Finding tan(pi/8) | How Many Ways Are There?

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27 сен 2024

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Комментарии : 18

@justabunga1 3 месяца назад

You can use the half angle formula if you rewrite tan(π/8)=tan((1/2)(π/4)). Since there are three tangent angle formulas, all three have the correct answer. The formula for those are tan(θ/2)=±√((1-cos(θ))/(1+cos(θ)))=sin(θ)/(1+cos(θ))=(1-cos(θ))/sin(θ). In order to get only one answer, we have to get rid of the ± sign. What we know is that π/8 is in the first quadrant, so we can reject the negative sign. To make things easier, I will go ahead and use the formula tan(π/8)=(1-cos(π/4))/sin(π/4)=(1-1/√(2))/(1/√(2)). We can clear the complex fraction by multiplying both the top and bottom by √(2), which is √(2)-1, or -1+√(2). If you want, you can also check the other 2 equations and will lead the same answer as before.

@NadiehFan 3 месяца назад

You don't need to check signs if you use the identity tan ½θ = sin θ/(1 + cos θ). We have sin θ/(1 + cos θ) = 2·sin(½θ)·cos(½θ)/(2·cos²(½θ)) = sin(½θ)/cos(½θ) = tan ½θ so the identity tan ½θ = sin θ/(1 + cos θ) is valid whenever cos θ ≠ −1, that is, θ ≠ (2k + 1)π, k ∈ ℤ, meaning that it is valid whenever tan ½θ is defined.

@wryanihad 3 месяца назад

4th method Tan(x)=sin(x)/[1+cos(x)] Put x=π/8 then tan(π/8)=sqr(2)-1

@roberttelarket4934 3 месяца назад

Not 2t since I hate tea. To coffee definitely.

@petrileskinen2988 3 месяца назад

My method of solving this might be a bit hard to explain with words but it is using the unit circle, a bit similar to your second method. First draw an angle of pi/4 in the first quadrant, so that it ends at the point (1/sqrt(2), 1/sqrt(2)). Next ,connect that point on the circle to the point (-1,0) remembering that the central angle is pi/4, and there for the inscribed angle at (-1,0) becomes a half of that: pi/8 e.g. what we were looking after. Therefore, looking at the angle at (-1,0), the opposite side has length 1/sqrt(2), and the adjacent side has lenght 1+1/sqrt(2). Finally we can calculate tan(pi/8) = (1/sqrt(2)) / (1+1/sqrt(2)) = ... = sqrt(2)-1 ... and this brings me to the end of my comment.

@NadiehFan 3 месяца назад

There is a diagram in the English Wikipedia article _Tangent half-angle substitution_ which illustrates exactly what you are doing here. Essentially, this boils down to using the identity tan ½θ = sin θ/(1 + cos θ) with the known values sin ¼π = cos ¼π = ½√2.

@AbouTaim-Lille 3 месяца назад

tg² Θ +1 = 1/cos² Θ So U need to calculate cos π/8. As you know cos2Θ = 2 cos² Θ -1. Thus cos² π/8 = ½(cos π/4 +1 ) = ½ (√2+1) So tg² π/8 = 2/(√2+1) - 1. The rest is just calculus.

@wasimvillidad3000 3 месяца назад

I often ask these two questions together in my mathematics classes - what is tan(π/8) and what is the mass of an 80 gsm A4 sheet of paper? If you know, you know.

@nicolascamargo8339 3 месяца назад

Genial

@anthonycheng1765 3 месяца назад

Use tan(x/2) = sin(x/2)/cos(x/2) = sqrt((1-cos(x))/2)/sqrt(1+cos(x))/2) = sqrt((1-cos(x))/(1+cos(x)), sqrt is positive as x is acute.

@LmaoDed-haha 3 месяца назад

Tan(x/2) = Sinx/(1+cosx) is way better .

@NadiehFan 3 месяца назад

@@LmaoDed-haha Exactly. No need to check signs here. We have sin θ/(1 + cos θ) = 2·sin(½θ)·cos(½θ)/(2·cos²(½θ)) = sin(½θ)/cos(½θ) = tan ½θ so the identity tan ½θ = sin θ/(1 + cos θ) is valid whenever cos θ ≠ −1, that is, θ ≠ (2k + 1)π, k ∈ ℤ, meaning that it is valid whenever tan ½θ is defined.

@NadiehFan 3 месяца назад

When at 10:23 you have the system a² − b² = 1 2ab = 1 you can use the identity (a² + b²)² = (a² − b²)² + 4a²b² to get a² + b² = √2 and from this and a² − b² = 1 we easily find a² = ½(√2 + 1) b² = ½(√2 − 1) This gives b²/a² = (√2 − 1)/(√2 + 1) = (√2 − 1)² and since tan ⅛π = b/a > 0 we immediately get tan ⅛π = √2 − 1. Fourth method: simply use the tangent half angle formula tan ½θ = sin θ/(1 + cos θ) and the known values sin ¼π = cos ¼π = ½√2 to get tan ⅛π = √2/(2 + √2) = √2·(2 − √2)/2 = (2√2 − 2)/2 = √2 − 1.

@RealQuInnMallory 3 месяца назад

{tanpi+e^xLoge^xtan8 ➖ }= { tanpi^4/tan64e^xLoge^xtanpi^4/tan64 } = {tan64/tanpie^xLoge^xtan64/tanpi^4}= {tanpi16/tanpi16e^xLoge^xtanpi16/tanpi16}= {tan4^2pi/tanpi4^2e^xLogtane^xpi4^2/tanpi^4^2}= {tanpi2^22^1/tanpi^2^2^2^1e^xLoge^xtanpi2^22^1/tanpi4^2}={ tanpi^1^1^1^/tanpi^1^1^1e^xLoge^xtanpi^1^1^1/tanpi4^2}= tanpi4^2.(tanpie^x ➖ 4/tanpi +2

@Chrisoikmath_ 3 месяца назад

2nd method is very smart and excellent!! You actually seldom use geometry to solve problems because you do not like it very much. My method: In triangle ABC you can take the bisector of B (let it be BD) and use the bisector theorem: AB/AC=AD/CD to find tan(π/8). 😊

@roberttelarket4934 3 месяца назад

Not 2t since I hate tea. To coffee definitely.