liamschumm
  • liamschumm
How do I find \[\cot10^\circ\cot30^\circ\cot50^\circ\cot70^\circ\]? (please give me small hints, do not do the problem for me)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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alekos
  • alekos
cot10 = cos10/sin10 Are you allowed to use a calculator?
liamschumm
  • liamschumm
No, and I know the definition of cot.
liamschumm
  • liamschumm
I tried using product to sum, but I did not get very far.

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More answers

alekos
  • alekos
It would not be easy lets start with sin30 = sin(20+10) = sin20cos10 + cos20sin10 = 2sin10cos10cos10 + [1-2sin^2(10)]sin10
liamschumm
  • liamschumm
Oh...
alekos
  • alekos
eventually we would get sin30 = 3sin10 - 4sin^3(10) which we would need to solve by making x = sin10 so we get 3x - 4x^3 = 1/2
liamschumm
  • liamschumm
Oh, and \[sin^2 x =1 - cos^2 x.\]
alekos
  • alekos
yes. i actually used cos^2x = 1 - sin^2x I'll leave it to you to solve the cubic
liamschumm
  • liamschumm
Hmm... I do not think this is how this problem is supposed to be solved in my class... We have been covering formulas and manipulations of trig functions.
alekos
  • alekos
Yeah you're probably right. that's the only thing I can think of at the moment. There must be a more efficient way of doing this. let me get back to you
liamschumm
  • liamschumm
Ok, thanks anyways!
anonymous
  • anonymous
\[\cot 10\cot 30\cot 50\cot 70=\cot 10*\frac{ \cos 30 }{ \sin 30 }\cot 50\cot 70\] \[=\frac{ \frac{ \sqrt{3} }{ 2 } }{ \frac{ 1 }{ 2 } }\left( \cot 70\cot 10 \right)\cot 50\]
liamschumm
  • liamschumm
Ok, gotten that far.
anonymous
  • anonymous
\[\frac{2 \cos 70\cos 10 }{ 2\sin 70\sin 10 }=\] \[\cos \left( A-B \right)+\cos \left( A+B \right)=2\cos A \cos B\]
anonymous
  • anonymous
\[\cos \left( A-B \right)-\cos \left( A+B \right)=2\sin A \sin B\]
liamschumm
  • liamschumm
So we have \[\frac{\cos 60 + \cos 80}{\cos 60 - \cos 80},\] right?
anonymous
  • anonymous
\[\frac{ 1+2\cos 80 }{ 1-2\cos 80 }*\frac{ \cos 50 }{ \sin 50 }\]
alekos
  • alekos
where to now?
alekos
  • alekos
I can't seem to find any way to simplify using trig identities, so we might have to go back to determining the value of each term
liamschumm
  • liamschumm
Oh, got it.
liamschumm
  • liamschumm
Using cos60+cos80 / cos60−cos80, like I said above. Thanks all!
alekos
  • alekos
We use a little known identity tanx*tan(60-x)*tan(60+x) = tan(3x) cot(10)*cot(30)*cot(50)*cot(70) = tan(80)*tan(60)*tan(40)*tan(20). Now tan(60) = sqrt(3), so we get = sqrt(3) * tan(20)*tan(40)*tan(80). Now use the identity tan(x)*tan(60 - x)*tan(60 + x) = tan(3x). and with x = 20 degrees we have tan(20)*tan(40)*tan(80) = tan(3*20) = tan(60) = sqrt(3), and so = sqrt(3)*sqrt(3) = 3.
alekos
  • alekos
I can show you the proof of the identity if you wish
anonymous
  • anonymous
\[2\cos 80\cos 50=\cos \left( 80+50 \right)+\cos \left( 80-50 \right)=\cos 130+\cos 30\] \[=\cos \left( 180-50 \right)+\frac{ \sqrt{3} }{ 2 }=-\cos 50+\frac{ \sqrt{3} }{ 2 }\]
alekos
  • alekos
@surjithayer i don't know why you posted that?
anonymous
  • anonymous
\[\cos 50+2\cos 80\cos 50=\cos 50-\cos 50+\frac{ \sqrt{3} }{ 2 }=\frac{ \sqrt{3} }{ 2 }\] i have solved the numerator . Similarly you can solve denominator.
alekos
  • alekos
but then the denominator comes out to be 2cos50 - sqrt3/2 doesnt get you anywhere
anonymous
  • anonymous
\[2 \cos 80\sin 50=\sin \left( 80+50 \right)-\sin \left( 80-50 \right)\] \[=\sin 130-\sin 30=\sin \left( 180-50 \right)-\frac{ 1 }{ 2 }=\sin 50-\frac{ 1 }{ 2 }\] \[\sin 50-2\cos 80\sin 50=\sin 50-\sin 50+\frac{ 1 }{ 2 }=\frac{ 1 }{ 2 }\]
anonymous
  • anonymous
\[\sin \left( A+B \right)-\sin \left( A-B \right)=2 \cos A \sin B\]
alekos
  • alekos
Yes. Thats a very sleek way of doing it and relies on the known trig identities. I prefer your method
anonymous
  • anonymous
thank you

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