1978 AHSME Problems/Problem 15
Revision as of 02:56, 13 February 2021 by Justinlee2017 (talk | contribs) (Created page with "==Solution== Squaring the equation, we get <cmath>\sin ^2 x + \cos ^2 x + 2 \sin x \cos x = \frac{1}{25} =\Rrightarrow 2\sin x \cos x = -\frac{24}{25} \Rrightarrow \sin x \cos...")
Solution
Squaring the equation, we get
![\[\sin ^2 x + \cos ^2 x + 2 \sin x \cos x = \frac{1}{25} =\Rrightarrow 2\sin x \cos x = -\frac{24}{25} \Rrightarrow \sin x \cos x = -\frac{12}{25}\]](http://latex.artofproblemsolving.com/7/b/1/7b1fe0d4477389dd2606f02659c08e63bbedcbbe.png)
Recall that 
, so
![\[(\sin x - \cos x)^2 = (\sin x + \cos x)^2 - 4 \sin x \cos x\]](http://latex.artofproblemsolving.com/c/5/f/c5fa134a4807a34a2c7d66a9e7bc1b812ade1bde.png)
![\[(\sin x - \cos x)^2 = \frac{1}{25} - 4 (- \frac{12}{25})\]](http://latex.artofproblemsolving.com/b/c/d/bcd276b8295fa79ba386e1b0633f50d87edb72b4.png)
![\[(\sin x - \cos x) = \frac{7}{5}\]](http://latex.artofproblemsolving.com/1/1/8/1187ddab8728f51f3b112ea3d2cac96d70be4fc0.png)
We can now solve for the values of 
and 
![\[\sin x + \cos x = \frac{1}{5}\]](http://latex.artofproblemsolving.com/2/1/2/21205e12e1631ca98e03493e275e714f909a2148.png)
![\[\sin x - \cos x = \frac{7}{5}\]](http://latex.artofproblemsolving.com/e/d/9/ed952606f3fea55ed8e87a6de4b4a7e0d7450170.png)
![\[2 \sin x = \frac{8}{5} \Rrightarrow \sin x = \frac{4}{5}, \cos x = -\frac{3}{5}\]](http://latex.artofproblemsolving.com/6/9/a/69a27d953e10f58d40249371ed57f892b21b021a.png)
Since 
, we have
![\[\tan x = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3}\]](http://latex.artofproblemsolving.com/7/d/1/7d1fc224a5744ebcf5df40031a04b63679368da9.png)
~JustinLee2017
