Trig identity pls help

Dec 2019
2
1
Asunción
Hello forum. This is my first post here. I need help with solving this identity. The exercise states as follows.
Verify the following identity:

8cos10cos20cos40=cot10

I really don't know what to do nor where to start from. Any help?
Thanks to everyone in advance!
 
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skipjack

Forum Staff
Dec 2006
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In the following, all angles are in degrees.

2sin(10)cos(40) = sin(50) - sin(30) = cos(40) - 1/2

2sin(10)cos(20) = sin(30) - sin(10) = 1/2 - sin(10)
4sin(10)cos(20)cos(40) = cos(40) - 2sin(10)cos(40) = 1/2

8cos(10)cos(20)cos(40) = cot(10)
 

mathman

Forum Staff
May 2007
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Step 1: 8sin(10)cos(20)cos(40)=1. $\cos(20)=\cos^2(10)-\sin^2(10)$ and $\cos(40)=\cos^2(20)-\sin^2(20)$.
Continue to get everything in terms of $10^\circ$.
 
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skipjack

Forum Staff
Dec 2006
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. . . get everything in terms of $10^\circ$.
One gets $8\sin(10^\circ)(1 - 2\sin^2(10^\circ))(8\sin^4(10^\circ) - 8\sin^2(10^\circ) + 1)$.
How would you show that this equals 1?
 

skipjack

Forum Staff
Dec 2006
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Without using an electronic calculator?

Another approach is to prove a useful identity and then use it.

$\begin{align*}4\cos(\theta)\cos(\theta + 60^\circ)\cos(\theta - 60^\circ) &= 2\cos(\theta)(\cos(2\theta) + \cos(120^\circ)) \\
&= 2\cos(\theta)(2\cos^2(\theta) - 3/2) \\
&= 4\cos^3(\theta) - 3\cos(\theta) \\
&= \cos(3\theta) \end{align*}$

For $\theta = 20^\circ$, $4\cos(20^\circ)\cos(80^\circ)\cos(40^\circ) = \cos(60^\circ) = 1/2$,
and.multiplying both sides of that by $2\cot(10^\circ)$ gives the original equation to be verified.