How to derive exact algebraic formulae for sin ( π 10 ) and cos ( π 10 )

Question

How to derive exact algebraic formulae for             sin              (                  π          10                )             and             cos              (                  π          10                )

Kylie Woodward · Accepted Answer

These roots can be expressed in trigonometric form as:

\cos (\frac{2 n π}{5}) + i \sin (\frac{2 n π}{5})

for

n = 0, 1, 2, 3, 4

The zeros of are as follows:

x^{5} - 1 = (x - 1) (x^{4} + x^{3} + x^{2} + x + 1)

x^{5} - 1 = (x - 1) x^{2} (x^{2} + x + 1 + \frac{1}{x} + \frac{1}{x^{2}})

x^{5} - 1 = (x - 1) x^{2} ({(x + \frac{1}{x})}^{2} + (x + \frac{1}{x}) - 1)

x^{5} - 1 = (x - 1) x^{2} ({((x + \frac{1}{x}) + \frac{1}{2})}^{2} - {(\frac{\sqrt{5}}{2})}^{2})

x^{5} - 1 = (x - 1) x^{2} (x + \frac{1}{x} + \frac{1}{2} - \frac{\sqrt{5}}{2}) (x + \frac{1}{x} + \frac{1}{2} + \frac{\sqrt{5}}{2})

x^{5} - 1 = (x - 1) (x^{2} + (\frac{1}{2} - \frac{\sqrt{5}}{2}) x + 1) (x^{2} + (\frac{1}{2} + \frac{\sqrt{5}}{2}) x + 1)

Thus, we discover:

\sin (\frac{π}{10}) + i \cos (\frac{π}{10}) = \cos (\frac{2 π}{5}) + i \sin (\frac{2 π}{5})

is a root of:

x^{2} + (\frac{1}{2} - \frac{\sqrt{5}}{2}) x + 1 = 0

We locate roots using the quadratic formula:

x = \frac{1}{4} (\sqrt{5} - 1) \pm i \frac{1}{4} \sqrt{10 + 2 \sqrt{5}}

Equating real and imaginary parts and choosing the appropriate sign, we find:

\sin (\frac{π}{10}) = \frac{1}{4} (\sqrt{5} - 1)

\cos (\frac{π}{10}) = \frac{1}{4} \sqrt{10 + 2 \sqrt{5}}

hordesowderneo8x7 · Accepted Answer

Let             π      10        =    A  , then       5    A    =          π      2       or       3    A    =          π      2        -    2    A  and             sin      3        A    =          sin              (                  π          2                -        2        A        )            i.e.       3          sin      A        -    4                  sin        3            A        =          sin      3        A    =          cos      2        A    =    1    -    2                  sin        2            A      or       4                  sin        3            A        -    2                  sin        2            A        -    3          sin      A        +    1    =    0              (              sin        A            -      1      )              (      4                        sin          2                A            +      2              sin        A            -      1      )        =    0   and dividing by             sin      A        -    1   we getor       4                  sin        2            A        +    2          sin      A        -    1    =    0  or             sin      A        =                  -        2        ±                                            2              2                        +            16                              8        =                  -        2        ±        2                  5                    8        =                  -        1        ±                  5                    4      but as             sin      A       cannot be negative,             sin      A        =          sin              (                  π          10                )              =                            5                -        1            4      and             cos              (                  π          10                )              =                  1        -                              (                                                            5                                -                1                            4                        )                    2                                        1        -                              6            -            2                          5                                16                      =                            10          +          2                      5                              4

How to derive exact algebraic formulae for sin(pi/10) and cos(pi/10)?

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