Question

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

Answer 1

cot10 = cos10/sin10

Are you allowed to use a calculator?

Answer 2

No, and I know the definition of cot.

Answer 3

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

Answer 4

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

Answer 5

Oh...

Answer 6

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

Answer 7

Oh, and \[sin^2 x =1 - cos^2 x.\]

Answer 8

yes. i actually used cos^2x = 1 - sin^2x

I'll leave it to you to solve the cubic

Answer 9

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.

Answer 10

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

Answer 11

Ok, thanks anyways!

Answer 12

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

Answer 13

Ok, gotten that far.

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

where to now?

Answer 19

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

Answer 20

Oh, got it.

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

but then the denominator comes out to be 2cos50 - sqrt3/2

doesnt get you anywhere

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

thank you