sine, maxima, responses

Yemsrach Woubbie

Let y=Sin(x), this equation has more than one points where the slope can be zero, it means it can have more than one maxima. y'=Cos(x)=0, implies that x can be one of the odd multiples of pi/2 or 90 degree.

Anuradha Tulachan

maxima would be where slope=0, where the value is neither increasing nor decreasing. or when d(sinx)/dx=0 or, cosx=0 ie when x=90,180,270......

Tasha Sakaguchi

To find the maximum of a sine graph, you would find it where the derivative fails to exist. Because the derivative of the sin graph is cos. where sin hits it's maximun, cosine is at it's minimum. So when you set cosine to zero, or it's minimum you would be finding sin's maximum. So at the point where cos is equal to 0, sin is at it's maximum or 1, on the unit circle. So at the point where the derivative fails to exist, where cos, is at it's minimum, sin is at it's maximum.


First, we take the derivative of y = sin(x); this, of course, works out to y' = cos(x). Now we find the critical cases on the whole real line, where y' = 0. This gives us the result that the maxima of y = sin(x) are at x = pi/2 and x = 3pi/2.

Alia Rahman-Khan

I am really not sure if this is correct. d/dx (sinx) = 2pi Since the sin function is periodic, meaning that is it repeats itself after evey can calculate the maxima and minima from only the 2pi and give it in the general form.

ingrid e. frau

i really have no idea because i wasn't there on friday but i can tell you that the maxima of cos(x) is 1 at 2k(pi)

Wanapon Techagaisiyavanit

dY/dX(sinX)= cosX Because when sin(X)is 0 at 0 degree ,cos(X) would be 1 Therefore the maxima of sin(X) = cos(X)

sarah fraser

the maximum of the sin(x) graph is x= 1.57 and y=1, I found this by using my calculator and asking the calc for the max by choosing two points.


To find the maxima of sin x, you take the derivative and set that function equal to zero to see where the slope of the original function is equal to zero. If y = sin x, then y'= cos x. Cos x = 0 at 1. So the maxima of sin x on the whole real line is 1.

collin hull

sin alternates between 1 and -1 so the max is 1 and occurs when sinx=1. so- at pi/2, 5pi/2, 9pi/2, etc.

Melanie LaFavre

1. the maximum is where the derivative vanishes or is equal to zero. The function is y=sin(x) so the derivative is y'=cos(x). To find where the derivate eqauls zero we set y to 0. 0=cos(x) intuetivly we know that cos equals zero at x=pie/2. so the answer is pie/2

Catherine Chan

Although the sine function has a lot of maxima; however, they are all periodic. If we just consider the maxima of one of the period between a particular interval, the graph of the sine function is made up of the same shape as the graph of Ax^2+Bx+C=0. It's very easy to find the maxima since the maxima occurs when the slope is zero. that means it's on the vertex. Once we get one value of maxima, it is very easy for us to find the maxima of the sine function of the whole real line because it is periodic.

Youngshin Cho

maxima of sin is when the graph is pointing the top of the graph. so it is 1

Youngshin Cho

maxima of sin is when the graph is pointing the top of the graph. so it is 1

hilary moore

on the whole real line, sin(x) 's maxima is the derivative. this being all real numbers as the limit, the maxima is cos(x) for the entire equation.

Bushra Husain

First you find the derivative of sin(x), which is cos(x). You then set cos(x) equal to zero. Cos(x) is equal to zero at 90 and 270 degrees. To find the answers, you look back at the unit circle. Cos(x) is 0 (in the x-axis) at 90 and 270 degrees.


y=sin(x) y'=cos(x) 0=cos(x) ; x=pi/2 since the graph of sin(x) is repeating, we may conclude that the maxima is at any point where x is divisible by pi/2. (i.e. 3pi/2, etc.)


To find candidates for the maxima of sin(x), we just take the derivative, set it equal to 0, and solve for x: cos(x)=0 means x=PI/2+N*PI, where PI=3.14159... and N is an integer. When N is even, sin(x)=1, and this is clearly the maximum value. When N is odd, sin(x)=-1, so we have (of course!) found the minima also. There are no endpoints of an interval in this problem, and the derivative of the very smooth sine function exists everywhere. We see that the usual procedure does indeed find all the maxima and minima. In particular, the maxima are at x=PI/2+2*n*PI, where n is an integer.