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Hello

 

I am having issue in finding the explanation on how to solve inverse trig funtion and expression with inverse trig funtions. I do not understand my school book and I was hoping that the software would have given me an extra help in understanding my school problems. 

 

Thank you very much

Regards,

Perla D'andrea

Hello,

After trigonometric manipulations in a mechanical problem, I can obtain the desired angles but defined with modulo 2Pi.

I would like to program or find a function which can do this operation :

While angle doesn't belong to [-Pi, Pi]

do 

  If angle > Pi then do angle = angle - 2Pi

  If angle < - Pi then do angle = angle + 2Pi

end

Is there an existing function which can do this operation ?

Otherwise, may you help me to program it ?

Thanks a lot for your help

Hello,

I would like to simplify a trigonometric equation that I obtain with a vectorial closure (in mechanics)

Here the equation that I would like to simplify 

eq_liaison := (-sin(p(t)+g(t))*cos(a(t))-sin(b(t))*sin(a(t))*cos(p(t)+g(t)))*l2[1]+((-sin(p(t)+g(t))*cos(a(t))-sin(b(t))*sin(a(t))*cos(p(t)+g(t)))*cos(th(t))+(-cos(p(t)+g(t))*cos(a(t))+sin(a(t))*sin(b(t))*sin(p(t)+g(t)))*sin(th(t)))*l3[1]+(-sin(a(t))*sin(g(t))*sin(b(t))+cos(a(t))*cos(g(t)))*xb[1]+sin(a(t))*cos(b(t))*yb[1]+(sin(a(t))*sin(b(t))*cos(g(t))+cos(a(t))*sin(g(t)))*zb[1]+x(t)-xp(t) = 0

Do you have ideas so as to simplify again this expression ?

This expression can still be simplified. You can find here the result expected :

 

I find surprising that I have so many difficulties to make trigonometric simplications with the trigonometric functions.

I attached the code 

Download TrigoTransformVectorialEquations.mwTrigoTransformVectorialEquations.mw

Thank you for your help

I have:

r:=t-><4cos2t+4sint,5cos3t+4cost,(sin3t+2sint)^2>

with(plots):

 

then I tried:

spacecurve(r(t), t = 0 .. 2*Pi, thickness = 2, color = black, axes = normal, labels = [x, y, z], numpoints = 150)

But I keep getting the error: Warning, unable to evalute the function to numeric values in the region. 

But I thought <4cos2t+4sint, 5cos3t+4cost, (sin3t+2sint)^2> was defined on the region t=0 to 2Pi??? 

I'm using Maple 2015 if that helps. 

hi,

here a comlicated formula,how i simplify

thanks  a lot.

``

f := (kappa*omega^2+omega^3)*(Y+(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(2*(kappa*omega^2+omega^3)))^2/(2*omega)+(-kappa*omega^2+omega^3)*(X+(sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(2*(-kappa*omega^2+omega^3)))^2/(2*omega)+(Omega*N*cos(theta[2])*omega+Omega*N*cos(theta[1])*omega-P__X^2*kappa+P__X^2*omega+P__Y^2*kappa+P__Y^2*omega)/(2*omega)-(sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)^2/(8*omega*(-kappa*omega^2+omega^3))-(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)^2/(8*omega*(kappa*omega^2+omega^3))

(1/2)*(kappa*omega^2+omega^3)*(Y+(-N^(1/2)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+N^(1/2)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(2*kappa*omega^2+2*omega^3))^2/omega+(1/2)*(-kappa*omega^2+omega^3)*(X+(N^(1/2)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+N^(1/2)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(-2*kappa*omega^2+2*omega^3))^2/omega+(1/2)*(Omega*N*cos(theta[2])*omega+Omega*N*cos(theta[1])*omega-P__X^2*kappa+P__X^2*omega+P__Y^2*kappa+P__Y^2*omega)/omega-(1/8)*(N^(1/2)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+N^(1/2)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)^2/(omega*(-kappa*omega^2+omega^3))-(1/8)*(-N^(1/2)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+N^(1/2)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)^2/(omega*(kappa*omega^2+omega^3))

(1)

``

(1/2)*(kappa*omega^2+omega^3)*(Y+(-N^(1/2)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+N^(1/2)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(2*kappa*omega^2+2*omega^3))^2/omega

(2)

``

    f is a complicated function,i want to make it more simplify,but i want to keep square style,

 let coefficients of X and Y keep one unit,and simplify terms  containd special symbol of omega

 

Download Q1119.mw

it what i wanted.

I want to reduce all solution of the equation sin(x)^2=1/4

restart:
sol:=solve(sin(x)^2=1/4, x, AllSolutions);

and

restart:
k:=combine((sin(x))^2);
sol:=solve(k=1/4, x, AllSolutions = true, explicit);
simplify(sol);

How can I reduce solution sol := -1/3*Pi*_B3+1/6*Pi+Pi*_Z3 ?

How can I get x= pi/6+k*pi and x= -pi/6+k*pi?

Hi guys.

I want to know how can I make the maple to give final response of a simple trigonometric function, e.g sin(pi/6). When I type sin(pi/6) in the command line and then press enter, maple give the same, I mean sin(pi/6) at the next line. I want 0.5 as the final response not sin(pi/6) again. Simplify command does not work for me in this case.

Thanks in advance.

When I do simplify(LegendreP(n, 1, cos(t))), Maple gives me -sqrt(1-cos(t))*sqrt(cos(t)+1). Isn't it the same thing as -sin(t)? How can I have Maple further convert/simplify it to -sin(t)? (I tried simplify(%, trig) but it didn't work). I am new on Maple. Thanks in advance for anyone's help!

 

I'm looking at Maple as a possible alternative to Mathcad (which I've been using for years, but is now very jaded compared to other options like Maple and Mathematica).  I'm a civil engineer and for what I do, one of the better features of Mathcad is the way it handles units.  For example, if I specify an angle in degrees (say phi=30 degrees) and then ask for sin(phi), I get 0.5.  At face value, I though Maple would do the same kind of thing.  However, this doesn't appear to be the case (see attached worksheet).  The only workaround that I can see is to specify the angle in degrees (but without assigning ) and then multiply the specified value by pi/180 (to convert to radians) before passing it to the sin function.  Which is all a bit messy and not at all an attractive solution.

Am I misunderstanding the way units work in Maple and is there a clean way of specifying angles in degrees (which is what engineers work with) and using these values directy in trig functions?

Thanks in anticipation,

Ian

  Solving trigonometry Equations  sin^2(2x)-cos^2(8x)=0.5cos(10x)

There are two ways of expressing the solution to a cubic equation, one of them uses cos and arccos [1]. How do I / is there a way to ask Maple to get this form?

More generally, can Maple be instructucted to solve equations using trig identities?

[1] http://en.wikibooks.org/wiki/Trigonometry/The_solution_of_cubic_equations

How to solve the equation
2^(sin(x)^4-cos(x)^2)-2^(cos(x)^4-sin(x)^2) = cos(2*x)
symbolically? The solve command produces a weird answer. Evalfing all its values, one sees
0.7853981634, -0.7853981634, 2.356194490, -2.356194490,

1.570796327 - 1.031718534 I, -1.570796327 + 1.031718534 I,

1.570796327 + 1.031718534 I, -1.570796327 - 1.031718534 I,

0.7853981634, -0.7853981634, 2.356194490, -2.356194490,

1.570796327 - 1.031718534 I, -1.570796327 + 1.031718534 I,

1.570796327 + 1.031718534 I, -1.570796327 - 1.031718534 I


The identify command
interprets the real solutions on -Pi..Pi as -3*Pi/4, -Pi/4, Pi/4, 3*Pi/4
(for example,
identify(2.356194490);

3*Pi/4 ).
Is it possible to obtain these with Maple in a simpler way?

PS. Mathematica 10 does the job.

PPS. So does even Mathematica 7.

Is it possible to find all the solutions of the equation

abs(tan(x)*tan(2*x)*tan(3*x))+abs(tan(x)+tan(2*x)) = tan(3*x)

which belong to the interval 0..Pi with Maple?

 

 

eq1:=
(1/4)*D^2*Pi-(1/4)*D^2*arccos((-D+2*h)/D)-(1/2)*sqrt(h)*sqrt(D-h)*D+h^(3/2)*sqrt
(D-h);
                          
eq2 :=
-(1/2)*sqrt(h)*sqrt(D-h)*D+h^(3/2)*sqrt(D-h)+(1/4)*D^2*arcsin((-D+2*h)/D)+(1/8)*
D^2*Pi;

These equations are the same. yet simplify(eq1-eq2,trig);
<>0
The Mathematica COMMAND FullSimplify[..] gets zero.

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