It is the same as -sin(24).
what is sine of 2004 degrees without using a calculator?
this was an extra credit problem given in an honors precalc class and i am curious on how to solve it. since it is precal, you can't use taylor polynomials. 2004 degrees is the same as 204 degrees but that is not one of the "special" angles like 30, 60, or 90.
Ok, so I have been thinking about this and it really seems to just boil down to solving a few LDEs.
. Now, obviously . So, now we must merely solve the LDE . Clearly we must have that . So, let's think about this. We only want values of that we can work with ( ). So, we automatically can throw out a few triplets considering, for example, . A little experimentation shows that works.
So we have that . A little simplification shows this equal to
But, . Once again, a little simplification shows that this is . Solving for gives
Subbing this back into and letting gives us . Which, with some effort (not going to bore you with the algebra) is equal to
Let such that . Substitution and subsequent simplification leads us to (look up depressed cubic)
So that we see that
Remark: After writing all of that I just realized that . Letting gives us the equivalent depressed cubic.
Pre-calc test, huh?
Since the sine function is periodic over , this is equivalent to
You can use the Taylor approximation series for the sine function :
Apply this to your problem :
Let's do this like ten times. We get that :
And you have your answer
EDIT : damn I forgot you weren't allowed to use Taylor polynomials
Oh my god. I didn't see that four. It makes it easier though.
. We make two subsequent observations
2. Solving for "nice" answers gives .
Combining these two we see that
It is fairly easy to verify that
Setting gives us
We may clear out the extraneous since it is clearly not equal to zero. We are left then with an equation of the form
Thus, by the quadratic formula,
But, since is increasing we must have that where it follows that
Also, we know that
We may now then conclude that