Items tagged with arctrig

Feed

I am curious whether anyone here can come up with a way to simplify the expression x1 to 15*Pi/32 in fewer exact, symbolic steps. The following was performed in Maple 2016.2 for Linux.

restart;

x1 := arcsin(1/2*(2+(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2));

arcsin((1/2)*(2+(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2))

x2 := evalc(convert(x1,expln));

arctan((2+(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2)/(2-(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2))

x3 := convert(x2, expln);

((1/2)*I)*(ln(1-I*(2+(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2)/(2-(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2))-ln(1+I*(2+(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2)/(2-(2+(2+2^(1/2))^(1/2))^(1/2))^(1/2)))

# non-Pro wolframalpha can simplify x3 to 15*Pi/32 (but not x2 or x1).
#

x4 := combine(x3);

((1/2)*I)*ln(-(I*2^(1/2)*((2+2^(1/2))*(2+(2+2^(1/2))^(1/2)))^(1/2)+I*(2*2^(1/2)+4)^(1/2)+I*2^(1/2)+(-1+I))/(I*2^(1/2)*((2+2^(1/2))*(2+(2+2^(1/2))^(1/2)))^(1/2)+I*(2*2^(1/2)+4)^(1/2)+I*2^(1/2)+1+I))

x5 := simplify(x4);

((1/2)*I)*(-ln(2)+ln(-(2+(2+2^(1/2))^(1/2))^(1/2)*(I*2^(1/2)*(2+2^(1/2))^(1/2)-I*2^(1/2)+1-I)))

x6 := expand(x5);

-((1/2)*I)*ln(2)+((1/4)*I)*ln(2+(2+2^(1/2))^(1/2))+((1/2)*I)*ln(-I*2^(1/2)*(2+2^(1/2))^(1/2)+I*2^(1/2)+(-1+I))

x7 := combine(x6);

-((1/4)*I)*(ln(-(1/2)*(2+2^(1/2))^(1/2)*(I*2^(1/2)+(-1-I)))+(2*I)*Pi)

simplify(x7,constant);

(15/32)*Pi

# It's a pity that last step worked while this next is inadequate.
simplify(x7);

-((1/4)*I)*ln(-(2+2^(1/2))^(1/2)*(I*2^(1/2)+(-1-I)))+((1/4)*I)*ln(2)+(1/2)*Pi

# Another way, using x7
simplify(evalc(x7));

(15/32)*Pi

# Another way, using x7
simplify(combine(expand(x7)));

(15/32)*Pi

 

Download simplify_example.mw

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

Page 1 of 1