What is the value of sin(pi/5) ?

The value of sin(pi/5) is (sqrt(2)sqrt(5-\sqrt{5)})/4

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sin(pi/5)

sin (\frac{π}{5})

\frac{2 5 - 5}{4}

Solution steps

sin (\frac{π}{5})

Show that:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

Show that:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

Add the following equations

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

((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) = (\frac{5}{2} + \frac{1}{2})

Refine

cos (\frac{π}{5}) = \frac{5 + 1}{4}

Square both sides

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

Use the following identity:

sin^{2} (x) = 1 - cos^{2} (x)

sin^{2} (\frac{π}{5}) = 1 - cos^{2} (\frac{π}{5})

Substitute

cos (\frac{π}{5}) = \frac{5 + 1}{4}

sin^{2} (\frac{π}{5}) = 1 - (\frac{5 + 1}{4})^{2}

Refine

sin^{2} (\frac{π}{5}) = \frac{5 - 5}{8}

Take the square root of both sides

sin (\frac{π}{5}) = \pm \frac{5 - 5}{8}

sin (\frac{π}{5})

cannot be negative

sin (\frac{π}{5}) = \frac{5 - 5}{8}

Refine

sin (\frac{π}{5}) = \frac{\frac{5 - 5}{2}}{2}

= \frac{\frac{5 - 5}{2}}{2}

\frac{\frac{5 - 5}{2}}{2} = \frac{2 5 - 5}{4}

= \frac{2 5 - 5}{4}

tan (x) = 0

arctan (2)

sin (\frac{8 π}{3})

tan (- \frac{π}{2})

sin (30 0^{\circ})

What is the value of sin(pi/5) ?
The value of sin(pi/5) is (sqrt(2)sqrt(5-\sqrt{5)})/4