problem stringlengths 17 4.31k | solution stringlengths 68 6.77k | answer stringlengths 0 159 |
|---|---|---|
Find the ordered pair $(a,b)$ of integers such that
\[\sqrt{9 - 8 \sin 50^\circ} = a + b \csc 50^\circ.\] | We write
\[9 - 8 \sin 50^\circ = \frac{9 \sin^2 50^\circ - 8 \sin^3 50^\circ}{\sin^2 50^\circ} = \frac{9 \sin^2 50^\circ - 6 \sin 50^\circ + 6 \sin 50^\circ - 8 \sin^3 50^\circ}{\sin^2 50^\circ}.\]By the triple angle identity,
\begin{align*}
6 \sin 50^\circ - 8 \sin^3 50^\circ &= 2 \sin (3 \cdot 50^\circ) \\
&= 2 \sin ... | (3,-1) |
Find the smallest positive integer $k$ such that $
z^{10} + z^9 + z^6+z^5+z^4+z+1
$ divides $z^k-1$. | First, we factor the given polynomial. The polynomial has almost all the powers of $z$ from 1 to $z^6,$ which we can fill in by adding and subtracting $z^2$ and $z^3.$ This allows us to factor as follows:
\begin{align*}
z^{10} + z^9 + z^6 + z^5 + z^4 + z + 1 &= (z^{10} - z^3) + (z^9 - z^2) + (z^6 + z^5 + z^4 + z^3 + ... | 84 |
If
\[\sin x + \cos x + \tan x + \cot x + \sec x + \csc x = 7,\]then find $\sin 2x.$ | Expressing everything in terms of $\sin x$ and $\cos x,$ we get
\[\sin x + \cos x + \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} + \frac{1}{\sin x} + \frac{1}{\cos x} = 7.\]Then
\[\sin x + \cos x + \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} + \frac{\sin x + \cos x}{\sin x \cos x} = 7,\]which becomes
\[\sin x + \cos x +... | 22 - 8 \sqrt{7} |
Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$, where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$. | By the product-to-sum formula,
\[\sin \frac{1}{2} \sin k = \frac{1}{2} \left[ \cos \left( k - \frac{1}{2} \right) - \cos \left( k + \frac{1}{2} \right) \right].\]Thus, we can make the sum in the problem telescope:
\begin{align*}
a_n &= \sum_{k = 1}^n \sin k \\
&= \sum_{k = 1}^n \frac{\sin \frac{1}{2} \sin k}{\sin \frac... | 628 |
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