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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)
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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alekos
 one year ago
Best ResponseYou've already chosen the best response.1cot10 = cos10/sin10 Are you allowed to use a calculator?

liamschumm
 one year ago
Best ResponseYou've already chosen the best response.1No, and I know the definition of cot.

liamschumm
 one year ago
Best ResponseYou've already chosen the best response.1I tried using product to sum, but I did not get very far.

alekos
 one year ago
Best ResponseYou've already chosen the best response.1It would not be easy lets start with sin30 = sin(20+10) = sin20cos10 + cos20sin10 = 2sin10cos10cos10 + [12sin^2(10)]sin10

alekos
 one year ago
Best ResponseYou've already chosen the best response.1eventually 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
 one year ago
Best ResponseYou've already chosen the best response.1Oh, and \[sin^2 x =1  cos^2 x.\]

alekos
 one year ago
Best ResponseYou've already chosen the best response.1yes. i actually used cos^2x = 1  sin^2x I'll leave it to you to solve the cubic

liamschumm
 one year ago
Best ResponseYou've already chosen the best response.1Hmm... 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
 one year ago
Best ResponseYou've already chosen the best response.1Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.1Ok, thanks anyways!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cot 10\cot 30\cot 50\cot 70=\cot 10*\frac{ \cos 30 }{ \sin 30 }\cot 50\cot 70\] \[=\frac{ \frac{ \sqrt{3} }{ 2 } }{ \frac{ 1 }{ 2 } }\left( \cot 70\cot 10 \right)\cot 50\]

liamschumm
 one year ago
Best ResponseYou've already chosen the best response.1Ok, gotten that far.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{2 \cos 70\cos 10 }{ 2\sin 70\sin 10 }=\] \[\cos \left( AB \right)+\cos \left( A+B \right)=2\cos A \cos B\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos \left( AB \right)\cos \left( A+B \right)=2\sin A \sin B\]

liamschumm
 one year ago
Best ResponseYou've already chosen the best response.1So we have \[\frac{\cos 60 + \cos 80}{\cos 60  \cos 80},\] right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1+2\cos 80 }{ 12\cos 80 }*\frac{ \cos 50 }{ \sin 50 }\]

alekos
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1Using cos60+cos80 / cos60−cos80, like I said above. Thanks all!

alekos
 one year ago
Best ResponseYou've already chosen the best response.1We use a little known identity tanx*tan(60x)*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
 one year ago
Best ResponseYou've already chosen the best response.1I can show you the proof of the identity if you wish

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[2\cos 80\cos 50=\cos \left( 80+50 \right)+\cos \left( 8050 \right)=\cos 130+\cos 30\] \[=\cos \left( 18050 \right)+\frac{ \sqrt{3} }{ 2 }=\cos 50+\frac{ \sqrt{3} }{ 2 }\]

alekos
 one year ago
Best ResponseYou've already chosen the best response.1@surjithayer i don't know why you posted that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos 50+2\cos 80\cos 50=\cos 50\cos 50+\frac{ \sqrt{3} }{ 2 }=\frac{ \sqrt{3} }{ 2 }\] i have solved the numerator . Similarly you can solve denominator.

alekos
 one year ago
Best ResponseYou've already chosen the best response.1but then the denominator comes out to be 2cos50  sqrt3/2 doesnt get you anywhere

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[2 \cos 80\sin 50=\sin \left( 80+50 \right)\sin \left( 8050 \right)\] \[=\sin 130\sin 30=\sin \left( 18050 \right)\frac{ 1 }{ 2 }=\sin 50\frac{ 1 }{ 2 }\] \[\sin 502\cos 80\sin 50=\sin 50\sin 50+\frac{ 1 }{ 2 }=\frac{ 1 }{ 2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin \left( A+B \right)\sin \left( AB \right)=2 \cos A \sin B\]

alekos
 one year ago
Best ResponseYou've already chosen the best response.1Yes. Thats a very sleek way of doing it and relies on the known trig identities. I prefer your method
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