## 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)

1. alekos

cot10 = cos10/sin10 Are you allowed to use a calculator?

2. liamschumm

No, and I know the definition of cot.

3. liamschumm

I tried using product to sum, but I did not get very far.

4. alekos

It would not be easy lets start with sin30 = sin(20+10) = sin20cos10 + cos20sin10 = 2sin10cos10cos10 + [1-2sin^2(10)]sin10

5. liamschumm

Oh...

6. 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

7. liamschumm

Oh, and $sin^2 x =1 - cos^2 x.$

8. alekos

yes. i actually used cos^2x = 1 - sin^2x I'll leave it to you to solve the cubic

9. 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.

10. 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

11. liamschumm

Ok, thanks anyways!

12. anonymous

$\cot 10\cot 30\cot 50\cot 70=\cot 10*\frac{ \cos 30 }{ \sin 30 }\cot 50\cot 70$ $=\frac{ \frac{ \sqrt{3} }{ 2 } }{ \frac{ 1 }{ 2 } }\left( \cot 70\cot 10 \right)\cot 50$

13. liamschumm

Ok, gotten that far.

14. anonymous

$\frac{2 \cos 70\cos 10 }{ 2\sin 70\sin 10 }=$ $\cos \left( A-B \right)+\cos \left( A+B \right)=2\cos A \cos B$

15. anonymous

$\cos \left( A-B \right)-\cos \left( A+B \right)=2\sin A \sin B$

16. liamschumm

So we have $\frac{\cos 60 + \cos 80}{\cos 60 - \cos 80},$ right?

17. anonymous

$\frac{ 1+2\cos 80 }{ 1-2\cos 80 }*\frac{ \cos 50 }{ \sin 50 }$

18. alekos

where to now?

19. 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

20. liamschumm

Oh, got it.

21. liamschumm

Using cos60+cos80 / cos60−cos80, like I said above. Thanks all!

22. 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.

23. alekos

I can show you the proof of the identity if you wish

24. anonymous

$2\cos 80\cos 50=\cos \left( 80+50 \right)+\cos \left( 80-50 \right)=\cos 130+\cos 30$ $=\cos \left( 180-50 \right)+\frac{ \sqrt{3} }{ 2 }=-\cos 50+\frac{ \sqrt{3} }{ 2 }$

25. alekos

@surjithayer i don't know why you posted that?

26. 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.

27. alekos

but then the denominator comes out to be 2cos50 - sqrt3/2 doesnt get you anywhere

28. anonymous

$2 \cos 80\sin 50=\sin \left( 80+50 \right)-\sin \left( 80-50 \right)$ $=\sin 130-\sin 30=\sin \left( 180-50 \right)-\frac{ 1 }{ 2 }=\sin 50-\frac{ 1 }{ 2 }$ $\sin 50-2\cos 80\sin 50=\sin 50-\sin 50+\frac{ 1 }{ 2 }=\frac{ 1 }{ 2 }$

29. anonymous

$\sin \left( A+B \right)-\sin \left( A-B \right)=2 \cos A \sin B$

30. alekos

Yes. Thats a very sleek way of doing it and relies on the known trig identities. I prefer your method

31. anonymous

thank you