how do i write sin^2(x) in matlab?

I am trying to plot sin^2(x) together with cos^2(x) between [0,2pi]

but cant get my matlab to accept sin^2(x).

here is what I wrote, what am i doing wrong?

x=0:0.01:2*pi
si=sin^2(x);
co=cos^2(x);
plot(si,co)

[manoj pokhriyal]

si=(sin(x))^.2

co=(cos(x))^.2

plot(si,co)

[manoj pokhriyal]

sorry my mistake

si=sin(x).^2

co=cos(x).^2

plot(si,co)

[Hammad]

how sin(-5) is equal to 0.9589 in matlab?

[Steven Lord]

That is the correct value for the sine of -5 radians. If you wanted to compute the sine of -5 degrees use the sind function instead of the sin function.

R1 = -5; % radians
D1 = rad2deg(R1) % -5 radians in degrees

D1 = -286.4789

sineOfMinus5Radians = [sin(R1); sind(D1)]

sineOfMinus5Radians = 2×1

0.9589 0.9589

D2 = -5; % degrees
R2 = deg2rad(D2) % -5 degrees in radians

R2 = -0.0873

sineOfMinus5Degrees = [sind(D2); sin(R2)]

sineOfMinus5Degrees = 2×1

-0.0872 -0.0872

[Dyuman Joshi]

@Hammad, the input of sin() is considered as radian, which can be seen from the documentation of sin (easily foundable by a simple search on the internet), thus the value is provided accordingly.

If you want to compute considering the input as degrees, use sind.

Also note that the output shown is a truncated value upto 4 digits after decimal, which is not what the actual value is.

Answer 1

[David Hill]

x=0:0.01:2*pi;
si=sin(x).^2;
co=cos(x).^2;
plot(x,si,x,co);
figure;
plot(si,co);%not sure which one you want

[Miri van de Kamp]

thanks

[David Hill]

sin(2*x);%sind(2*x) for x in degrees

[Sam Chak]

@윤선 이, no worries, I teach you.

x = 0:0.01:pi;
y = sin(2*x);
plot(x, y)

[Image Analyst]

An alternative to specifying the spacing is to specify the number of elements in the vector with linspace(), like

numElements = 2000; % Should be enough to fit all the way across your screen.
x = linspace(0, pi, numElements);
y = sin(2 * x);
plot(x, y, 'b-', 'LineWidth', 2);
grid on;
xlabel('x');
ylabel('y');
title('Y vs X');
xlim([min(x), max(x)]);

Answer 2

[hemin ahmad]

sin(x)

[Walter Roberson]

No, this is not correct. As explained above, sin(x).^2 is needed.