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An image of six graphs. Each graph has an x axis that runs from -2 pi to 2 pi and a y axis that runs from -2 to 2. The first graph is of the function “f(x) = sin(x)”, which is a curved wave function. The graph of the function starts at the point (-2 pi, 0) and increases until the point (-((3 pi)/2), 1). After this point, the function decreases until the point (-(pi/2), -1). After this point, the function increases until the point ((pi/2), 1). After this point, the function decreases until the point (((3 pi)/2), -1). After this point, the function begins to increase again. The x intercepts shown on the graph are at the points (-2 pi, 0), (-pi, 0), (0, 0), (pi, 0), and (2 pi, 0). The y intercept is at the origin. The second graph is of the function “f(x) = cos(x)”, which is a curved wave function. The graph of the function starts at the point (-2 pi, 1) and decreases until the point (-pi, -1). After this point, the function increases until the point (0, 1). After this point, the function decreases until the point (pi, -1). After this point, the function increases again. The x intercepts shown on the graph are at the points (-((3 pi)/2), 0), (-(pi/2), 0), ((pi/2), 0), and (((3 pi)/2), 0). The y intercept is at the point (0, 1). The graph of cos(x) is the same as the graph of sin(x), except it is shifted to the left by a distance of (pi/2). On the next four graphs there are dotted vertical lines which are not a part of the function, but act as boundaries for the function, boundaries the function will never touch. They are known as vertical asymptotes. There are infinite vertical asymptotes for all of these functions, but these graphs only show a few. The third graph is of the function “f(x) = csc(x)”. The vertical asymptotes for “f(x) = csc(x)” on this graph occur at “x = -2 pi”, “x = -pi”, “x = 0”, “x = pi”, and “x = 2 pi”. Between the “x = -2 pi” and “x = -pi” asymptotes, the function looks like an upward facing “U”, with a minimum at the point (-((3 pi)/2), 1). Between the “x = -pi” and “x = 0” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (-(pi/2), -1). Between the “x = 0” and “x = pi” asymptotes, the function looks like an upward facing “U”, with a minimum at the point ((pi/2), 1). Between the “x = pi” and “x = 2 pi” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (((3 pi)/2), -1). The fourth graph is of the function “f(x) = sec(x)”. The vertical asymptotes for this function on this graph are at “x = -((3 pi)/2)”, “x = -(pi/2)”, “x = (pi/2)”, and “x = ((3 pi)/2)”. Between the “x = -((3 pi)/2)” and “x = -(pi/2)” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (-pi, -1). Between the “x = -(pi/2)” and “x = (pi/2)” asymptotes, the function looks like an upward facing “U”, with a minimum at the point (0, 1). Between the “x = (pi/2)” and “x = (3pi/2)” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (pi, -1). The graph of sec(x) is the same as the graph of csc(x), except it is shifted to the left by a distance of (pi/2). The fifth graph is of the function “f(x) = tan(x)”. The vertical asymptotes of this function on this graph occur at “x = -((3 pi)/2)”, “x = -(pi/2)”, “x = (pi/2)”, and “x = ((3 pi)/2)”. In between all of the vertical asymptotes, the function is always increasing but it never touches the asymptotes. The x intercepts on this graph occur at the points (-2 pi, 0), (-pi, 0), (0, 0), (pi, 0), and (2 pi, 0). The y intercept is at the origin. The sixth graph is of the function “f(x) = cot(x)”. The vertical asymptotes of this function on this graph occur at “x = -2 pi”, “x = -pi”, “x = 0”, “x = pi”, and “x = 2 pi”. In between all of the vertical asymptotes, the function is always decreasing but it never touches the asymptotes. The x intercepts on this graph occur at the points (-((3 pi)/2), 0), (-(pi/2), 0), ((pi/2), 0), and (((3 pi)/2), 0) and there is no y intercept.
The six trigonometric functions are periodic.

Just as with algebraic functions, we can apply transformations to trigonometric functions. In particular, consider the following function:

f ( x ) = A sin ( B ( x α ) ) + C .

In [link] , the constant α causes a horizontal or phase shift. The factor B changes the period. This transformed sine function will have a period 2 π / | B | . The factor A results in a vertical stretch by a factor of | A | . We say | A | is the “amplitude of f . ” The constant C causes a vertical shift.

An image of a graph. The graph is of the function “f(x) = Asin(B(x - alpha)) + C”. Along the y axis, there are 3 hash marks: starting from the bottom and moving up, the hash marks are at the values “C - A”, “C”, and “C + A”. The distance from the origin to “C” is labeled “vertical shift”. The distance from “C - A” to “A” and the distance from “A” to “C + A” is “A”, which is labeled “amplitude”. On the x axis is a hash mark at the value “alpha” and the distance between the origin and “alpha” is labeled “horizontal shift”. The distance between two successive minimum values of the function (in other words, the distance between two bottom parts of the wave that are next to each other) is “(2 pi)/(absolute value of B)” is labeled the period. The period is also the distance between two successive maximum values of the function.
A graph of a general sine function.

Notice in [link] that the graph of y = cos x is the graph of y = sin x shifted to the left π / 2 units. Therefore, we can write cos x = sin ( x + π / 2 ) . Similarly, we can view the graph of y = sin x as the graph of y = cos x shifted right π / 2 units, and state that sin x = cos ( x π / 2 ) .

A shifted sine curve arises naturally when graphing the number of hours of daylight in a given location as a function of the day of the year. For example, suppose a city reports that June 21 is the longest day of the year with 15.7 hours and December 21 is the shortest day of the year with 8.3 hours. It can be shown that the function

h ( t ) = 3.7 sin ( 2 π 365 ( x 80.5 ) ) + 12

is a model for the number of hours of daylight h as a function of day of the year t ( [link] ).

An image of a graph. The x axis runs from 0 to 365 and is labeled “t, day of the year”. The y axis runs from 0 to 20 and is labeled “h, number of daylight hours”. The graph is of the function “h(t) = 3.7sin(((2 pi)/365)(t - 80.5)) + 12”, which is a curved wave function. The function starts at the approximate point (0, 8.4) and begins increasing until the approximate point (171.8, 15.7). After this point, the function decreases until the approximate point (354.3, 8.3). After this point, the function begins increasing again.
The hours of daylight as a function of day of the year can be modeled by a shifted sine curve.

Sketching the graph of a transformed sine curve

Sketch a graph of f ( x ) = 3 sin ( 2 ( x π 4 ) ) + 1 .

This graph is a phase shift of y = sin ( x ) to the right by π / 4 units, followed by a horizontal compression by a factor of 2, a vertical stretch by a factor of 3, and then a vertical shift by 1 unit. The period of f is π .

An image of a graph. The x axis runs from -((3 pi)/2) to 2 pi and the y axis runs from -3 to 5. The graph is of the function “f(x) = 3sin(2(x-(pi/4))) + 1”, which is a curved wave function. The function starts decreasing from the point (-((3 pi)/2), 4) until it hits the point (-pi, -2). At this point, the function begins increasing until it hits the point (-(pi/2), 4). After this point, the function begins decreasing until it hits the point (0, -2). After this point, the function increases until it hits the point ((pi/2), 4). After this point, the function decreases until it hits the point (pi, -2). After this point, the function increases until it hits the point (((3 pi)/2), 4). After this point, the function decreases again.
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Describe the relationship between the graph of f ( x ) = 3 sin ( 4 x ) 5 and the graph of y = sin ( x ) .

To graph f ( x ) = 3 sin ( 4 x ) 5 , the graph of y = sin ( x ) needs to be compressed horizontally by a factor of 4, then stretched vertically by a factor of 3, then shifted down 5 units. The function f will have a period of π / 2 and an amplitude of 3.

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Key concepts

  • Radian measure is defined such that the angle associated with the arc of length 1 on the unit circle has radian measure 1. An angle with a degree measure of 180 ° has a radian measure of π rad.
  • For acute angles θ , the values of the trigonometric functions are defined as ratios of two sides of a right triangle in which one of the acute angles is θ .
  • For a general angle θ , let ( x , y ) be a point on a circle of radius r corresponding to this angle θ . The trigonometric functions can be written as ratios involving x , y , and r .
  • The trigonometric functions are periodic. The sine, cosine, secant, and cosecant functions have period 2 π . The tangent and cotangent functions have period π .

Key equations

  • Generalized sine function
    f ( x ) = A sin ( B ( x α ) ) + C

For the following exercises, convert each angle in degrees to radians. Write the answer as a multiple of π .

For the following exercises, convert each angle in radians to degrees.

Evaluate the following functional values.

sin ( 3 π 4 )

2 2

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sin ( π 12 )

3 1 2 2

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For the following exercises, consider triangle ABC , a right triangle with a right angle at C. a. Find the missing side of the triangle. b. Find the six trigonometric function values for the angle at A . Where necessary, round to one decimal place.

Questions & Answers

Good morning,,, how are you
Harrieta Reply
d/dx{1/y - lny + X^3.Y^5}
mogomotsi Reply
How to identify domain and range
Umar Reply
hello
Akpevwe
He,,
Harrieta
hi
Dr
hello
velocity
I only talk to girls
Dr
women are smart then guys
Dr
Smarter
Adri
sorry
Dr
hi adri ana
Dr
:(
Shun
was up
Dr
hello
Adarsh
is it chatting app?.. I do not see any calculus here. lol
Adarsh
Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
mukul Reply
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
ade
show that lim f(x) + lim g(x)=m+l
BARNABAS Reply
list the basic elementary differentials
Chio Reply
Differentiation and integration
Okikiola Reply
yes
Damien
proper definition of derivative
Syed Reply
the maximum rate of change of one variable with respect to another variable
Amdad
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
Inembo Reply
what is calculus?
BISWAJIT Reply
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
Geoffrey Reply
what is x and how x=9.1 take?
Pravin Reply
what is f(x)
Inembo Reply
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
Adri
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
Adri
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
ayo Reply
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
Ahmad
by using integration product formula
Roha
find derivative f(x)=1/x
Mul Reply
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
all real numbers . except zero
Roha
please try to guide me how?
Meher
what do u want to ask
Roha
?
Roha
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no. (-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.
Mckenzie
👍
Roha
Roha are u A Student
Lutf
yes
Roha
Practice Key Terms 4

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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