Proof that sin6 > 1/10 and sin10 > 1/6?

Question

How do u proove the above? Steps will be appreciated. Thank you.

in degrees

Answer 1

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The function

     sin(πx/180) - x/60

is 0 when x = 0 and x = 30, and can be shown, using derivatives, to have a maximum between 0 and 30

∴ sin(πx/180) > x/60 if 0 < x < 30

∴ sin(xº) > x/60 if 0 < x < 30

Putting x = 6 and 10 get

sin 6º > 1/10
sin 10º > 1/6

and similarly, also get

sin 1º > 1/60
sin 2º > 1/30
sin 3º > 1/20
sin 4º > 1/15
sin 5º > 1/12
sin 12º > 1/5
sin 15º > 1/4
sin 20º > 1/3

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Alternatively, prove the special cases as follows.

Proof that sin 6º > 1/10
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See Source 1 to calculate:

     sin 36º = ¼√( 2 ( 5 - √5 ) )
     cos 36º = ¼√5

     sin 30º = ½ and cos 30º = ½√3

Use difference formula to get

     sin 6º = sin( 36º - 30º ) = sin 36º cos 30º - sin 30º cos 36º

Simplify to get (and check with Source 2):

     sin 6º = [√6 √( 5 - √5) - (√5 + 1 ) ] / 8

To prove this > 1/10 rearrange, square, etc to get

     30 - 6√5 > 206 / 25 + 18 / 5 √5

     √5 < 34 / 15

     1125 < 1156 which is true.

A simpler method is similar to the following for 10º. Use the multiple angle formula for sin (5 × 6º) and show that when s = 1/10

     32 s^5 - 40 s³ + 10 s - 1 < 0

Proof that sin 10º > 1/6
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Use multiple angle formula (Source 3) to get

     sin 30º = 3 cos² 10º sin 10º - sin³ 10º

Let s = sin 10º and simplify to

     f(s) = 8s³ - 6s + 1 = 0

Use calculus to show that this function decreases from 0 to ½ then increases. So sin 10º < 1/6 if f(1/6) > 0 which is true.

Answer 2

But is your "6" 6 degrees or 6 radians? It's true if
you mean 6 degrees, but patently false in the other
case. The sin of 6 radians is in the 4th quadrant
and is negative.

Answer 3

Use Maclaurin expansion:
sin x = x-x^3/3!+x^5/5! - ... +(-1)^(m+1)[x^(2m-1)]/(2m-1)! .....

For x =6, this gives sin 6 = .1016
For x =10 it gives sin 10 = .1736