CODEWITHSUNDEEP
matlabmath
user3603752
user3603752

Reputation: 5

Math, 2 equations 2 unknowns, and matlab, plot

i am trying to solve these 2 equations with 2 unknowns. i want to find the angle of a1(alpha1), a2(alpha2) in radians. however when i solve it in maple i get 0.112 for a2 and 0.089 for a1. but some other guys used matlab and they get something different with plot and 2 angles.. can someone tell me if i am right or he is? MY equations: equation 1: 4/(2*pi)*(cos(2*a1)+cos(2*a2))=0 equation 2: 4/(10*pi)*(cos(10*a1)+cos(10*a2))=0

his matlab:

j=0;
for i=0:pi/100:1
j=j+1;
a2(j)=i;
a1_1(j)=acos(-cos(2*i))/2;
a1_2(j)=acos(-cos(10*i))/10;
end

plot(a2,a1_1,'-k',a2,a1_2,'-b','LineWidth',1.4);

I wanna plot like him..but i am not sure his Equations a1_1 is correct? by the way the main equation it came from: bn = 4/(npi)(cos(na1)+cos(na2))=0 bn = 0 and n is 2th and 10th harmonics to be eliminated

Upvotes: 0

Views: 137

Answers (1)

Lutz Lehmann
Lutz Lehmann

Reputation: 25972

cos(A)+cos(B) = 0

happens for 

A + B = pi + 2*k*pi  or A - B = pi + 2*k*pi

Both of the two original equations could resolve the the same of the two variants or to different variants. Resolving to the same equation, say the first, will produce 

7*(a1+a2) = pi + 2*k7*pi  and  13*(a1+a2) = pi + 2*k13*pi

which can only hold if 

0 = 6*pi + 2*(13*k7-7*k13)  <=>  3 = 7*k13-13*k7

which implies a1+a2 = pi + 2*m*pi. .

In the case of different to different variant, one obtains (modula sign variations of a1, a2)

7*(a1+a2) = pi + 2*k7*pi  and  13*(a1-a2) = pi + 2*k13*pi

and thus

a1 = (pi/7+pi/13)/2 + k7*pi/7 + k13*pi/13
a2 = (pi/7-pi/13)/2 + k7*pi/7 - k13*pi/13

and all possible sign variations, since cosine is an even function. However, simultaneous sign variations are also covered by variations of k7 and k13. Thus you obtain additionally (up to) 2*7*13=182 solutions to your problem.

Upvotes: 2

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