Prove the following:
- .
- .
- .
- Proof. We use the triple angle identity of the sine function,
Then, since , we have,
For the cosine part, we use the triple angle identity for the cosine (the second version in the exercise),
Thus,
In solving both of these cubics, we used that and which we know since we determined all real for which sine and cosine are zero
- Here we use part (a) and the co-relations (Thm 2.3, part d),
Similarly,
- Proof. First, by the double angle identity for sine,
And by the double angle identity for the cosine,
Using the second equation first, we have,
Then, using this value in the first equation,
Thus,
I think in proof a) you missed the explanation on how you discarded the -1 solution to the equation.
sin (π/6) ≠ -1 .
since in the previous problem [2-c] we proved that
sin(x) = -1 for x = 3π/2 + 2nπ
and (π/6) ≠ 3π/2 + 2nπ for any n∈Z.