Finding tan(5) in terms of tan(19)

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #TrigonometricExpressions

EXPLORE 😎:

PLAYLISTS 🎵 :

[luggepytt]

When it comes to deriving formulas for the sine, cosine, or tangent of a multiple of an angle, I like to use complex arithmetic. Thus I can avoid the “gigantic monstrous expression” @5:59.

We have defined x = tan 19°
Now consider the complex number z = 1 + ix

If we look at the polar representation of z, we see that the angle, also known as Arg(z) = 19°, because tan 19° = x.

If we raise z to the 5th power, the angle will be multiplied by 5, in other words:
Arg(z⁵) = 5×19° = 95°

So let’s calculate z⁵ using the binomial formula:
z⁵ = (1 + ix)⁵
= 1 + 5ix + 10i²x² + 10i³x³ + 5i⁴x⁴ + i⁵x⁵
= 1 + 5ix - 10x² - 10ix³ + 5x⁴ + ix⁵
= 1 - 10x² + 5x⁴ + i(5x - 10x³ + x⁵)
= 5x⁴ - 10x² + 1 + i(x⁵ - 10x³ + 5x)

To get tan 95°, we simply divide the imaginary part of z by the real part.
tan 95° = (x⁵ - 10x³ + 5x) / (5x⁴ - 10x² + 1)

And, of course, we also get the general formula:
tan 5α = (tan⁵ α - 10 tan³ α + 5 tan α) / (5 tan⁴ α - 10 tan² α + 1)

[YossiSirote]

Theoretically you can get tan 5 exactly without using anything else. Tan 15 can be obtained as tan(45-30)= (sqrt(3)-1)^2 / 2. And you can use the same triple tan(3 z) you derived above. Problem is you get a cubic equation.

[WahranRai]

Another way could be to use multiple angle formula :
we have : tan(5*19) = tan(95) = -cotan(5) = -1/tan(5)
combining the expanding formula of tan(2x) and tan(3x) gives with tan(x) = t:
tan(5x) = (t^5 -10t^3 + 5t) / (5t^4 - 10t^2 + 1)
implies for x=19 and tan(19) = t
tan(95) = - 1/tan(5) = (t^5 -10t^3 + 5t) / (5t^4 - 10t^2 + 1)
tan(5) = - (5t^4 - 10t^2 + 1) / (t^5 -10t^3 + 5t)
Rem : the CAS Maxima online with trigexpand(tan(5*x)) will give you the expanded formula

[yurihung]

so the first part is really finding how to get 5 from playing around 19 or its multiples.

[txikitofandango]

I haven't watched this yet, but 19 * 5 - 5 = 90. I'm sure that could be useful

[math]

very well done bro, thanks for sharing

[vishalmishra]

Let x = tan(19) and y = tan(5). We need y as a function of x.
Let's use 19 x 5 = 95 = 90 + 5 and tan(90-t) = cot(t) = 1 / tan(t). Therefore, -1/y = 1/tan(-5) = cot(-5) = tan(90+5) = tan(19x5)
How to calculate tan(nT) ? Use (cosT + i sinT)^n and the binomial co-efficients from n'th row of Pascal's Triangle.
Tan(5T) = sin(5T) / cos(5T) and (C+iS)^5 = [ 1 5 10 10 5 1 ] dot [ even and odd terms = C^5 C^3S^2 CS^4 and C^4S C^2S^3 S^5] = ([5 -10 1] [T T^3 T^5]) / ([1 -10 5] [1 T^2 T^4 ])
Therefore, if T = tan(t), then tan(5t) = (5T - 10T^3 + T^5) / (1 - 10T^2 + 5T^4)
Hence, -1/y = tan(95) = (5x - 10x^3 + x^5) / (1 - 10x^2 + 5x^4)

[Pythagoriko]

Maybe this technique can be generalized. Having tan(x) and tan(y), I suppose it will depends from the lcm(x, y). The generalization can be an excellent argument for a next video.

[satyapalsingh]

Very nice problem . You are a great mathematician .

[klementhajrullaj]

How can find it the "sin 23°" without calculater? Thank you! ...

[michaelempeigne]

Note that the roots of the numerator are reciprocals of the roots of the denominator

[ManuelRuiz-xibt]

It's basically all about deriving the formula for tan(5a) = t(5 - 10 tt + tttt) / (1 - 10tt + 5tttt) with t = tan(a).

[manojsurya]

Cool. I first found tan 38 with double angle formula n since tan(38)=cot(52), n get tan 52 then with triple formula, we get tan 57, so after that I evaluated 5 as 57-52, and surprisingly I didn't mess up in the algebra lol😆

[txikitofandango]

Okay yeah, my way worked although it wasn't pretty.

tan(95) = tan(5x) = (5x - 10x^3 + x^5)/(1 - 10x^2 + 5x^4) through repeated tan addition formula. Call this rational function r(x).

Then, tan(90) = tan(95-5) = (r(x) + tan 5) / (1 - r(x)tan5) = infinity

Then 1 - r(x)tan5 = 0
tan 5 = 1/r(x) = (1 - 10x^2 + 5x^4) / (5x - 10x^3 + x^5)

I know it's not pretty

[ashmitsarkar]

We can find tan (5 theta) and then proceed by putting theta = 19° and then we get -cot 5° and then tan 5°

[doontz]

In general, given a sin/cos/tan of one angle, can you find the sin/cos/tan of any other angle?

And if so, how can you prove it?

[spiderwings]

tan5 = tan(arctan(x)-14). something hard next time please
(/s)

[fredbiper]

That is interesting. If You express tg5 as tg(19-14) where tg14 =1/4. Is another way to solve this problem, is not?