What is the exact value (not decimal) of the sin of a 40 degree angle of a right triangle?
If A=40, sin120=3sin40-4sin^3(40)=sqrt(3)/2.
Put x=sin40: 3x-4x^3-sqrt(3)/2=0 or 8x^3-6x+sqrt(3)=0, which reduces to: x^3-(3/4)x+sqrt(3)/8=0, so one solution of this cubic will be sin40. The cubic will also give sin20 as a solution, because sin120=sin60=sin(3*20)=sqrt(3)/2, and sin(-80) (=-sin80), because sin(-240)=sqrt(3)/2 (quadrant II), the same as sin120. sin40 is the most positive solution, of course, and it is an irrational number so cannot be represented as a fraction a/b where a and b are integers.
The cubic factorises: (x-sin20)(x-sin40)(x+sin80)=0=x^3-x^2(sin20+sin40-sin80)+x(sin20sin40-sin20sin80-sin40sin80)+sin20sin40sin80. Comparing this with the cubic coefficients, sin20+sin40-sin80=0 because there is no x^2 coefficient, and sin40=sin80-sin20 (check: sin80-sin20=2cos((80+20)/2)sin((80-20)/2)=2cos50sin30=2sin40sin30=sin40, because sin30=1/2). Similarly, we can equate expressions for -3/4 and sqrt(3)/8, relating sin40 to sin20 and sin80.
sin40=sin80-sin20 can be written sin(2*20)=sin(2*40)-sin20;
2cos20=4cos20(2cos^2(20)-1)-1; 8cos^3(20)-6cos20-1=0. Writing y=cos20: 8y^3-6y-1=0, which contains no irrational numbers, but is otherwise similar to the cubic obtained earlier. Although the solution includes cos20, sin40 can easily be calculated from it. Other solutions are cos100 (cos(180-80)) and cos140 (cos(180-40)).
If the solution to the cubic equations can be expressed in terms of square roots, or other roots, I'll update this answer later. Read More: ...