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Prove that $\sin\left(\frac{\pi}{3}\right)+\cos\left(\frac{\pi}{4}\right)$ is algebraic.

Evaluating value of this, it is sum of two irrational numbers. How to find if it is algebraic?

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  • $\begingroup$ Are you familiar with the result that the sum and the product of two algebraic numbers are both algebraic? $\endgroup$ – Jyrki Lahtonen May 30 '18 at 12:12
  • $\begingroup$ yes, I know that $\endgroup$ – Epsilon Delta May 30 '18 at 12:13
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    $\begingroup$ So what is the problem? You do know the values of these trig functions? $\endgroup$ – Jyrki Lahtonen May 30 '18 at 12:15
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Hint:

Let $$s:=\sqrt3+\sqrt2.$$

Then $$s^2=5+2\sqrt6$$ and $$(s^2-5)^2=24.$$

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You can find a polynomial $p\in \mathbb{Q}[x]$ with that number as a root.

$\sin(\pi/3)=\frac{\sqrt{3}}{2}$

$\cos(\pi/4)=\frac{\sqrt{2}}{2}$

So the sum is $x=\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}$. To avoid denominators I'm going to consider $2x=\sqrt{3}+\sqrt{2}$. Then

$ 2x-\sqrt{3}=\sqrt{2}\Rightarrow (2x-\sqrt{3})^2=2\Rightarrow 4x^2+3-4\sqrt{3}=2\Rightarrow 4x^2+1=4\sqrt{3}\Rightarrow 16x^4+8x^2+1=48 $

Hence, a polynomial with this root is

$p(x)=16x^4+8x^2-47$.

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$$Sin(\pi /3)= \frac{\sqrt{3}}{2} = Root[4x^2-3==0] $$ $$ Cos(\pi /4)=\frac{\sqrt{2}}{2} = Root[2x^2==1] $$

So, both numbers are algebraic and sum of finte number of algebraic numbers is algebraic

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