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I tried to compute the value of $\sin 75^\circ$ using the sine of standard values $(30^\circ, 45^\circ...)$ and did it by two ways. One, by expanding $\sin (45^\circ+30^\circ)$ and the other by computing the half of $\sin 150^\circ$ using basic identities. It gave me these two answers respectively: $$\frac{\sqrt{3}+1}{2\sqrt{2}}\ ,\ \ \ \frac{\sqrt{2+\sqrt{3}}} {2}$$ When I first did it, I was worried that I had got one of them wrong as I couldn't think of a way to show them equal to each other. I evaluated them in the calculator and indeed, they are equal (and to $\sin 75^\circ$) which made me think of how does one show expressions like these to be equal.

So is there any way one could show that these two expressions are equal to each other? Thanks.

Alraxite
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4 Answers4

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You want to show that $$\frac{\sqrt{3}+1}{2\sqrt{2}} = \frac{\sqrt{2+\sqrt{3}}} {2}.$$ Rearrange this into $$2(\sqrt{3} + 1) = 2\sqrt{2}\sqrt{2+\sqrt{3}}.$$ Since both sides are evidently positive, this is equivalent to $$(2(\sqrt{3} + 1))^2 = (2\sqrt{2}\sqrt{2+\sqrt{3}})^2,$$ which simplifies to $$4(4 + 2\sqrt{3}) = 8(2+\sqrt{3}),$$ which is of course true.

HJSprime
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  • Thanks! I have just one doubt in the line "Since both sides are evidently positive...". If both sides were negative, one could still square them and proceed with their argument so writing that was redundant, right? – Alraxite Oct 17 '12 at 13:06
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    @Alraxite Nope, it is a necessary line. For, otherwise, you could conclude $-4 = 4$ from the fact that $(-4)^2 = 4^2$. More generally, $x^2 = y^2$ implies $x = \pm y$; to conclude $x=y$ from $x^2=y^2$, you need that $x$ and $y$ have the same sign. – HJSprime Oct 17 '12 at 13:08
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    Oh, I see. It's kind of there to maintain the "if and only if" part of the proof. If I read your proof backwards, that line does seem necessary. – Alraxite Oct 17 '12 at 13:15
  • Yep, thats exactly right. – HJSprime Oct 17 '12 at 13:16
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$$ \frac{\sqrt{2+\sqrt{3}}}{2}=\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}=\frac{\sqrt{(\sqrt{3})^2+2\sqrt{3}+1}}{2\sqrt{2}}=\frac{\sqrt{(\sqrt{3}+1)^2}}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}} $$

blindman
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$$\frac{\sqrt{3}+1}{2\sqrt{2}}=x$$

$$\left[\frac{\sqrt{3}+1}{2\sqrt{2}}\right]^2=x^2$$

$$\frac{4+2\sqrt{3}}{8}=x^2$$

$$4+2\sqrt{3}=8x^2$$

$$x=\sqrt{\frac{4+2\sqrt{3}}{8}}$$

$$x=\frac{\sqrt{2+\sqrt{3}}} {2}$$

DonAntonio
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Salech Alhasov
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$$ \frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{\sqrt{2+\sqrt{3}}} {2}\\ \frac{(\sqrt{3}+1)^2}{2}={2+\sqrt{3}} \\ 3+2\sqrt3 +1 =2({2+\sqrt{3}}) \\ 4+2\sqrt3 =4+2\sqrt{3} \\ $$

draks ...
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