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liamschumm

  • one year ago

How do I find \[\cot10^\circ\cot30^\circ\cot50^\circ\cot70^\circ\]? (please give me small hints, do not do the problem for me)

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  1. alekos
    • one year ago
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    cot10 = cos10/sin10 Are you allowed to use a calculator?

  2. liamschumm
    • one year ago
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    No, and I know the definition of cot.

  3. liamschumm
    • one year ago
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    I tried using product to sum, but I did not get very far.

  4. alekos
    • one year ago
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    It would not be easy lets start with sin30 = sin(20+10) = sin20cos10 + cos20sin10 = 2sin10cos10cos10 + [1-2sin^2(10)]sin10

  5. liamschumm
    • one year ago
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    Oh...

  6. alekos
    • one year ago
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    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

  7. liamschumm
    • one year ago
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    Oh, and \[sin^2 x =1 - cos^2 x.\]

  8. alekos
    • one year ago
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    yes. i actually used cos^2x = 1 - sin^2x I'll leave it to you to solve the cubic

  9. liamschumm
    • one year ago
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    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.

  10. alekos
    • one year ago
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    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

  11. liamschumm
    • one year ago
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    Ok, thanks anyways!

  12. anonymous
    • one year ago
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    \[\cot 10\cot 30\cot 50\cot 70=\cot 10*\frac{ \cos 30 }{ \sin 30 }\cot 50\cot 70\] \[=\frac{ \frac{ \sqrt{3} }{ 2 } }{ \frac{ 1 }{ 2 } }\left( \cot 70\cot 10 \right)\cot 50\]

  13. liamschumm
    • one year ago
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    Ok, gotten that far.

  14. anonymous
    • one year ago
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    \[\frac{2 \cos 70\cos 10 }{ 2\sin 70\sin 10 }=\] \[\cos \left( A-B \right)+\cos \left( A+B \right)=2\cos A \cos B\]

  15. anonymous
    • one year ago
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    \[\cos \left( A-B \right)-\cos \left( A+B \right)=2\sin A \sin B\]

  16. liamschumm
    • one year ago
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    So we have \[\frac{\cos 60 + \cos 80}{\cos 60 - \cos 80},\] right?

  17. anonymous
    • one year ago
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    \[\frac{ 1+2\cos 80 }{ 1-2\cos 80 }*\frac{ \cos 50 }{ \sin 50 }\]

  18. alekos
    • one year ago
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    where to now?

  19. alekos
    • one year ago
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    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

  20. liamschumm
    • one year ago
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    Oh, got it.

  21. liamschumm
    • one year ago
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    Using cos60+cos80 / cos60−cos80, like I said above. Thanks all!

  22. alekos
    • one year ago
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    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.

  23. alekos
    • one year ago
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    I can show you the proof of the identity if you wish

  24. anonymous
    • one year ago
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    \[2\cos 80\cos 50=\cos \left( 80+50 \right)+\cos \left( 80-50 \right)=\cos 130+\cos 30\] \[=\cos \left( 180-50 \right)+\frac{ \sqrt{3} }{ 2 }=-\cos 50+\frac{ \sqrt{3} }{ 2 }\]

  25. alekos
    • one year ago
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    @surjithayer i don't know why you posted that?

  26. anonymous
    • one year ago
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    \[\cos 50+2\cos 80\cos 50=\cos 50-\cos 50+\frac{ \sqrt{3} }{ 2 }=\frac{ \sqrt{3} }{ 2 }\] i have solved the numerator . Similarly you can solve denominator.

  27. alekos
    • one year ago
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    but then the denominator comes out to be 2cos50 - sqrt3/2 doesnt get you anywhere

  28. anonymous
    • one year ago
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    \[2 \cos 80\sin 50=\sin \left( 80+50 \right)-\sin \left( 80-50 \right)\] \[=\sin 130-\sin 30=\sin \left( 180-50 \right)-\frac{ 1 }{ 2 }=\sin 50-\frac{ 1 }{ 2 }\] \[\sin 50-2\cos 80\sin 50=\sin 50-\sin 50+\frac{ 1 }{ 2 }=\frac{ 1 }{ 2 }\]

  29. anonymous
    • one year ago
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    \[\sin \left( A+B \right)-\sin \left( A-B \right)=2 \cos A \sin B\]

  30. alekos
    • one year ago
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    Yes. Thats a very sleek way of doing it and relies on the known trig identities. I prefer your method

  31. anonymous
    • one year ago
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    thank you

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