Whats-is-the-value-of-sin-6-

Question Number 68495 by Maclaurin Stickker last updated on 12/Sep/19

Whats is the value of \sin (6 °) ?

Commented by Pk1167156@gmail.com last updated on 12/Sep/19

soln

\sin 6 °

= \sin (36 - 30)

= \sin 36 ° \cos 30 ° - \cos 36 ° \sin 30 °

= \frac{\sqrt{10 - 2 \sqrt{5}}}{4} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{5} + 1}{4} \times \frac{1}{2}

= \frac{\sqrt{30 - 6 \sqrt{5}}}{8} - \frac{\sqrt{5 + 1}}{8}

= \frac{\sqrt{30 - 6 \sqrt{5}} - (\sqrt{5} + 1)}{8}

Answered by MJS last updated on 12/Sep/19

we can calculate \sin 15 ° by using the formula

\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} with x = 30 ° \Rightarrow

\Rightarrow \sin 15 ° = \frac{\sqrt{6} - \sqrt{2}}{4}; \cos 15 ° = \frac{\sqrt{6} + \sqrt{2}}{4}

\sin 18 ° = \cos 72 ° is known from the regular

pentagon

\sin 18 ° = \frac{\sqrt{5} - 1}{4}; \cos 18 ° = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}

now

\sin 3 ° = \sin (18 ° - 15 °) = \sin 18 ° \cos 15 ° - \cos 18 ° \sin 15 ° =

= \frac{2 (1 - \sqrt{3}) \sqrt{5 + \sqrt{5}} - (1 - \sqrt{5}) (1 + \sqrt{3}) \sqrt{2}}{16}

\cos 3 ° = \sqrt{1 - \sin^{2} 3 °}

\sin 6 ° = \sin 2 \times 3 ° = 2 \sin 3 ° \cos 3 ° =

\dots

= \frac{\sqrt{9 - \sqrt{5} - \sqrt{30 + 6 \sqrt{5}}}}{4} \approx . 104528

Commented by MJS last updated on 12/Sep/19

generally we can give exact values for

trigonometric functions of n \times 3 ° with n \in Z

Commented by Maclaurin Stickker last updated on 12/Sep/19

Thank you, sir!

Commented by malwaan last updated on 13/Sep/19

g r e a t w o r k

t h a n k y o u ♡

Commented by Maclaurin Stickker last updated on 12/Sep/19

Or we can use the arc subtraction \sin (6 °) = \sin (36 ° - 30 °) \Rightarrow \sin (36 °) \cos (30 °) - \sin (30 °) \cos (36 °) \Rightarrow \frac{\sqrt{10 - 2 \sqrt{5}}}{4} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{\sqrt{5} + 1}{4} \Rightarrow \frac{\sqrt{30 - 6 \sqrt{5}} - \sqrt{5} - 1}{8} = 0.104528 \cdot \cdot \cdot