sempteim245

2022-03-31

I have to simplify and evaluate this :
${\mathrm{cos}70}^{\circ }+4{\mathrm{cos}70}^{\circ }$

zalutaloj9a0f

The easy way :
$\frac{{\mathrm{cos}70}^{\circ }+4{\mathrm{cos}70}^{\circ }{\mathrm{sin}70}^{\circ }}{{\mathrm{sin}70}^{\circ }}=\frac{{\mathrm{sin}20}^{\circ }+2{\mathrm{sin}40}^{\circ }}{{\mathrm{cos}20}^{\circ }}$
$=\frac{{\mathrm{sin}20}^{\circ }+{\mathrm{sin}40}^{\circ }+{\mathrm{cos}50}^{\circ }}{{\mathrm{cos}20}^{\circ }}$
Sum to product formula gives :
$=\frac{{\mathrm{cos}10}^{\circ }+{\mathrm{cos}50}^{\circ }}{{\mathrm{cos}20}^{\circ }}$
Again using the formula gives $2{\mathrm{cos}30}^{\circ }=\sqrt{3}\approx 1.732$

Leonardo Mcpherson

First $70=90-20$
We can express all in terms of $\mathrm{cos}\left(20\right)$ and use that $\frac{1}{2}=\mathrm{cos}\left(60\right)=4{\mathrm{cos}}^{3}\left(20\right)-3\mathrm{cos}\left(20\right)$
Let's write $x=\mathrm{cos}\left(20\right),y=\mathrm{sin}\left(20\right)$ to write less.
So, $4{x}^{3}-3x-\frac{1}{2}=0$
We square your expression such that we don't have to write radicals, but we can go without it too if we wanted.
${\left(\mathrm{cot}\left(70\right)+4\mathrm{cos}\left(70\right)\right)}^{2}={\left(\frac{\mathrm{cos}\left(90-20\right)+4\mathrm{cos}\left(90-20\right)\mathrm{sin}\left(90-20\right)}{\mathrm{sin}\left(90-20\right)}\right)}^{2}$
$={\left(\frac{y+4xy}{x}\right)}^{2}$
$={y}^{2}\frac{16{x}^{2}+8x+1}{{x}^{2}}$
But
$\frac{1}{x}=8{x}^{2}-6$
$=\left(1-{x}^{2}\right)\left(16{x}^{2}+8x+1\right){\left(8{x}^{2}-6\right)}^{2}$
$=-1024{x}^{8}-512{x}^{7}+2496{x}^{6}+1280{x}^{5}-1952{x}^{4}-1056{x}^{3}+444{x}^{2}+288x+36$
Now divide this polynomial by $4{x}^{3}-3x-\frac{1}{2}$, which is zero.
$=\left(-256{x}^{5}-128{x}^{4}+432{x}^{3}+192{x}^{2}-180x-66\right)\cdot \left(4{x}^{3}-3x-\frac{1}{2}\right)+3$
The remainder gives you the value 3.

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