Can we find the exact value(not numerical/approximation) of sin 1? I tried to do so by solving a cubic equation using Cardano formula but I ended up with complex nested radicals): I was told to use other methods but that won't yield exact values
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$\begingroup$sin(1) is a Transcendental number.
ie ,sin(1) is not a solution a non-zero polynomial equation with rational coefficients.In other words sin(1) can't be expressed in closed form only using fractions and radicals. Note: sin(a), where a is algebraic(not transcendental) ,is transcendental.
Does that help you?
$\endgroup$ 4 $\begingroup$Series[Sin[x], {x, 0, 14}]$$\sin (x) = x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}-\frac{x^{11}}{39916800}+\frac{x^{13}}{6227020800}+O\left(x^{15}\right)$$
DeleteCases[CoefficientList[Series[Sin[x], {x, 0, 14}], x], 0]$$\sin (1) = 1-\frac{1}{6}+\frac{1}{120}-\frac{1}{5040}+\frac{1}{362880}-\frac{1}{39916800}+\frac{1}{6227020800}-...$$
$$1-\frac{1}{6}+\frac{1}{120}-\frac{1}{5040}+\frac{1}{362880}-\frac{1}{39916800}+\frac{1}{6227020800}-... = 0.84147098480866...$$
$$\sin (1) = 0.84147098480790...$$
Table[(2*n - 1)!, {n, 1, 7}]{1, 6, 120, 5040, 362880, 39916800, 6227020800}
$\endgroup$ 1 $\begingroup$If $1$ is degrees here's 2 closed forms:
$$\frac{1}{2}ie^{-\frac{i\pi}{180}} - \frac{1}{2}ie^{\frac{i\pi}{180}}$$
$$-\frac{1}{2}(\sqrt[180]{-1} - 1)(\sqrt[180]{-1} + 1)(-1)^{\frac{89}{180}}$$
If $1$ is radians here's one closed form:
$$\frac{i e^{-i}}{2}-\frac{i e^i}{2}$$
$\endgroup$ $\begingroup$Yes, we can. However the radicals will be nested as well as repeating. Using Precise-Rewritten method, exact value of Sin 1 degree is: $$√(2 - √(2 + √(2 \overline {+ √(2 + √(2 + √(2 + √(2 - √(2 - √(2 - √(2 + √(2 - √(2 - √(2 + √(2 - √(2}]/2$$
Please care the overline (not fraction) and closing brackets have collapsed by']'.
This method gives exact values of Sine of all integer angles. Please visit for the list.
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