Write a function called sindeg that takes a matrix input called deg. The function returns a matrix of the same size as deg with elements containing the sine of the corresponding element of deg and mean of the sine.we are nor allowed builtin functions

function [sine,mean_sine] = sindeg(deg)
sine = (exp(i*deg) - exp(-i*deg))/(2i)
mean_sine=sine/length(deg)
I am not getting right value. instead I am getting this. sine = 0.1060 -0.2624 0.5514 0.6702 0.5514 0.9866 -0.9483 mean_sine = 0.0151 -0.0375 0.0788 0.0957 0.0788 0.1409 -0.1355 ans = 0.1060 -0.2624 0.5514 0.6702 0.5514 0.9866 -0.9483

1 Kommentar

Whenever you write "not getting the right values", mention what the rigth values are and why you assume this.

Melden Sie sich an, um zu kommentieren.

Antworten (2)

Jan
Jan am 17 Mai 2017
Bearbeitet: Jan am 17 Mai 2017
mean_sine = sum(sine) / length(deg);
% ^^^
The mean is the sum of the elements divided by the number of elements.
we are nor allowed builtin functions
This is nonsense. I know that teachers ask such questions, but:
a = b
calls the builtin function subsassgn already.
"deg" sounds like the input is in degrees. Then convert the values to radians at first.
Append semicolons to the lines to suppress the output and call your function like
[sine,mean_sine] = sindeg(deg)
I think what they were talking about with this question is that you can't use sind the function that allows for inputs of degrees which easily solves the question. I'm slightly confused by the 2nd line of you code because you don't actually take the sin of anything.

2 Kommentare

Please don't answer a question if you only have a comment. Hint: There are comments when you want to comment.
@Isaac: Compare:
deg = 123.4;
rad = deg * pi / 180;
sin1 = (exp(1i*rad) - exp(-1i*rad)) / (2i)
sin2 = sin(rad)
exp(1i * x) = cos(x) + 1i * sin(x)

Melden Sie sich an, um zu kommentieren.

Gefragt:

am 17 Mai 2017

Bearbeitet:

Jan
am 4 Jan. 2018

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by