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I found the value of $\cos{15}^\circ$ using 2 methods.

Method 1: using $\cos{(a-b)} = \cos{a}\cos{b}+\sin{a}\sin{b}$

$\cos{(45-30)^\circ} = \cos{45^\circ}\cos{30^\circ}+\sin{45^\circ}\sin{30^\circ}$

$\cos{15^\circ} = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\frac{1}{2}$

$\cos{15^\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

Method 2: using $\cos{2\alpha} = 2\cos^2{\alpha}-1$

$\cos{(2_\dot{}15)^\circ} = 2\cos^2{15^\circ}-1$

$\cos{30^\circ} = 2\cos^2{15^\circ}-1$

substituting $\cos{15^\circ}$ with $x$ and $\cos{30^\circ}$ with $\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2} = 2x^2-1$

$2x^2-1-\frac{\sqrt{3}}{2} = 0$

solving for $x$, we get $x = \frac{\pm\sqrt{\sqrt{3}+2}}{2}$ and therefore $\cos{15^\circ} = \frac{\pm\sqrt{\sqrt{3}+2}}{2}$

Note: Only the positive value out of the two values we have got can be the correct value of $\cos{15\circ}$ since $\cos{15\circ}$ cannot be a negative value. So going forward I shall only refer to the positive value obtained in Method 2 as the solution from Method 2.

Now my question is: Are both answers equal? If yes, why do we even get these two answers?

The reason I ask equal and not correct is because the universal answer everywhere on the internet and in books is that $\cos{15^\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

So I am confused as to how you can get completely different-looking answers. Now a quick use of the calculator reveals that the decimal approximation of both answers is the same i.e. 0.96592... . So if that is the case then why do we get such different-looking solutions?

Prithvidiamond
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To see that $$\sqrt {\sqrt 3 +2}=\frac {\sqrt 3 +1}{\sqrt 2}$$

simply square both sides.

Since both terms are positive we see that our desired equality is equivalent to $$\sqrt 3+ 2 =\frac {3+1+2\sqrt 3}{2}=2+\sqrt 3$$

and we are done.

As a general remark, algebraic relations can be very complicated and unintuitive. It is certainly a good idea to compare numerical expressions, as you did, as sanity checks.

lulu
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  • And right here is the reason why I fail all my math exams! I know how to equate and square stuff, but I didn't do it. WHY?! WHY AM I SO DUMB?!

    Anyways, sorry for the digression, but yeah thanks! But apart from just mentioning that "algebraic relations can be very complicated and unintuitive", is there a definitive reason why the two forms look so different? Because I couldn't find a way to convert one solution into the other and equivalent values should always be inter-convertible from one form to the other, right? Is it just not possible because of the inconvenient square root?

     – Prithvidiamond Mar 20 '21 at 14:19
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    @Prithvidiamond They are indeed interconvertible, you just need to be a bit ingenious by noting that $2+\sqrt 3 = \frac 12 (4+2\sqrt 3) = \frac 12 (\sqrt 3 +1)^2$. – Vishu Mar 20 '21 at 14:27
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    These things take practice. Your instinct, to check the values numerically, is the right one. Once you see that they match numerically you can be quite confident that no blunder was made and that they are really the same...then you can figure out how to prove it. There are, I should say, weirder equalities than this one. See, for instance, Nested Radicals – lulu Mar 20 '21 at 14:27
  • @Tavish Yeah, I knew they had to be interconvertible, just that as you mentioned, I didn't know how to do it. Thanks for showing me how to. – Prithvidiamond Mar 20 '21 at 14:30
  • @lulu Ok then. Also, Thanks for the link. Hopefully, a read of it will get me better at this stuff. – Prithvidiamond Mar 20 '21 at 14:32
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    @Prithvidiamond You’re welcome. – Vishu Mar 20 '21 at 19:34