tan(theta) equal to?
sin(theta)/cos(theta)

what is cot(theta) equal to?
cos(theta)/sin(theta) or the reciprocal of the tan value.

what does the sin graph look like?
it intersects the origin.

what does the cos graph look like
it intersects (0,1)

if a cartesian graph (regular xy plane graph) has asymptotes, what does this meen? and which trig functions can this possibly represent?
asymptotes mean that the graph is discontinuous there. in terms of trig functions it means that it is equal to something over zero. it can only include such functions as tan(x),cot(x),csc(x), or sec(x).

where are the asymptotes of y = sec(x) located at?
well sec(x) is the same as 1/cos(x) so you are asking where cos(x) is zero, because that is where it is undefined hence the asymptotes.
cos(x) is zero at angles pi/2 and pi/2 radiants or 90 degrees and 90 degrees. therefore that is where sec(x) has asymptotes in a cartesian graph.

what are y = csc(x) asymptotes located?
csc(x) = 1/sin(x) so asymptotes are located where sin(x) = 0. which is at 0 degrees and 180 degrees of 0 and pi/2 radiants.

where are the asymptotes of y = tan(x)
tan(x) = sin(x)/cos(x) therefore it's asymptotes are located where cos(x) = 0. which is at 90 degrees and 270 degrees or 90 degrees. which is the same as pi/2 and 3pi/2 or pi/2 in radiants. (looking at a polor coordinate system)

Okay so you see a graph wih asymptotes so you narrow the trig functions it can possibly be to tan(x), cot(x), csc(x), or sec(x). What's next?
 (1) for tan(x) and cot(x) the graph looks like an 'S'.
 (2) for csc(x) and sec(x) the graph looks like 'U' so that you can furthur narrow your choices.
 Check where the graph is positive or negative. Use polar coordinate system to tell where each would be positive or negative.
 csc(x) is positive in the I and II quadrants because that's where the y values are positive.
 sec(x) is positive in the I and IV quadrants

If you are given sin(2arctan(1)) what do you do?
 This is where you use the double angle formula sin(2A) = 2sinAcosA
 in this problem A = arctan(1) which is the same as inverse tangent by the way. This means that tanA = 1. which means that A = 45 degrees or pi/4 radiants.
 so now you solve sin(2A) by finding sin(45) and cos(45) and plugging into the above formula.

