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Glossary |
Starting from the equations:
e = 2d2 - 1 and d = -2e2 + 1.
if we add them together, we obtain:
e + d = 2d2 - 1 + -2e2 + 1 = 2(d2 - e2) = 2(d - e)(d + e)
This implies that, if e + d ¹ 0, we can divide e + d from both sides to get 1 = 2(d - e), so that e = d - 1/2. This means that 2d2 - 1 = e = d - 1/2, or 2d2 - d - 1/2 = 0, which we can solve by the quadratic formula to obtain d = , and e = d - 1/2 = .
On the other hand, if e + d = 0, so that e = -d, then 2d2 - 1 = e = -d or 2d2 + d - 1 = 0. This has solutions d = -1 and 1/2, with e = -d = 1, -1/2. In general, we have the four pairs of solutions:
d
= -1 and e = 1
d = 1/2 and e = -1/2
d =
and e =
Go back to The Algebra of Trigonometric Functions
| Table of Contents | Send questions or
comments to jwicks@northpark.edu |
Glossary |