Wikipedia:Reference desk/Archives/Mathematics/2013 February 10

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February 10

How to prove

How to prove that tan20⋅tan30=tan10⋅tan50 ? 117.227.40.234 (talk) 03:32, 10 February 2013 (UTC)

Are these in degrees ? If so, why not just multiply them out ? StuRat (talk) 03:34, 10 February 2013 (UTC)

Ya, they are in degrees. How I can I do it using tanA = cot(90-A)? 117.227.145.29 (talk) 03:47, 10 February 2013 (UTC)

Or some any other basic identities. 117.227.18.193 (talk) 06:06, 10 February 2013 (UTC)

((( My browser doesn't show the symbol used between the tans here, can someone say what it is please (* / + - ???) -- SGBailey (talk) 09:49, 10 February 2013 (UTC) )))

Hello, I think the symbol is meant to show multiplication, like latex's '\cdot' (in a hex editor it comes up as 0xB7 which is Unicode (UTF-16)'s 'MIDDLE DOT') so it's

$\tan(20)\cdot \tan(30)=\tan(10)\cdot \tan(50)$.77.86.3.26 (talk) 10:44, 10 February 2013 (UTC)

There may be simpler ways, but here's an outline of mine. Multiply both sides by cos 20 cos 30 cos 10 cos 50 so that the entire identity deals only with sin and cos. Now using the product-to-sum formulas at Trigonometric_identity#Product-to-sum_and_sum-to-product_identities, you can get rid of the 20's and 30's entirely and only be left with angles of 10 and 50. Using the double-angle formulas, you can then replace some of the 50's with 100's. The 100's can in turn be replaced with 10's using the formulas for shifting by 90. Now it's not too hard to prove the identity. You'll still need an angle addition formula, though. 96.46.195.35 (talk) 13:01, 10 February 2013 (UTC)

The identity $\tan(10)=\tan(20)\cdot \tan(30)\cdot \tan(40)$ is mentioned at Trigonometric identities#Identities without variables (though not with a clear explanation, unfortunately) and this identity is identical to the one above by recognising that $\tan(40)=\cot(90-40)=\cot(50)={\frac {1}{\tan(50)}}$. So, the identity the IP mentions is certainly true, provided that WP page is correct. EdChem (talk) 13:03, 10 February 2013 (UTC)

Yes, I just noticed that. See also Morrie's law, which relates directly to the OP's identity in the same way. 96.46.195.35 (talk) 13:08, 10 February 2013 (UTC)

How about this: we can write $\tan(30^{\circ })=1/{\sqrt {(}}3)$, then we have:

$\tan(20)/{\sqrt {(}}3)=\tan(10)\cdot \tan(50)$.

Using your "tanA = cot(90-A)" we have

$\tan(20)\cdot \tan(40)/{\sqrt {(}}3)=\tan(10)$.

From List_of_trigonometric_identities#Double-angle.2C_triple-angle.2C_and_half-angle_formulae we have:

$\tan(2A)={\frac {2\tan(A)}{1-\tan ^{2}(A)}}$, so we have:

$\tan(20)\cdot {\frac {\tan(20)}{1-\tan ^{2}(20)}}/{\sqrt {(}}3)=\tan(10)$.

Using the half-angle formula:

$\tan(A/2)={\frac {\tan(A)}{1+{\sqrt {1+\tan ^{2}(A)}}}}$ we have

$\tan(20)\cdot {\frac {\tan(20)}{1-\tan ^{2}(20)}}/{\sqrt {(}}3)={\frac {\tan(20)}{1+{\sqrt {1+\tan ^{2}(20)}}}}$.

What do people think?77.86.3.26 (talk) 13:15, 10 February 2013 (UTC)

You dropped a factor of 2 along the way. Also, although if you restore the 2 the calculations are correct, the argument is still incomplete. It's something special about the number tan(20) that makes the last equation true - for tan(21) it would be false.96.46.195.35 (talk) 14:42, 10 February 2013 (UTC)

Thanks for spotting my mistake. One possible way to go is by expanding the above into an equation for $\tan(20^{\circ })$ and noting that:

$\tan(3x)={\frac {3\tan x-\tan ^{3}x}{1-3\tan ^{2}x}}$, so

$\tan(60^{\circ })={\frac {3\tan(20^{\circ })-\tan ^{3}(20^{\circ })}{1-3\tan ^{2}(20^{\circ })}}={\sqrt {3}}$

and comparing? 77.86.3.26 (talk) 16:28, 10 February 2013 (UTC)

Sorry that's no help, one equation is quartic and the other cubic. 77.86.3.26 (talk) 16:35, 10 February 2013 (UTC)

Complete proof

Recalling the double angle formula for the sine function and rearranging, we can state that, for any angle α:

$\cos(\alpha )={\frac {\sin(2\alpha )}{2\sin(\alpha )}}$     and similarly that     $\cos(2\alpha )={\frac {\sin(4\alpha )}{2\sin(2\alpha )}}$     and     $\cos(4\alpha )={\frac {\sin(8\alpha )}{2\sin(4\alpha )}}$

On multiplying these together, we get that

$\cos(\alpha )\cdot \cos(2\alpha )\cdot \cos(4\alpha )={\frac {\sin(2\alpha )}{2\sin(\alpha )}}\cdot {\frac {\sin(4\alpha )}{2\sin(2\alpha )}}\cdot {\frac {\sin(8\alpha )}{2\sin(4\alpha )}}={\frac {1}{2\sin(\alpha )}}\cdot {\frac {1}{2}}\cdot {\frac {\sin(8\alpha )}{2}}={\frac {\sin(8\alpha )}{8\sin(\alpha )}}$

Setting α = 20°, we get

$\cos(20^{\circ })\cdot \cos(40^{\circ })\cdot \cos(80^{\circ })={\frac {\sin(160^{\circ })}{8\sin(20^{\circ })}}$

and since $\sin(20^{\circ })=\sin(180^{\circ }-20^{\circ })=\sin(160^{\circ })$

$\cos(20^{\circ })\cdot \cos(40^{\circ })\cdot \cos(80^{\circ })={\frac {1}{8}}$     . .     . .     . .     (1)

This is the so-called Morrie's law.

Now, recalling the sums-to-products formula for the sine function we can state that, for any angles α and β:

$\sin(\alpha )\cdot \sin(\beta )={\frac {\cos(\alpha -\beta )-\cos(\alpha +\beta )}{2}}$

Taking α = 80° and β = 40°:

$\sin(80^{\circ })\cdot \sin(40^{\circ })={\frac {\cos(80^{\circ }-40^{\circ })-\cos(80^{\circ }+40^{\circ })}{2}}={\frac {\cos(40^{\circ })-\cos(120^{\circ })}{2}}$

Then, nothing that   $\cos(120^{\circ })=-\cos(180^{\circ }-120^{\circ })=-\cos(60^{\circ })={\frac {-1}{2}}$:

$\sin(80^{\circ })\cdot \sin(40^{\circ })={\frac {\cos(40^{\circ })+{\frac {1}{2}}}{2}}$

On multiplying by sin(20°), we get:

$\sin(20^{\circ })\cdot \sin(40^{\circ })\cdot \sin(80^{\circ })=(\cos(40^{\circ })+{\frac {1}{2}})\cdot {\frac {\sin(20^{\circ })}{2}}={\frac {\cos(40^{\circ })\cdot \sin(20^{\circ })}{2}}+{\frac {\sin(20^{\circ })}{4}}$

Now, invoking the sums-to-products formula for the product of a sine and a cosine function, for any angles α and β:

$\sin(\alpha )\cdot \cos(\beta )={\frac {\sin(\alpha +\beta )+\sin(\alpha -\beta )}{2}}$

with α = 20° and β = 40°:

$\sin(20^{\circ })\cdot \cos(40^{\circ })={\frac {\sin(20^{\circ }+40^{\circ })+\sin(20^{\circ }-40^{\circ })}{2}}={\frac {\sin(60^{\circ })+\sin(-20^{\circ })}{2}}$

Now, as $\sin(60^{\circ })={\frac {\sqrt {3}}{2}}$     and     $\sin(-20^{\circ })=-\sin(20^{\circ })$

$\sin(20^{\circ })\cdot \cos(40^{\circ })={\frac {{\frac {\sqrt {3}}{2}}-\sin(20^{\circ })}{2}}$

and it follows that:

$\sin(20^{\circ })\cdot \sin(40^{\circ })\cdot \sin(80^{\circ })={\frac {\frac {{\frac {\sqrt {3}}{2}}-\sin(20^{\circ })}{2}}{2}}+{\frac {\sin(20^{\circ })}{4}}={\frac {\sqrt {3}}{8}}-{\frac {\sin(20^{\circ })}{4}}+{\frac {\sin(20^{\circ })}{4}}={\frac {\sqrt {3}}{8}}$     . .     . .     . .     (2)

Now, divind equation (2) by equation (1), we get:

${\begin{aligned}{\frac {\sin(20^{\circ })\cdot \sin(40^{\circ })\cdot \sin(80^{\circ })}{\cos(20^{\circ })\cdot \cos(40^{\circ })\cdot \cos(80^{\circ })}}={\frac {\frac {\sqrt {3}}{8}}{\frac {1}{8}}}\\\tan(20^{\circ })\cdot \tan(40^{\circ })\cdot \tan(80^{\circ })={\sqrt {3}}\\\end{aligned}}$

Recalling that one of the properties of the tangent function is that:

$\tan(90^{\circ }-\theta )=\cot \theta$     and applying it at θ = 50° and θ = 10° and recalling that     $\cot(30^{\circ })={\sqrt {3}}$     and substituting, we find:

$\tan(20^{\circ })\cdot \tan(90^{\circ }-50^{\circ })\cdot \tan(90^{\circ }-10^{\circ })=\cot(30^{\circ })$

$\tan(20^{\circ })\cdot \cot(50^{\circ })\cdot \cot(10^{\circ })=\cot(30^{\circ })$

$\tan(20^{\circ })\cdot {\frac {1}{\tan(50^{\circ })}}\cdot {\frac {1}{\tan(10^{\circ })}}={\frac {1}{\tan(30^{\circ })}}$

$\tan(20^{\circ })\cdot \tan(30^{\circ })=\tan(10^{\circ })\cdot \tan(50^{\circ })$

as required. EdChem (talk) 06:22, 11 February 2013 (UTC)

Alternate proof

Here is an alternate proof, loosely modelled on the outline above. We compute cos² 50 + tan 10 sin 50 cos 50 = (1 + cos 100)/2 + tan 10 (sin 100 /2 ) = (1 - sin 10)/2 + tan 10 (cos 10 / 2) = 1/2 - (sin 10)/2 + (sin 10)/2 = 1/2 = cos 60 = cos 10 cos 50 - sin 10 sin 50.

Dividing the result by cos 50, we find cos 50 + tan 10 sin 50 = cos 10 - sin 10 tan 50.

We rearrange to obtain cos 10 - cos 50 = tan 10 sin 50 + sin 10 tan 50 = tan 10 tan 50 cos 10 + tan 10 cos 10 tan 50 = tan 10 tan 50 (cos 10 + cos 50).

Now we divide by cos 10 + cos 50 and use the identity for a product of tangents to find tan 10 tan 50 = (cos 10 - cos 50)/(cos 10 + cos 50) = tan 20 tan 30. 64.140.122.50 (talk) 08:14, 12 February 2013 (UTC)

Extension

Above, an IP noted that the result is equivalent (after re-adding a lost 2) to the statement that:

$\tan(20)\cdot {\frac {2\tan(20)}{1-\tan ^{2}(20)}}/{\sqrt {3}}={\frac {\tan(20)}{1+{\sqrt {1+\tan ^{2}(20)}}}}$

which is the same as the polynominalin t:

${\frac {2t}{1-t^{2}}}={\frac {\sqrt {3}}{1+{\sqrt {1+t^{2}}}}}$

Solving this for t in the range 0 < t < 1 will give the exact value for tan 20°. EdChem (talk) 06:22, 11 February 2013 (UTC)

Letting t = tan 20, we have sqrt(3) = tan 60 = tan 3t = (t^3 - 3t)/(3t^2 - 1), hence t^3 -3sqrt(3)t^2 - 3t + sqrt(3) = 0.

Multiplying this equation by t^3 + 3sqrt(3)t^2 - 3t -sqrt(3), we find that t^6 - 33t^4 + 27t^2 - 3 = 0.

By Eisenstein's criterion, the polynomial on the left is irreducible over the rational numbers, so it is not possible to find a polynomial equation of lower degree with rational coefficients of which t is a root. 64.140.122.50 (talk) 08:13, 12 February 2013 (UTC)

Multiplying by a uniformly-distributed random variable

If I multiply random variable A by random variable B, and random variable B has a uniform distribution, will random variable A*B then have the same type of distribution as what random variable A had? Thorstein90 (talk) 22:29, 10 February 2013 (UTC)

No. I don't think it will ever be the same, and it can be very different, as for example if you multiply a narrow Gaussian r.v. by a broad uniform r.v. Looie496 (talk) 00:40, 11 February 2013 (UTC)

But what I mean is if you multiply a narrow Gaussian by a broad uniform, would it still be (maybe a broader) Gaussian? Thorstein90 (talk) 04:51, 11 February 2013 (UTC)

Clearly not, consider the Gaussian N(1,0). (That's the constant 1.) McKay (talk) 05:54, 11 February 2013 (UTC)

A zero-variance Gaussian? Thorstein90 (talk) 06:48, 11 February 2013 (UTC)

Yes, a zero-variance Gaussian is just a limiting case of a Gaussian as the variance goes to zero (it has all its probability mass at the mean). If something (such as preservation of the class of distributions) were true for all members of the class, it would be true in the limiting case as well. But here, with a unit-mean, zero-variance Gaussian (or any other distribution of unit mean and zero variance), the product with a uniform distribution is simply the same uniform distribution. Duoduoduo (talk) 15:20, 11 February 2013 (UTC)

See Normal distribution#Zero-variance limit. Duoduoduo (talk) 15:50, 11 February 2013 (UTC)

So how do you get the answer to be "yes, it will be random", the same as xoring by a random one time pad? You have a good source of randomness so is there some way to "xor" by it in the same way, to get an equally random result as the "OTP" what you're xoring by? 91.120.48.242 (talk) 07:42, 11 February 2013 (UTC)

My interpretation if 91.120's question is this: A one-time pad (OTP) is uniformly distributed over some fixed range of integers. Is there any random variable A, somehow distributed, that we could multiply by the OTP distribution to get another distribution that is also uniformly distributed? This is a different question from the original one, which wanted the product to be distributed like A rather than like the uniform. This question is also different because one-time pads are discretely distributed. My own answer is that I doubt it, but I'm not sure. Duoduoduo (talk) 15:32, 11 February 2013 (UTC)

What you probably want is (modular) addition, not multiplication. If A is a random variable in [0,1], and B is independent and uniformly distributed in [0,1], then $C=(A+B)\ \mod 1$ is uniformly distributed in [0,1], independent of A, and using B and C you can reconstruct A. This can be extended to (bounded) ranges which are not [0,1]. -- Meni Rosenfeld (talk) 13:44, 12 February 2013 (UTC)