Question

What is the exact value (not decimal) of the sin of a 40 degree angle of a right triangle?

Answer 1

sin3A=sin(2A+A)=sin2AcosA+cos2AsinA=2sinAcos^2A+sinA(1-2sin^2A)=

 2sinA(1-sin^2A)+sinA-2sin^3A=2sinA-2sin^3A+sinA-2sin^3A=3sinA-4sin^3A.

 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;

2sin20cos20=2sin40cos40-sin20;

2sin20cos20=4sin20cos20(2cos^2(20)-1)-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.

Comments

I’ve tried this problem, and I don’t believe there is an exact value either.

I used Rod’s cubic function with x = sin(40). I ended up with the sum of two cube roots of complex conjugates. The only way I could get rid of the imaginary components was to create an infinite series for each cube root (via Binomial expansion) which cancelled out the imaginary bits. But that, in effect, leaves us with an approximation and not an exact value.

Another bummer for finding an exact value is according to Gauss!

We are looking for sin(40) and 40 = 2π/9 = π/(4.5) . Now Gauss proved (see here, search on “Do we know all the angles”) that when φ = π/n then sin(φ) is expressible in surd form only if n is a power of 2, or n is a Fermat prime (no pun intended). i.e. n = 3, 5, 17, 257, 65537.

So the only answer I could give is: sin(40) = (1/4){qrt(C+) + qrt(C-)}

Where qrt(x) = cube root of x, and C+ = -4sqrt(3) + 4i, and C- = -4sqrt(3) – 4i.

It’s exact(!), but it’s not Real-ly a solution (pun intended).

Answer 2

I don't understand your question... a right triangle can be 45 45 90 or 30 60 90 you can also have 40 90 which if u add them equals 130 then 180-130 = 50  40 50 90 well the sin(40) = opposite/hypetunese are you given the sides?

Comments

I'm puzzled, too, because I think the question anticipates or assumes there will be a Pythagorean triangle (like 3-4-5) in which there is an angle of 40 degrees. I'm pretty sure there isn't one, but it may still be possible to find sin40 in terms of irrational numbers like square roots, cube roots, etc., without having to write out a decimal approximation.