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Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$

From a question I asked before this, I have trouble actually with the numbers manipulating part.

Using trigo identity, $\sin^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{12} = 1$ so , $\cos^2 \frac{\pi}{12} = 1- \sin^2 \frac{\pi}{12}$

To find $\cos \frac{\pi}{12} = \sqrt{1- \sin^2 \frac{\pi}{12}}$

$\sin^2 \frac{\pi}{12} = (\frac{\sqrt{3} -1}{2 \sqrt{2}})^2 = \frac{(\sqrt{3}-1)^2}{(2\sqrt{2})^2} = \frac{2- \sqrt{3}}{4}$

$\cos \frac{\pi}{12} = \sqrt{1-(\frac{\sqrt{3} -1}{2 \sqrt{2}})^2} $

$\cos \frac{\pi}{12} = \sqrt{1- \frac{2-\sqrt{3}}{4}}$

$\cos \frac{\pi}{12} = \frac{\sqrt{2+\sqrt{3}}}{2}$

What is wrong with my steps?

user307640
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  • do me a favor, type in two more lines expanded out, $\sin^2 \frac{\pi}{12}$ and $\cos^2 \frac{\pi}{12}$ They add to one, they are also very similar in appearance. – Will Jagy Sep 26 '22 at 17:21
  • There is nothing wrong with your answer. There is however a simpler form, that does not involve taking square root of an expression involving a square root. – Andrei Sep 26 '22 at 17:25
  • Maybe this will help. $(\sqrt3-1)^2+ ?^2 = (2\sqrt 2)^2 = 8$. – Ted Shifrin Sep 26 '22 at 17:30
  • Your answer is correct. But you may also simplify like this:$\cos \frac{\pi}{12} = \frac{\sqrt{2+\sqrt{3}}}{2}=\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt2}=\frac{\sqrt{\sqrt 3^2+1+2\sqrt{3}}}{2\sqrt2}=\frac{\sqrt{(\sqrt 3+1)^2}}{2\sqrt2}$. – Koro Sep 26 '22 at 17:34

4 Answers4

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Another way would be to use the double angle formula

$$\sin 2\theta=2\sin\theta\cos\theta$$

With $\theta=\frac {\pi}{12}$, then

$$\sin 2\theta=\sin\frac {\pi}6=\frac 12$$

Thus

\begin{align*} \cos\frac {\pi}{12} & =\frac {\sin 2\theta}{2\sin\theta}\\ & =\frac 1{4\sin\theta}\\ & =\frac 1{\sqrt 2(\sqrt 3-1)} \end{align*}

Which is numerically equal to the answer you found. To get it into the form Wolfram Alpha produces, multiply the fraction by $\sqrt 3+1$.

Frank W
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Let $$\sqrt{2+\sqrt{3}}=\sqrt{x}+\sqrt{y}\implies x+y=2. xy=3/4 \implies x=3/2, y=1/2.$$ So $$\cos(\pi/12)=\frac{\sqrt{3}+1}{2\sqrt{2}}$$ OP is right.

MathDona
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Nothing wrong.

If you prefer, we can do some simplification.

Let me focus on $\sqrt{2+\sqrt3}$.

Let $$\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3}=x$$

$$2+\sqrt3+2-\sqrt3+2=x^2$$

Hence, we have $$\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3}=\sqrt6.$$

Similarly, we have $$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$

Hence $$\sqrt{2+\sqrt3}=\frac{\sqrt6+\sqrt2}{2}=\frac{\sqrt3+1}{\sqrt2}.$$

$$\cos \frac{\pi}{12}=\frac{\sqrt3+1}{2\sqrt2}$$

Siong Thye Goh
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You can also use $$\cos(x)+\sin(x)=\sqrt{2}\sin(x+\frac{\pi}4)$$

In this instance with $x=\frac{\pi}{12}$ you get to calculate $\sin(\frac{\pi}3)$ which is known.

The advantage is that you don't get nested root, nor have to rationalize $\sqrt{3}-1$ on denominator, just fraction addition.

zwim
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