Unit 4
how to find the range of a cscθ graph
(-∞, valley] u [peak,∞) -(opposite of the sin function)
how to make a coordinate of a unit circle
(length, height) -length is x value & height is y value
Basic Identities/Pythagorean Identities
(they're all version of the Pythagorean trig. identities. EX: when you divide 1st equation by sin^2(θ) you get 3rd equation & when you divide 1st equation by cos^2(θ) you get 2nd equation)
half angle formulas 1)for sin of a given angle 2)for cos of a given angle 3)for tan of a given angle
-a is the given angle times 2(so when you divide a by 2 you get the given angle) -for sin & cos equation you have to decide whether it's positive or negative, which depends on the ORIGINAL given angle, so find the quadrant which the og angle is in to determine if the tri is pos./neg, -for the tan equation both equations get the same answer
what is the degree(& give reference angle) 1)π/6 2)π/4 3)π/3 4)2π/3 5)3π/4 6)5π/6 7)7π/6 8)5π/4 9)4π/3 10)5π/3 11)7π/4 12)11π/6
-any radian w/ the denominator 6 is 30°, w/the denominator 3 is 60°, w/ the denominator 4 is 45° -remember that when the numerator & denominator has the same factor (EX: 6π/6 or 3π/3) the degree is 180° or if the numerator's factor is double the denominator (EX: 12π/3 & 6π/3) the degree is 360°, so if you based on the denominator you can just add or minus the reference angle to 180° or 360° (EX: 7π/6 is π/6 than 6π/6, so add 30° to 180°, which makes 7π/6=210°)(EX: 11π/6 is π/6 less than 360°, so 360°-30°=330°) 1)30° 2)45° 3)60° 4)120°(60°) 5)135°(45°) 6)150°(30°) 7)210°(30°) 8)225°(45°) 9)240°(60°) 10)300°(60°) 11)315°(45°) 12)330°(30°)
What does 1)SOH 2)CAH 3)TOA stand for?
-hypotenuse is opposite of 90 degree
Double angle formulas
-θ is the given angle divided by 2(so when θ is multiplied by 2 it will equal the given angle) -sin(2θ) & cos(2θ) could also be solved for using sum & difference formulas sin(θ+θ) & cos(θ+θ) -(under cos(2θ) all the equations are interchangeable, so memorize them all bc they'll be helpful in verifying identities)
co-function identity
-π/2 is 90° -x is the given angle
when does 1)sinθ=0 2)cosθ=0
1)0 & π (& the coterminals: 2π...) 2)π/2 & 3π/2 (& the coterminals: 5π/2...)
Sum & Difference Formulas for Cos of a given angle
1)cos(A+B)=(cosA)(cosB)-(sinA)(sinB) 2)cos(A-B)=(cosA)(cosB)+(sinA)(sinB)
what are the reciprocal of 1)sine (sin) -what does this equal? 2)cosine (cos) -what does this equal? 3)tangent (tan) -what does this equal?
1)cosecant (csc) -1/sinθ= hypotenuse/opposite 2) secant (sec) -1/cosθ= hypotenuse/adjacent 3) cotangent(cot) -1/tan = adjacent/opposite
how to graph cscθ
1)graph sinθ 2)draw dotted lines at every points (from the graph of sinθ) that are on the midline -they're going to be asymptotes for the graph of cscθ 3)graph cscθ: draw the vertex of a parabola at each peak & valley, then extend the parabola's asymptotes along the dotted lines
what is(remember to imagine a unit circle's special right triangle): 1)cos(30) 2)sin(30) 3)cos(60) 4)sin(60) 5)cos(45) 6)sin(45)
1)√3/2 2)1/2 3)1/2 4)√3/2 5)√2/2 6)√2/2
how many equations can a trig graph have?
Multiple -EX: a curve could be both a sin or cos graph bc everything is the same but sin starts on midline & cos starts on amplitude
what do you need to do, so that an inverse graph is a function?
Restrict the domain 4.11 Edpuzzle
trigonometric functions
SOH(sin=opposite/hypotenuse), CAH, TOA, csc equation, sec equation, & cot equation
what is the reference angle for tanX=√3
X= 60° or π/3 bc remember tan=sin/cos, so when sine is divided by cosine it should equal √3, so the only options in the unit circle would be sin=√3/2 & cos=1/2 since (√3/2)/(1/2)=(√3/2)x(2) = √3
Law of Cosines
a^2=b^2+c^2-2bccos...
what is sin^-1(2) equal to? what is sin(2) equal to ?
all NA bc 2 is not on the unit circle; all trig functions(sin, csc, sin^-1, cos, sec, cos^-1) are in between 1 and -1 -the trig. functions where it's possible for the answer to not be in between 1 & -1 are the variations of tan: tan, tan^-1, cot & cot^-1
when is cotX undefined
at 0 & π -bc undefined means the denominator is 0, so: cotX=cos/sin=cos/0=(find when sin, aka y-value, is 0) 0 & π
when does sinX=0
at 0 & π -use this to find when cscX & cotX is undefined bc they have cos as a denominator
how to find the important points(the points you will graph) of a cos/sin graph
divide the period by 4 & those are your x-values so find the y-values that accompany that
difference betweeen solving for the angle of sin & arcsin (aka sin^-1) equation?
for sin ur solving for X in sin(X)which is the the input(x-value) while for sin^-1 ur solving for the the (y-value) output arcsin(√3/2)=X -for sin the variable in parentheses is the angle, while for sin^-1 the # in parentheses is the sin of the reference angle -this^ relationship applies to cos, sin, & tan with their inverse cos^-1, sin^-1, & tan^-1
where are the possible answers for the reference angle of arccos & arcsec
from 0 to π (the top half of the unit circle) -Q1 & Q2
where are the possible answers for the reference angle of arctan & arcot -what quadrants?
from π/2 to -π/2 (aka 3π/2) (the left side of the unit circle)(same as sin) -Q1 & Q4
when does a curve start for a cosine curve(what is the y-value of the 1st point of a parent curve, which is the point you will draw the rest of the graph based on)
it always starts at a peak(if the coefficient in the very front of the right side of the equation is pos.) or valley(if the coefficient in the very front of the right side of the equation is neg.)
when does a curve start for a sine curve(what is the y-value of the 1st point of a parent curve, which is the point you will draw the rest of the graph based on)
it always starts at the midline
reciprocal
one of a pair of numbers whose product is 1: the reciprocal of 2/3 is 3/2
a reference angle is always...
positive, acute, & formed w/ the x-axis -while coterminal or given angle could be neg. or pos.
how to graph secθ
same as how you graph a cscθ except you 1st graph cosθ instead of sinθ
law of sines
sinA/a=sinB/b=sinC/c or a/sinA=b/sinB=c/sinC -the one u r trying to solve goes on top -this is for non right triangles -EX: Synch 4.7, #10
Pythagorean identity
sin²θ + cos²θ = 1 -same as Pythagorean theorem except we known that the hypotenuse is always 1 bc it's a unit circle(remember that cosine θ=x value on unit circle & sineθ=y value on unit circle)`
quotient identity
tan=sin/cos & cot=cos/sin -this makes sense bc in a unit circle cosine θ=x value & sineθ=y value -remember a quotient=a result obtained by dividing one quantity by another
what does it show on a graph when tanx=undefined (so if 0 was the denominator)
that is on the vertical asymptote
reference angle
the acute angle formed by the terminal side IN the quadrant that the angle terminates and the x-axis -remember angles of a circle form anti-clockwise, so the terminal side is the one further in an anticlockwise clock -if a given angle is in the (positive, positive) quadrant, then the reference angle is the same as the given angle
how many possible answers are there for the angle of a trig. function(EX: what is sinx= -1/2)
there are an infinite amount of answers bc there are an infinite amount of coterminals
how many possible answers are there for the angle of an inverse trig. function(EX: what is sin^-1 x= -1/2)
there's only 1 answer -bc we restricted the domain & range, so that there's only 1 y-value for each x-value, which caused there to only be 1 x-value when solving for it
general domain angles
when given a trig. function with a missing angle, the general domain angles are the infinite amount of possible angles (includes coterminal angles)
restricted domain angles
when given a trig. function with a missing angle, the restricted domain angles are the finite amount of possible angles (no coterminal angles); the angles between (0, 360°]
what is tan(60degrees) or tan(pi/3)
√3
cotθ graph -difference between tanθ & cotθ graph
-in the tanθ graph, (0,0) is the 1st parent point, but for cotθ graph, line 0 is the 1st parent asymptote
what reference angle does tanX=1
45 degrees (or π/4) -bc tan=opposite/adjacent=(√2/2)/(√2/2)=1
the 2 ways to measure angles of a circle
degrees(°) or radians(π) -1 π is half of a circle or 180°
how dyk if a graph shows a function
if the same x-value has only 1 or no y-value -(the vertical line test: if you draw a vertical line & it passes through more than 1 point, then it's not a function)
when do you use Sum & Difference Formulas
when the angle given is not on the unit circle, but 2 angle numbers which could add or subtract to that given angle are on the unit circle(EX: cos15 = cos(45-30))
standard position
when the initial point of a vector is at the origin
zero product property
when the product of two or more factors is zero, one of these factors must equal zero (EX: if ab=0, then a=0 and/or b=0)
how to find the equation for the midline points algebraically (you won't be required to do this)
x=the 1st positive midline point(make y=0 in the given trig. equation & solve for x) + periodn -EX y=2tan4(x+π)-1: 0=2tan4(x+π)-1, x=, x=smthn
how to graph a sin or cos function
1)draw midline 2) find the 1st point of the parent curve -sin curves start at the midline -cos curves start at the amplitude(if "a" is pos then starts an amplitude above midline, but if "a" is neg then it starts an amplitude below midline) 3) find & graph all the important point
Sum & Difference Formulas for Sin of a given angle
1)sin(A+B)=(sinA)(cosB)+(cosA)(sinB) -A & B are angles from the unit circle that should add to the given angle 2)sin(A-B)=(sinA)(cosB)-(cosA)(sinB)
Sum & Difference Formulas for Tan of a given angle
1)tan(A+B)=(tanA+tanB)/(1-tanAtanB) 2)tan(A-B)=(tanA-tanB)/(1+tanAtanB)
inverse -what's the inverse of sin? what's the reciprocal of sin?
"reverses" another function: if the function(f) has an input x & gives a result of y, then its inverse function(g) is where y gives the result x, and vice versa -often inverse & reciprocal are the same, but not for trig. functions (EX: reciprocal of sin is 1/sin or csc, while inverse of sin is sin^-1 or arcsin)
when graphing, what is the 1)x-value 2)y-value of a cosine curve
1)the given angle 2)the x-coordinate from unit circle(the cos of the given angle)
amplitude
the distance from the peaks/valleys to the midline -(how high/low we go from the midline)
1) how to find what part of a cos curve is the parent curve? 2) how to find what part of a sin curve is the parent curve?
1) the 1st point(the point at the x-value of the phase shift) to the next peak(if the 1st point is a peak) or to the next valley(if the 2st point is a valley) 2) the 1st point(the midpoint at the phase shift), passing 1 valley & 1 hill to the next midpoint
1) how to find restricted domain of a tan graph 2) how to find restricted range of a tan graph
1) the x-value of the asymptotes surrounding the 1st point of a parent curve (the phase shift is the x-value of the 1st point of a parent curve) -same answer as sin except use PARENTHESES instead of hard brackets bc we're using asymptotes 2)the y-value of the asymptotes surround the 1st point of a parent curve-always (-∞,∞)
1) how to find restricted domain of a cos graph 2) how to find restricted range of a cos graph
1)[the x-value of the the 1st point of a parent curve(the phase shift), the next important point that's not a midline point(would be a peak if the 1st point of a parent curve was a valley & vice versa)] 2)[the y-value of the peak, the y-value of the valley] that surrounds the 1st point of a parent curve
1) how to find restricted domain of an inverse trig function 2) how to find restricted range of an inverse trig function
1)the restricted range of the noninverted function 2)the restricted domain of the noninverted function -bc when you graph the inverse of a function, you switch the x & y values
1)how to solve for missing angle in the domain(0, 2π] when tan(x)=0 (yes zero 0, not theta θ) 2)what about when tan(3x)=0
1)x= 1 & -1 -bc tan=sin/cos=y/x, so you look for whenever y=0 which would be at x=1 & -1 2)x=1/3 & -1/3 -bc the angle 3x=1 & -1, so x= 1/3 & -1/3
when to use half angle formulas
given angles that are not on the unit circle & cannot be subtracted/added up by unit circle angles(adding 45, 30, 60 or 90 degrees does not equal that given angle), so you must divide an angle on the unit circle by 2, which equals the given angle & use the half angle formulas
when does cosX=0
at π/2 & 3π/2 -use this to find when secX & tanX is undefined bc they have cos as a denominator
from which point do you start graphing 1) sine curve 2) cosine curve (after this 1st point you keep going 1 amplitude up or 1 amplitude down)
y = asinb(x − c) + d & y= acosb(x − c) + d 1) (c, d) (starts on midline) 2) (c, d+a) (start on a peak/valley)
how to find reference angle when the angle is in the 1)1st quadrant(0° to 90°) 2)2nd quadrant(90° to 180°) 3)3rd quadrant(180° to 270°) 4)4th quadrant(270° to 360°)
1)reference angle = angle 2)reference angle = 180° - angle 3)reference angle = angle - 180° 4)reference angle = 360° - angle
what is arcsin?
the inverse of sin (sin^-1) -arctan=sin^-1, arcsos=cos^-1, arccsc=csc^-1, arcsec=sec^-1
what is tan(30degrees) or tan(π/6)
√3/3
Sum & Difference Formulas for Csc of a given angle -what about sec & cot?
(it's the Sum & Difference Formulas for Sin of a given angle OVER 1) 1)sin(u+v)=1/[(sinu)(cosv)+(cosv)(sinv)] -u & v should add to the given angle 2)sin(u-v)=1/[(sinu)(cosv)-(cosv)(sinv)] -Sum & Difference Formulas for csc, sec, & cot are just Sum & Difference Formulas for sin, cos, & tan but over 1
what are all the unit circle angles
0°(or 360°) 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° -the closest angle to the y & x-axis is the axis +/- 30° -the angle between the x-axis is the axis +/- 45° OR the closest angle +/- 30°
what do you need to remember when solving for the (non-general) angles of a trig function that includes half, double, triple, etc. angles?
1) find your general angle answers ignoring the coefficient infront of the x -EX Final review: tan3x=0 -> find when tanx=0 ignoring the 3 -> 3x=0 & pi -> 3x=pi n 2) divide your general answers by the coefficient -3x=pi n -> x=(pi n)/3 3) back and add to each answer the thing you were adding to each angle in the general solution until you get to 2π for each angle from the general solution(or 360 degrees) -x=pi/3, 2pi/3, pi, 4pi/3, 5pi/3 (we're not including 2pi bc the interval is [0, 2pi)) -EX: 16:30 on Asynch 4.3 Day 2 Edpuzzle) -EX: practice quiz 1, #8
How to find the general solution of possible angles given a trig function
1)Simplify the equation if needed -factor -if the variables are different(cos^2x + sinx-1=0) use basic identities to make them the same variable(like turn cos^2x into 1-sin^x) -EX: if ur given 0=cosX(csc^2X-2), then you'll need to separate the factors (cosX=0 & csc^2X-2=0 & solve X separately) 2)Find what possible quadrant the angle could be in -(EX: if it's sin(x)=-1 it must be in quadrant 2 or 3 bc those are the quadrants with negative sine values) 3)draw the sides in the quadrant(s) to find the angle -EX: sin(x)=√3/2, so the opposite side of x is √3/2, so the angle x must be 60 degrees 4)add 2πn(n is the number of terms) -EX(final answer of sin(x)=√3/2): x=√3/2 + 2πn -bc there are an infinite amount of conterminal angles so you can keep adding 2πn -if the the 2 possible angles from 2 quadrants are directly across from eachother, forming 1 line, then the general equation can be the smaller angle plus πn(EX: 4.3 Solving Trig Equations, 14:54) -if the 4 possible angles from 4 quadrants are 45 degrees, then the general equation can be the smallest angle plus (π/2)n(EX: 4.3 Solving Trig Equations, 20:13)
1) how to find restricted domain of a sin graph 2) how to find restricted range of a sin graph
1)[the x-value of the valley, the x-value of the pear] that surround the 1st midpoint(often found at the origin) of a parent curve of a sin(non inverse) graph -(or vice versa [peak, valley] if the amplitude was negative) (the phase shift is the x-value of the 1st point of a parent curve) 2)[the y-value of the valley, the y-value of the peak] that surround the 1st point of a parent curve -(or vice versa [peak, valley] if the amplitude was negative) -talked about in beginning of 11.4 edpuzzle -restricted domain & range means how we restrict the domain of the normal sin graph to use it for inverse sin graph -restricted domain are NOT just the same values as the parent curve
How to find the equation for tan or cot graph
1)draw the asymptotes 2)draw the midline(the y line where all the inflection points are) 3)find amplitudes(the y-value distance between the inflection points & right before the line seems like it's going to go infinitely towards the asymptote) example on 4.10 edpuzzle, 19:37 4)find "b" value/the coefficient infront of the angle(b=π/b; different from sin/cos/csc/sec) 5)find phase shifts
How to find the equation for csc or sec graph
1)draw the sin graph for csc & the cos graph for sec -the vertex of the parabolas are the peaks & valleys of the sin/cos graph -the midline is between the y-value of the parabola's vertex -the x-value midline points are in between the x-values of the porabola's vertex(aka the new valleys & peaks) 2)write the sin or cos equation for the sin or cos graph you drew, but replace "sin" w/ "csc" & "cos" w/ "sec"
how to find the midline from a cos graph & sin graph
1)find the y-value distance between a peak & valley point 2)divide that distance by 2(this is the amplitude) 3)substract that distance from the peak y-value point(or add that to a valley) & that is your midline
how to find the equation for the asymptote equation of any trig. graph(remember that only sec, csc, cot, & tan have asymptotes) algebraically
1)find when the trig. function is undefined(when the denominator is 0) -EX: secX is always undefined when cosX=0(bc secX=1/cos=1/0), which is always at π/2 & 3π/2 2) make the whole angle(w/ coefficients & shift)& equal that to when the trig. function is undefined (this will find one asymptote) -EX: given y=-3sec(2x-π)+1 make 2x-3=π/2 (or equal 3π/2... both work) x=3π/4 (this is one asymptote) 3) a. for sec & csc: add nperiod/2 to the asymptote you found b. for cot: nperiod to the asymptote you found -EX: x=3π/4 + π/2n
how to solve for the angle of an inverse trig. ? -EX: sin^-1(√3/2)
1)find which quadrant the angle will be in based on if it's pos. or neg. (disregard that the function is an inverse & just treat it as a normal trig. function, but limit the possible quadrants) -EX: for sin^-1(√3/2) & √3/2 is positive, so it's in Q1 or Q2 but any sin^-1 can only be in Q1 or Q4, so it's in Q1 2) find the angle in the unit circle that gives the trig. function outcome -EX: in Q1 the angle π/6 is where sin=√3/2
difference between finding cotθ graph & tanθ graph
1)tanθ's 1st parent point is on the midline, while cotθ's 1st parent "point" is the amplitude - look at previous 2 slides to compare cot & tan grahs 2) the important point above the midline for tanθ is on the right of the midline important point, but on cotθ is on the left of the midline important point 3) the important point below the midline for tanθ is on the left of the midline important point, but on cotθ is on the right of the midline important point -for cotθ the x-ints & asymptotes are switched from the tanθ graph, but everything else is the same -bc cotθ=cosθ/sinθ, while tanθ=sinθ/cosθ -so where there used to be an x-int for tanθ, replace that w/ a asymptote & vice versa
when graphing, what are the 1)x-values 2)y-values of a sin curve
1)the given angle(θ) 2)the y-coordinate from unit circle(the sin of the given angle)
how to find phase shifts for 1) tan graphs 2) cot graphs
1)the x-value of any inflection point(aka midpoint) -bc a normal tan graph w/ no P.S. has a midpoint at (0, y) 2)the x-value of any asymptote -bc a normal tan graph w/ no P.S. has an asymptote at x=0
guidelines for verifying identities
1)work w/ 1 side at a time(the side which is complex, so you need to simplify & if both sides look complex, then choose the side that looks like there's a good step to help you solve) 2)look for chances to factor, add fractions, square binomials, or create 1 term denominators 3)Look for chances to use basic identities 4)If nothing else works turn everything into sines & cosines
what coordinate does 1) cosA 2) sinA 3) tanA equal in a unit circle
1)x coordinate 2)y coordinate 3)y coordinate/x coordinate (bc sine/cosine) -makes sense bc in a unit circle the hypotenuse is 1, so when sin=opposite/hypotenuse & cos=adjacent/hypotenuse the denominator(hypotenuse) is always 1, so the value is just the y value(opposite) or x value(adjacent)
what reference angle does tanX=√3/3
30 degrees (or π/6) -bc tan=opposite/adjacent=(1/2)/(√3/2)=√3/3 -so the angle of -tanX=√3/3 would be 11π/6 or 5π/6 bc they both have a reference angle of π/6 & they are in quadrants where tan is negative
what reference angle does tanX=√3
60 degrees (or π/3) -bc tan=opposite/adjacent=(1/2)/(√3/2)=√3
coterminal angle(define) 1)how to find coterminal angle? 2)how many coterminal angles does a given angle have?
angles with the same terminal side -in the pic: so if 120 degrees is the given angle, then -240 degrees is a coterminal angle 1)-find it by adding or substracting 360 to the given angle 2)infinite amounts bc you can keep adding or subtracting 360 degrees to the given angle
when is secX undefined
at π/2 & 3π/2 -bc undefined means the denominator is 0, so: secX=1/cos=1/0=(find when cos, aka x-value, is 0) π/2 & 3π/2
how to find the coefficient infront of the angle for a cos or sin graph (the "b" in y = asinb(x − c) + d or y = acosb(x − c) + d)
b=2π/period (plug in the period & solve for b) -bc a period=2π/b
where are the possible answers for the reference angle of arcsin & arccsc(what are the answers restricted to for sin^-1) -what quadrants?
from π/2 to -π/2 (aka 3π/2) (the left side of the unit circle) -Q1 & Q4 -EX: -π/4 is the reference angle for sin^-1, but 7π/4 is not bc 7π/4 requires you to go through the other sections of the unit circle that are no just between π/2 to -π/2, so it's not possible
for tan & cot graphs how to find: 1) period 2) phase shift 3) vertical shift 4) midpoints 5) x-intercepts 6) asymptotes 7) domain 8) range
given y = AtanB(x − C) + D 1) period=π/b (while in cos & sin it's 2π/b) -Important points are still the period/4 2) C -if it's Bx-C instead of B(x-C) make sure to factor out the B 3) D 4) phase shift + period n 5) (turn the y in the equation into 0 & solve for x) + period n -EX: 4.9 Asynch 5d 6) [graph the curve to find the 1st asymptote]+ period n 7) all real numbers, x≠asymptotes -all real numbers except when x=asymptotes(insert asymptote equation you found from part 5) 8) (-∞,∞)f
what is the inverse of a trig. function?
it's the trig. function to the power of negative one or place "arc" before it: sin^-1, cos^-1, tan^-1 aka arcsin, arccos, arctan -not csc or sec, but sin^-1 & cos^-1 bc: csc/sec is the reciprocal of sin/cos (cscX∙sinX=1), while sin^-1/cos^-1 is the inverse (sin^-1∙sinX= 0.0087≠1)
difference between finding cotθ graph & -tanθ graph
negative tanθ means that the point are reflected so the above midline points are now on the left of the midline point, which is just like cotθ, but the phase shift indicated on the cotθ gives the 1st asymptote, while the P.S. on the -tanθ graph gives the 1st midline point, so there's just a shift difference between the 2
one period of a sine curve
the curve/graph formed when the x-values being plotted are all from the first 360 degrees(or 2radian) -they repeat after this -to find the period, it's the x-value distance from a midline point then passes through a peak & valley, to the other midline point
how to find the period(per) from any trig. graph
the distance between any 2 points that are equally spaced out throughout the graph & that have the same y-value (but remember that when using 2 midlines points' x-value, those 2 points should have a valley & a peak in between) -a period should start at a midline point, include one valley & one peak, then end on the next midline point(for csc & sec when you draw the sin or cos graph when finding the equation & the same applies) -or to find the distance of a period it could start at 1 peak(or 1 valley) & end at the next peak(or valley) (this is OK to calculate the distance of a period, but is not officially the definition of a period) -EX: x-value distance between adjacent midpoint & midpoint, OR valley & valley, OR peak & peak
midline -what is for cosine & sine parent graphs
the line which the valleys & peaks always go back to -always y=0(or x-axis) for cosine & sine parent graphs
which trig functions are positive in quadrant 1, 2,3, & 4
the memory aid is A(ll) S(tudents) T(ake) C(alculus) 1) All trig functions are positive 2)only Sin & its reciprocal(csc) 3)only Tan & its reciprocal(cot) 4)only Cos & its reciprocal(sec)
when is the answer no solution when trying to find the angle of a sin trig function?
when the sinX is not between -1 & 1 -same applies to cosX, cscX, & secX -EX: sinX=-4/3 is no solution
how to find when the asymptotes of a cscθ or a secθ graph occur
x= the 1st, smallest positive asymptote +/- 1/2period -1/2 a period is the distance between each midline point -also all the asymptotes are when cscθ=undefined, so when the denominator is 0 -0 is a positive -you can use any asymptote, but it's easier to just use the 1st, smallest positive asymptote
in the equation for a trig function where is the 1)amplitude 2)vertical shift(where our midline is) 3)helps determine the period in a trig function 4)phase/horizontal shift 5) domain of a trig function graph(sin, cos, tan, sec, csc, cot) 6) range of a trig function graph(sin, cos, tan, sec, csc, cot)
y = asinb(x − c) + d 1)coefficient infront of trig function(a) -for sin: after the 1st point of a curve(on the midline) the next important point(right) is 1 amplitude up, but if "a" is negative then the next important point is 1 asymptote down the midline -for cos: a parent curve usually starts from above the midline, but if "a" is negative then you start curve below midline -EX: y=2cos would start the parent curve at a y-value of 2, but y=-2cos would start the parent curve at y=-2 2)the # being added(d) -y=d 3)coefficient infront of the angle(b) -usually a period is 2π, then the coefficient infront of the angle would make the period 2π/b -there should never be a coefficient infront of x in the angle (x − c), so take that coefficient out of the parentheses 4)horizontal(phase shift; moves graph left of right)(c) -x MINUS c means you're going right c -this is where your parent curve starts(starts at c instead of the y-axis) 5) for csc, cot, sec, tan: R, x≠asymptote for sin, cos:(-∞,∞) -means all real numbers except for when x= asymptotes 6)for csc, cot, sec, tan: (-∞,∞) for sin, cos: [y-value of the valleys, y-valley of the peaks]
how to find the phase shift(P.S.) from a 1) cos graph 2) sin graph
y = asinb(x − c) + d or y = acosb(x − c) + d 1) -if you make "a" pos: the x-value of ANY valley(they all work) -if you make "a" neg: the x-value of ANY peak 2) -if you make "a" pos: the x-value of ANY point on the midline where the peak is directly after that point -if you make "a" neg: the x-value of ANY point on the midline where the valley is directly after that point
how to graph tanθ(in 7 steps)
y = atanb(x − c) + d 1)state the amplitude(a), period(π/b), phase shift(c) 2)plot the 1st parent point -on the midline, 1 phase shift 3)plot points on the midline: the next point is 1 period away from the previous one 4)draw asymptotes: in the middle of the midline points 5)draw the main points above the midline: in between a midline point & the closest asymptote, the y-value goes up from the midline by 1 amplitude(a) 6)draw the important points below the midline: in between a midline point & asymptote, the y-value goes down from the midline by 1 amplitude(a) 7) connect points & use asymptotes!
