Difference between revisions of "2024 AMC 12A Problems/Problem 23"
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<cmath>=(\tan x+\tan (\frac{\pi}{2}-x))^2-2</cmath> | <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 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 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{\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{1}{\cos x \sin x}\right)^2-2</cmath> | ||
<cmath>=\left(\frac{2}{\sin 2x}\right)^2-2</cmath> | <cmath>=\left(\frac{2}{\sin 2x}\right)^2-2</cmath> | ||
| Line 28: | Line 28: | ||
<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> | <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> | ||
| − | <cmath>=(\frac{4}{\sin^2 \frac{\pi}{8}}-2)(\frac{4}{\sin^2 \frac{3\pi}{8}}-2)</cmath> | + | <cmath>=\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)</cmath> |
Note that | Note that | ||
| − | <cmath>\sin^2 \frac{\pi}{8}=\frac{1-\cos frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}</cmath> | + | <cmath>\sin^2 \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}</cmath> |
and | and | ||
| − | <cmath>\sin^2 \frac{3\pi}{8}=\frac{1-\cos frac{3\pi}{4}}{2}\frac{2+\sqrt{2}}{4}</cmath> | + | <cmath>\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\pi}{4}}{2}=\frac{2+\sqrt{2}}{4}</cmath> |
Hence, | Hence, | ||
| − | <cmath>(\frac{4}{\sin^2 \frac{\pi}{8}}-2)(\frac{4}{\sin^2 \frac{3\pi}{8}}-2)</cmath> | + | <cmath>\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)</cmath> |
| − | <cmath>=(\frac{16}{2-\sqrt{2}}-2)(\frac{16}{2+\sqrt{2}}-2)</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>=(14+8\sqrt{2})(14-8\sqrt{2})</cmath> | ||
| Line 48: | Line 48: | ||
<cmath>=68</cmath> | <cmath>=68</cmath> | ||
| − | Therefore, the answer is \fbox{\textbf{(B) } 68}. | + | Therefore, the answer is <math>\fbox{\textbf{(B) } 68}</math>. |
~tsun26 | ~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 19:52, 8 November 2024
Problem
What is the value of ![\[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?\]](http://latex.artofproblemsolving.com/2/d/2/2d248ce9756f0f09265dd538de3b16b22a944c46.png)
Solution 1 (Trigonometric Identities)
First, notice that
![\[=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})*(\tan^2\frac{3\pi}{16}+\tan^2 \frac{4\pi}{16})\]](http://latex.artofproblemsolving.com/1/5/3/15311b0a684668146fdf061ff3a07f0f84196cbb.png)
Here, we make use of the fact that
![\[\tan^2 x+\tan^2 (\frac{\pi}{2}-x)\]](http://latex.artofproblemsolving.com/d/9/c/d9cc31bca74f57bf2838b0b6598bc1f00dff7571.png)
![\[=(\tan x+\tan (\frac{\pi}{2}-x))^2-2\]](http://latex.artofproblemsolving.com/b/6/7/b6793d18f913669880bf136ac07b85de29437588.png)
![\[=\left(\frac{\sin x}{\cos x}+\frac{\sin (\frac{\pi}{2}-x)}{\cos (\frac{\pi}{2}-x)}\right)^2-2\]](http://latex.artofproblemsolving.com/7/9/3/7933b5b30ffa9897d781738ab273c23ef1bb28c8.png)
![\[=\left(\frac{\sin x \cos (\frac{\pi}{2}-x)+\sin (\frac{\pi}{2}-x) \cos x}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\]](http://latex.artofproblemsolving.com/1/b/2/1b2030fb90cbf1e3c35d71c2058af8d5ba22dff7.png)
![\[=\left(\frac{\sin \frac{\pi}{2}}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\]](http://latex.artofproblemsolving.com/c/f/7/cf70707a0104f210582ad6d9d6c6102e9d9c65a3.png)
![\[=\left(\frac{1}{\cos x \sin x}\right)^2-2\]](http://latex.artofproblemsolving.com/7/0/9/709c6b08ef0737e8b6bea68c03278d9c9c9c1f5b.png)
![\[=\left(\frac{2}{\sin 2x}\right)^2-2\]](http://latex.artofproblemsolving.com/6/7/a/67a0346eedaba73caf1bd1565dade45e9f18032f.png)
![\[=\frac{4}{\sin^2 2x}-2\]](http://latex.artofproblemsolving.com/6/6/c/66cf0347f5720950827b3ccc7539a5cd1ff39a18.png)
Hence,
![\[(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})*(\tan^2\frac{3\pi}{16}+\tan^2 \frac{4\pi}{16})\]](http://latex.artofproblemsolving.com/e/7/b/e7b1453ec17c755e9e1a165ca1d7e88d40001437.png)
![\[=\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\]](http://latex.artofproblemsolving.com/d/e/b/debb6db041d8ae12049f00e2a1e502bcdf1fd554.png)
Note that
![\[\sin^2 \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}\]](http://latex.artofproblemsolving.com/c/0/d/c0d937d7c96c598596abf879a2858dfe03279c78.png)
and
![\[\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\pi}{4}}{2}=\frac{2+\sqrt{2}}{4}\]](http://latex.artofproblemsolving.com/a/9/0/a90264ad7a0531c9c468da3648ae7f189b84c632.png)
Hence,
![\[\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\]](http://latex.artofproblemsolving.com/c/6/4/c64ea214245a92d0b7e7ed3ec29d9a7b9a4f4b4e.png)
![\[=\left(\frac{16}{2-\sqrt{2}}-2\right)\left(\frac{16}{2+\sqrt{2}}-2\right)\]](http://latex.artofproblemsolving.com/0/4/4/044f7e8a8a39685da59a37f351e39327160c09b5.png)
![\[=(14+8\sqrt{2})(14-8\sqrt{2})\]](http://latex.artofproblemsolving.com/f/2/c/f2c6a5cb2a1fb1fde4bb562119d2e31685b8a08e.png)
![\[=68\]](http://latex.artofproblemsolving.com/5/a/d/5ad6af812b7642452ebc0775651efbcd868b54aa.png)
Therefore, the answer is 
.
~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
