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    Question :
    Q.Show that cos 1?, sin 1?, and tan 1? are irrational numbers.
    Category Puzzles Interview Questions
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    Added on 2/6/2012
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    We will use appropriate multiple angle formulae to show that, if cos 1?, sin 1?, or tan 1? is rational, then the cosine or tangent of some multiple of 1? must also be rational. By choosing a multiple of 1? for which a suitable trigonometric function is known to be irrational, we can derive a contradiction and thereby establish each result.
    cos 1?

    De Moivre's theorem states that for any real number x and any integer n,

    cos nx + i sin nx = (cos x + i sin x)n

    Expanding the right-hand side using the binomial theorem, and equating real parts, we have

    cos nx = cosnx − [n(n − 1)/2] cosn−2x sin2x + ...

    Given that sin2x = 1 − cos2x, we can thereby express cos nx as a polynomial in cos x, with integer coefficients.
    Hence, cos x rational implies cos nx rational.
    Equivalently, cos nx irrational implies cos x irrational.

    Taking n = 30 and x = 1?, since we know cos 30? = root 3/2 is irrational, it follows that cos 1? is irrational.
    sin 1?

    Building on the above result, since cos 2x = 1 − 2 sin2x, we have

    cos 2nx irrational implies cos 2x irrational implies sin2x irrational implies sin x irrational.

    Taking n = 15 and x = 1?, it follows that sin 1? is irrational.
    tan 1?

    The standard addition formula for tangents
    tan(a+b) = (tan a + tan b)/(1 - tan a tan b)

    tells us that, if tan a and tan b are rational, then tan(a + b) is rational. (Of course, we must also have tan a, tan b, and tan(a + b) defined, so that tan a tan b not equal to 1.)

    We know that tan 30? = 1 /root 3 is irrational.
    Since 30? can be built up as a series of binary sums, beginning with 1? and 1?, it follows, by contradiction, that tan 1? is irrational.

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