Difference between revisions of "1981 AHSME Problems/Problem 24"
Math piggy (talk | contribs) m (→Solution) |
Math piggy (talk | contribs) m (→Solution) |
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Because <math>\cos</math> is even and <math>\sin</math> is odd: | Because <math>\cos</math> is even and <math>\sin</math> is odd: | ||
− | + | \begin{align*} | |
− | + | &=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta) \\ | |
− | + | &=\boxed{\textbf{2\cos(n\theta)}}, | |
+ | \end{align*} | ||
which gives the answer <math>\boxed{\textbf{D}}.</math> | which gives the answer <math>\boxed{\textbf{D}}.</math> |
Revision as of 15:20, 6 July 2021
Problem
If is a constant such that
and
, then for each positive integer
,
equals
Solution
Multiply both sides by and rearrange to
. Using the quadratic equation, we can solve for
. After some simplifying:
Substituting this expression in to the desired gives:
Using DeMoivre's Theorem:
Because is even and
is odd:
\begin{align*}
&=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta) \\
&=\boxed{\textbf{2\cos(n\theta)}},
\end{align*}
which gives the answer