What Is the Exact Value of Sin 345.5 Degrees?

[songoku]

Homework Statement

Find the exact value of sin 345.5^o

Homework Equations

Trigonometry Identities

The Attempt at a Solution

Don't know where to start.

Tried sin 345.5^o = - sin 14.5^o but stuck. Also tried multiply 345.5 with positive integer to get sin 2θ or sin 3θ or sin 4θ but also stuck

Thanks

 

[Ray Vickson]

Homework Statement

Find the exact value of sin 345.5^o

Homework Equations

Trigonometry Identities

The Attempt at a Solution

Don't know where to start.

Tried sin 345.5^o = - sin 14.5^o but stuck. Also tried multiply 345.5 with positive integer to get sin 2θ or sin 3θ or sin 4θ but also stuck

Thanks

You should definitely start with 2*345.5 = 691. Now try to reduce 691 to an angle < 90 degrees, by subtracting suitable multiples of 180 or 90.

 

[songoku]

Sorry for taking long time to reply

sin 691 = sin 331 = - sin 29.

You mean finding sin 29 through the link you gave in other thread and using double angle formula?

Thanks

 

[willem2]

This is impossible. You can't get an exact value of the sign for any angle in degrees that's not divisible by 3. You won't ever get rid of the factor 3 by halving/doubling adding or subtracting angles.

 

[SteamKing]

This is impossible. You can't get an exact value of the sign for any angle in degrees that's not divisible by 3. You won't ever get rid of the factor 3 by halving/doubling adding or subtracting angles.

I think you mean sine. The trig function is called the 'sine'.

 

[BvU]

http://www.intmath.com/blog/mathematics/how-do-you-find-exact-values-for-the-sine-of-all-angles-6212

 

[Ray Vickson]

This is impossible. You can't get an exact value of the sign for any angle in degrees that's not divisible by 3. You won't ever get rid of the factor 3 by halving/doubling adding or subtracting angles.

No, that is incorrect. A paper giving exact algebraic formulas for the sine of all angles from 1 to 90 degrees, in 1 degree increments, has been published on-line (with proofs included). It was done as a retirement project by an ex-professor of mathematics; for a precise citation, see one of my responses in the previous thread by user 'songoku' on a related topic.

Note added in edit: I see that BvU has already dealt with this issue, in a post that appeared on my screen only after I pressed the 'enter' key.

 

[haruspex]

http://www.intmath.com/blog/mathematics/how-do-you-find-exact-values-for-the-sine-of-all-angles-6212

To go further, I believe it should be possible to find an exact representation of cos(pi/n) only involving square roots if and only if there exists a ruler and compass construction for a regular n-sided polygon. As is well known, that is possible whenever n is the product of a power of 2 and distinct Fermat primes. That is enough to get all multiples of 1 degree as well as pi/17 etc.

But if we allow other surds then there are more possibilities. Since cos(nx) can be expanded as an nth order polynomial in cos(x), and cos(x)=-cos(pi-x), we can expand cos(4pi/7)=-cos(3pi/7) to obtain a quartic in cos(pi/7).
I feel there should be a generalization of the Fermat primes that corresponds to roots up to cubic and quartic, but I'm not aware of such.

 

[SammyS]

°

To go further, I believe it should be possible to find an exact representation of cos(pi/n) only involving square roots if and only if there exists a ruler and compass construction for a regular n-sided polygon. As is well known, that is possible whenever n is the product of a power of 2 and distinct Fermat primes. That is enough to get all multiples of 1 degree as well as pi/17 etc.

Doesn't that only get you down to multiples of 3° ?

 

[haruspex]

°

Doesn't that only get you down to multiples of 3° ?

Sorry, yes, 3°. To get to 1° you need to use the cos(3x) expansion or similar, so does involve cube roots.