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Evaluate $\cos 36 ^\circ - \cos 72 ^\circ$

Here is my method:

First, $\cos 72 ^\circ$ can be written as $\cos^2 36 ^\circ - \sin^2 36 ^\circ$ via the double angle formula. Then to get rid of the $\sin^2 36^\circ$ I used the Pythagorean Identity and said $\sin^2 36 ^\circ = 1 - \cos^2 36 ^\circ$.

So then our equation becomes $-2 \cos^2 36 ^\circ + \cos 36 ^\circ +1$ (After a little algebra). I noticed this was a quadratic and therefore I set $\cos 36 ^\circ = y$ and solved for $y$ where I got roots $\frac {1} {2}, -1$. $\frac {1} {2}$ is right answer to the equation but since I was solving for $y$ I believe it shouldn't be correct. Can someone help explain where I got it wrong and how to solve the equation using my method.

José Carlos Santos
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Pichoosee_1.9035
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  • Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. – José Carlos Santos Feb 01 '22 at 08:12
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    "solved for y" $;-;$ What exactly did you solve for $,y,$? What you have there is an expression that you need to evaluate, not an equation. – dxiv Feb 01 '22 at 08:13
  • @JoséCarlosSantos put the title in for you. – Pichoosee_1.9035 Feb 01 '22 at 08:13
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    @Pichoosee_1.9035 That is not an equation, an equation is of the form $a=b$. For example, $x^2+2$ is not an equation, $x^2+2=4$ is. You cannot solve $x^2+2$, but you can solve (for $x$) $x^2+2=4$. – Gary Feb 01 '22 at 08:14
  • @Gary however then should I denote the answer as $t$ or some other letter? Just asking. – Pichoosee_1.9035 Feb 01 '22 at 08:20
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    The idea is that you should put the question in the body for everyone, rather than just for me. – José Carlos Santos Feb 01 '22 at 08:21
  • @Pichoosee_1.9035 You showed that $\cos (36^ \circ ) - \cos (72^ \circ )$ is the same as $- 2\cos ^2 (36^ \circ ) + \cos (36^ \circ ) + 1$. Now this latter thing is a number, you have to tell what number it is. – Gary Feb 01 '22 at 08:22
  • @Gary Yes but since I do not know what $\cos 36 ^\circ$ is off the top of my head I need to solve for it. – Pichoosee_1.9035 Feb 01 '22 at 08:23
  • @Pichoosee_1.9035 Solve for it what? There is no equation here. – Gary Feb 01 '22 at 08:25
  • @Pichoosee_1.9035 Right, but for that you first need to find an equation that $\cos 36^\circ$ satisfies, which you can then solve to determine it. The special thing about $36^\circ$ is that it's $180^\circ / 5$, so you have to use that in some way - see the duplicate question and the other one linked from it for several possible approaches. – dxiv Feb 01 '22 at 08:26

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