I have to simplify and evaluate this :
$$\cot 70^\circ+4\cos 70^\circ$$
On evaluating it, the answer comes out to be $1.732$, or $\sqrt 3$ .
I tried to get everything in $\sin$ and $\cos$, but it doesn't go any further. Any hints?
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what are you trying to do exactly? – danimal Mar 23 '15 at 14:51
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6to show that the square of this equals 3 exactly. – Empy2 Mar 23 '15 at 14:52
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Related : http://math.stackexchange.com/questions/10661/find-the-value-of-displaystyle-sqrt3-cdot-cot-20-circ-4-cdot-cos – lab bhattacharjee Mar 24 '15 at 16:12
4 Answers
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The easy way :
$$\dfrac{\cos70^\circ+4\cos70^\circ\sin70^\circ}{\sin70^\circ} = \dfrac{\sin20^\circ+2\sin40^\circ}{\cos20^\circ} = \dfrac{\sin20^\circ+\sin40^\circ+\cos50^\circ}{\cos20^\circ} $$
Sum to product formula gives :
$$\dfrac{\cos10^\circ+\cos50^\circ}{\cos20^\circ} $$
Again using the formula gives $2\cos30^\circ=\sqrt3\approx1.732$
lab bhattacharjee
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antman
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First $70=90-20$
We can express all in terms of $\cos(20)$ and use that $\frac{1}{2}=\cos(60)=4\cos^3(20)-3\cos(20)$.
Let's write $x=\cos(20), y=\sin(20)$ to write less.
So, $4x^3-3x-\frac{1}{2}=0$.
We square your expression such that we don't have to write radicals, but we can go without it too if we wanted.
$$\begin{align} \left(\cot(70)+4\cos(70)\right)^2&=\left(\frac{\cos(90-20)+4\cos(90-20)\sin(90-20)}{\sin(90-20)}\right)^2\\ &=\left(\frac{y+4xy}{x}\right)^2\\ &=y^2\frac{16x^2+8x+1}{x^2}\\&=??\end{align}$$
But
$$\frac{1}{x}=8x^2-6.$$
$$\begin{align}??&=(1-x^2)(16x^2+8x+1)(8x^2-6)^2\\&=-1024 x^8-512 x^7+2496 x^6+1280 x^5-1952 x^4-1056 x^3+444 x^2+288 x+36\end{align}$$
Now divide this polynomial by $4x^3-3x-1/2$, which is zero.
$$??=(-256 x^5-128 x^4+432 x^3+192 x^2-180 x-66)\cdot(4 x^3-3 x-1/2) + 3$$
The remainder gives you the value $3$.
We could have worked too with $3\cdot(70)=270-60$ to get smaller numbers in the coefficients, but well, I already wrote it with $20$. You can try this technique with $\cos(70)$ directly to check that you understood how it works.
Nathanson
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Great method . Only one that works without prior knowledge of the answer . But I dont understand why cos(70) would have been easier ? I am getting a similar expression(x is swapped with y) for "??" – antman Mar 23 '15 at 16:36
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2@antman It is just my guess. I didn't try it. Another thing you can do to get smaller polynomials is to use the long division before expanding the whole multiplication. In any case these are only cosmetic changes. What is important is to reduce to a single algebraic unknown and use its minimal polynomial and long division to reduce the expression and finally compute it. – Nathanson Mar 23 '15 at 16:51
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Multiply by $\sin70$ and you want to show $$\cos70+4\cos70\sin70=\sqrt{3}\sin70=2\sin60\sin70$$
Empy2
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1Do you know formulas for $\sin\alpha\sin\beta$, and $\sin(a+b)$? You should use those formulas. I think it works out, but I would not have found this method without knowing your answer $\sqrt{3}$. – Empy2 Mar 23 '15 at 15:10
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Like Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $,
If $2\sin3x=-1$
$$\cot x+4\cos x=4\cos x-2\sin3x\cot x$$
$$=\dfrac{4\cos x\sin x-2\sin3x\cos x}{\sin x}$$
$$=\dfrac{2\sin2x-(\sin2x+\sin4x)}{\sin x}$$
$$=-2\cos3x\text{ as }\sin x\ne0$$
Here $x=70^\circ$
Generalization:
$$\sin3x=-\dfrac12\implies3x=180^\circ n+(-1)^n(-30^\circ)\text{ where $n$ is any integer}$$
$\implies x=60^\circ n+(-1)^n(-10^\circ)$ where $n\equiv-1,0,1\pmod3$
$n=-1\implies x=-50^\circ$
$n=0\implies x=-10^\circ$
$n=1\implies x=70^\circ$
lab bhattacharjee
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