Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2024 AMC 12A Problems/Problem 23"

(→‎Solution 1 (Trigonometric Identities))
(→‎Solution 1 (Trigonometric Identities))
Line 4: Line 4:
 
<math>\textbf{(A) } 28 \qquad \textbf{(B) } 68 \qquad \textbf{(C) } 70 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 84</math>
 
<math>\textbf{(A) } 28 \qquad \textbf{(B) } 68 \qquad \textbf{(C) } 70 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 84</math>
  
−
==Solution 1 (Trigonometric Identities)==
+
==Solution 1 (Trigonometric Identities)==
  
 
First, notice that
 
First, notice that
−
\begin{align*}
 
−
&\tan^2 \frac{\pi}{16} \cdot \tan^2 \frac{3\pi}{16} + \tan^2 \frac{\pi}{16} \cdot \tan^2 \frac{5\pi}{16} + \tan^2 \frac{3\pi}{16} \cdot \tan^2 \frac{7\pi}{16} + \tan^2 \frac{5\pi}{16} \cdot \tan^2 \frac{7\pi}{16} \\
 
−
&= \left( \tan^2 \frac{\pi}{16} + \tan^2 \frac{7\pi}{16} \right) \cdot \left( \tan^2 \frac{3\pi}{16} + \tan^2 \frac{5\pi}{16} \right)
 
−
\end{align*}
 
  
−
Here, we use the identity
+
<cmath>\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?</cmath>
−
\begin{align*}
+
 
−
\tan^2 x + \tan^2 \left( \frac{\pi}{2} - x \right) &= \left( \tan x + \tan \left( \frac{\pi}{2} - x \right) \right)^2 - 2 \\
+
 
−
&= \left( \frac{\sin x}{\cos x} + \frac{\sin \left( \frac{\pi}{2} - x \right)}{\cos \left( \frac{\pi}{2} - x \right)} \right)^2 - 2 \\
+
<cmath>=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})*(\tan^2\frac{3\pi}{16}+\tan^2 \frac{4\pi}{16})</cmath>
−
&= \left( \frac{\sin x \cos \left( \frac{\pi}{2} - x \right) + \sin \left( \frac{\pi}{2} - x \right) \cos x}{\cos x \cos \left( \frac{\pi}{2} - x \right)} \right)^2 - 2 \\
+
 
−
&= \left( \frac{\sin \frac{\pi}{2}}{\cos x \cos \left( \frac{\pi}{2} - x \right)} \right)^2 - 2 \\
+
 
−
&= \left( \frac{1}{\cos x \sin x} \right)^2 - 2 \\
+
Here, we make use of the fact that
−
&= \left( \frac{2}{\sin 2x} \right)^2 - 2 \\
+
 
−
&= \frac{4}{\sin^2 2x} - 2
+
<cmath>\tan^2 x+\tan^2 (\frac{\pi}{2}-x)</cmath>
−
\end{align*}
+
<cmath>=(\tan x+\tan (\frac{\pi}{2}-x))^2-2</cmath>
 +
<cmath>=\left(\frac{\sin x}{\cos x}+\frac{\sin (\frac{\pi}{2}-x)}{\cos (\frac{\pi}{2}-x)}\right)^2-2</cmath>
 +
<cmath>=\left(\frac{\sin x \cos (\frac{\pi}{2}-x)+\sin (\frac{\pi}{2}-x) \cos x}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2</cmath>
 +
<cmath>=\left(\frac{\sin \frac{\pi}{2}}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2</cmath>
 +
<cmath>=\left(\frac{1}{\cos x \sin x}\right)^2-2</cmath>
 +
<cmath>=\left(\frac{2}{\sin 2x}\right)^2-2</cmath>
 +
<cmath>=\frac{4}{\sin^2 2x}-2</cmath>
  
 
Hence,
 
Hence,
−
\begin{align*}
 
−
\left( \tan^2 \frac{\pi}{16} + \tan^2 \frac{7\pi}{16} \right) \cdot \left( \tan^2 \frac{3\pi}{16} + \tan^2 \frac{5\pi}{16} \right) &= \left( \frac{4}{\sin^2 \frac{\pi}{8}} - 2 \right) \left( \frac{4}{\sin^2 \frac{3\pi}{8}} - 2 \right)
 
−
\end{align*}
 
  
−
Note that
+
<cmath>(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})*(\tan^2\frac{3\pi}{16}+\tan^2 \frac{4\pi}{16})</cmath>
−
\begin{align*}
+
<cmath>=\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)</cmath>
−
\sin^2 \frac{\pi}{8} &= \frac{1 - \cos \frac{\pi}{4}}{2} = \frac{2 - \sqrt{2}}{4} \\
+
 
−
\sin^2 \frac{3\pi}{8} &= \frac{1 - \cos \frac{3\pi}{4}}{2} = \frac{2 + \sqrt{2}}{4}
+
Note that
−
\end{align*}
+
 
 +
<cmath>\sin^2 \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}</cmath>
 +
 
 +
and
 +
 
 +
<cmath>\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\pi}{4}}{2}=\frac{2+\sqrt{2}}{4}</cmath>
 +
 
 +
Hence,
 +
 
 +
<cmath>\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)</cmath>
 +
 
 +
<cmath>=\left(\frac{16}{2-\sqrt{2}}-2\right)\left(\frac{16}{2+\sqrt{2}}-2\right)</cmath>
 +
 
 +
<cmath>=(14+8\sqrt{2})(14-8\sqrt{2})</cmath>
 +
 
 +
<cmath>=68</cmath>
  
−
Thus,
+
Therefore, the answer is <math>\fbox{\textbf{(B) } 68}</math>.
−
\begin{align*}
 
−
\left( \frac{4}{\sin^2 \frac{\pi}{8}} - 2 \right) \left( \frac{4}{\sin^2 \frac{3\pi}{8}} - 2 \right) &= \left( \frac{16}{2 - \sqrt{2}} - 2 \right) \left( \frac{16}{2 + \sqrt{2}} - 2 \right) \\
 
−
&= (14 + 8\sqrt{2})(14 - 8\sqrt{2}) \\
 
−
&= 68
 
−
\end{align*}
 
  
−
Therefore, the answer is \(\boxed{\textbf{(B) } 68}\).
+
~tsun26
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=22|num-a=24}}
 
{{AMC12 box|year=2024|ab=A|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:03, 8 November 2024

Problem

What is the value of \[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?\]

$\textbf{(A) } 28 \qquad \textbf{(B) } 68 \qquad \textbf{(C) } 70 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 84$ 

==Solution 1 (Trigonometric Identities)==

First, notice that

\[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?\]


\[=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})*(\tan^2\frac{3\pi}{16}+\tan^2 \frac{4\pi}{16})\]


Here, we make use of the fact that

\[\tan^2 x+\tan^2 (\frac{\pi}{2}-x)\] \[=(\tan x+\tan (\frac{\pi}{2}-x))^2-2\] \[=\left(\frac{\sin x}{\cos x}+\frac{\sin (\frac{\pi}{2}-x)}{\cos (\frac{\pi}{2}-x)}\right)^2-2\] \[=\left(\frac{\sin x \cos (\frac{\pi}{2}-x)+\sin (\frac{\pi}{2}-x) \cos x}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\] \[=\left(\frac{\sin \frac{\pi}{2}}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\] \[=\left(\frac{1}{\cos x \sin x}\right)^2-2\] \[=\left(\frac{2}{\sin 2x}\right)^2-2\] \[=\frac{4}{\sin^2 2x}-2\]

Hence,

\[(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})*(\tan^2\frac{3\pi}{16}+\tan^2 \frac{4\pi}{16})\] \[=\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\]

Note that

\[\sin^2 \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}\]

and

\[\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\pi}{4}}{2}=\frac{2+\sqrt{2}}{4}\]

Hence,

\[\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\]

\[=\left(\frac{16}{2-\sqrt{2}}-2\right)\left(\frac{16}{2+\sqrt{2}}-2\right)\]

\[=(14+8\sqrt{2})(14-8\sqrt{2})\]

\[=68\]

Therefore, the answer is $\fbox{\textbf{(B) } 68}$.

~tsun26

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12A_Problems/Problem_23&oldid=232261"