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Popular Trigonometry >

tan^2(a)=((2tan(a)))/((1-tan^2(a)))

  • Pre Algebra
  • Algebra
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Solution

tan2(a)=(1−tan2(a))(2tan(a))​

Solution

a=πn,a=−0.98930…+πn
+1
Degrees
a=0∘+180∘n,a=−56.68315…∘+180∘n
Solution steps
tan2(a)=(1−tan2(a))(2tan(a))​
Solve by substitution
tan2(a)=1−tan2(a)2tan(a)​
Let: tan(a)=uu2=1−u22u​
u2=1−u22u​:u=0,u≈−1.52137…
u2=1−u22u​
Multiply both sides by 1−u2
u2=1−u22u​
Multiply both sides by 1−u2u2(1−u2)=1−u22u​(1−u2)
Simplifyu2(1−u2)=2u
u2(1−u2)=2u
Solve u2(1−u2)=2u:u=0,u≈−1.52137…
u2(1−u2)=2u
Move 2uto the left side
u2(1−u2)=2u
Subtract 2u from both sidesu2(1−u2)−2u=2u−2u
Simplifyu2(1−u2)−2u=0
u2(1−u2)−2u=0
Factor u2(1−u2)−2u:−u(u3−u+2)
u2(1−u2)−2u
Apply exponent rule: ab+c=abacu2=uu=uu(−uu+1)−2u
Factor out common term u=u(u(−u2+1)−2)
Factor u(−u2+1)−2:−(u3−u+2)
u(−u2+1)−2
u(−u2+1)=−u(u+1)(u−1)
u(−u2+1)
Factor −u2+1:−(u+1)(u−1)
−u2+1
Factor out common term −1=−(u2−1)
Factor u2−1:(u+1)(u−1)
u2−1
Rewrite 1 as 12=u2−12
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)u2−12=(u+1)(u−1)=(u+1)(u−1)
=−(u+1)(u−1)
=−u(u+1)(u−1)
=−u(u+1)(u−1)−2
Expand −u(u+1)(u−1)−2:−u3+u−2
−u(u+1)(u−1)−2
Expand −u(u+1)(u−1):−u3+u
Expand (u+1)(u−1):u2−1
(u+1)(u−1)
Apply Difference of Two Squares Formula: (a+b)(a−b)=a2−b2a=u,b=1=u2−12
Apply rule 1a=112=1=u2−1
=−u(u2−1)
Expand −u(u2−1):−u3+u
−u(u2−1)
Apply the distributive law: a(b−c)=ab−aca=−u,b=u2,c=1=−uu2−(−u)⋅1
Apply minus-plus rules−(−a)=a=−u2u+1⋅u
Simplify −u2u+1⋅u:−u3+u
−u2u+1⋅u
u2u=u3
u2u
Apply exponent rule: ab⋅ac=ab+cu2u=u2+1=u2+1
Add the numbers: 2+1=3=u3
1⋅u=u
1⋅u
Multiply: 1⋅u=u=u
=−u3+u
=−u3+u
=−u3+u
=−u3+u−2
=−u3+u−2
Factor −u3+u−2:−(u3−u+2)
−u3+u−2
Factor out common term −1=−(u3−u+2)
=−(u3−u+2)
=u(−(u3−u+2))
Refine=−u(u3−u+2)
−u(u3−u+2)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u=0oru3−u+2=0
Solve u3−u+2=0:u≈−1.52137…
u3−u+2=0
Find one solution for u3−u+2=0 using Newton-Raphson:u≈−1.52137…
u3−u+2=0
Newton-Raphson Approximation Definition
f(u)=u3−u+2
Find f′(u):3u2−1
dud​(u3−u+2)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u3)−dudu​+dud​(2)
dud​(u3)=3u2
dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=3u3−1
Simplify=3u2
dudu​=1
dudu​
Apply the common derivative: dudu​=1=1
dud​(2)=0
dud​(2)
Derivative of a constant: dxd​(a)=0=0
=3u2−1+0
Simplify=3u2−1
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=−2:Δu1​=1
f(u0​)=(−1)3−(−1)+2=2f′(u0​)=3(−1)2−1=2u1​=−2
Δu1​=∣−2−(−1)∣=1Δu1​=1
u2​=−1.63636…:Δu2​=0.36363…
f(u1​)=(−2)3−(−2)+2=−4f′(u1​)=3(−2)2−1=11u2​=−1.63636…
Δu2​=∣−1.63636…−(−2)∣=0.36363…Δu2​=0.36363…
u3​=−1.53039…:Δu3​=0.10597…
f(u2​)=(−1.63636…)3−(−1.63636…)+2=−0.74530…f′(u2​)=3(−1.63636…)2−1=7.03305…u3​=−1.53039…
Δu3​=∣−1.53039…−(−1.63636…)∣=0.10597…Δu3​=0.10597…
u4​=−1.52144…:Δu4​=0.00895…
f(u3​)=(−1.53039…)3−(−1.53039…)+2=−0.05393…f′(u3​)=3(−1.53039…)2−1=6.02629…u4​=−1.52144…
Δu4​=∣−1.52144…−(−1.53039…)∣=0.00895…Δu4​=0.00895…
u5​=−1.52137…:Δu5​=0.00006…
f(u4​)=(−1.52144…)3−(−1.52144…)+2=−0.00036…f′(u4​)=3(−1.52144…)2−1=5.94435…u5​=−1.52137…
Δu5​=∣−1.52137…−(−1.52144…)∣=0.00006…Δu5​=0.00006…
u6​=−1.52137…:Δu6​=2.92858E−9
f(u5​)=(−1.52137…)3−(−1.52137…)+2=−1.74069E−8f′(u5​)=3(−1.52137…)2−1=5.94378…u6​=−1.52137…
Δu6​=∣−1.52137…−(−1.52137…)∣=2.92858E−9Δu6​=2.92858E−9
u≈−1.52137…
Apply long division:u+1.52137…u3−u+2​=u2−1.52137…u+1.31459…
u2−1.52137…u+1.31459…≈0
Find one solution for u2−1.52137…u+1.31459…=0 using Newton-Raphson:No Solution for u∈R
u2−1.52137…u+1.31459…=0
Newton-Raphson Approximation Definition
f(u)=u2−1.52137…u+1.31459…
Find f′(u):2u−1.52137…
dud​(u2−1.52137…u+1.31459…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=dud​(u2)−dud​(1.52137…u)+dud​(1.31459…)
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(1.52137…u)=1.52137…
dud​(1.52137…u)
Take the constant out: (a⋅f)′=a⋅f′=1.52137…dudu​
Apply the common derivative: dudu​=1=1.52137…⋅1
Simplify=1.52137…
dud​(1.31459…)=0
dud​(1.31459…)
Derivative of a constant: dxd​(a)=0=0
=2u−1.52137…+0
Simplify=2u−1.52137…
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=−0.65729…:Δu1​=1.65729…
f(u0​)=12−1.52137…⋅1+1.31459…=0.79321…f′(u0​)=2⋅1−1.52137…=0.47862…u1​=−0.65729…
Δu1​=∣−0.65729…−1∣=1.65729…Δu1​=1.65729…
u2​=0.31119…:Δu2​=0.96849…
f(u1​)=(−0.65729…)2−1.52137…(−0.65729…)+1.31459…=2.74663…f′(u1​)=2(−0.65729…)−1.52137…=−2.83597…u2​=0.31119…
Δu2​=∣0.31119…−(−0.65729…)∣=0.96849…Δu2​=0.96849…
u3​=1.35459…:Δu3​=1.04339…
f(u2​)=0.31119…2−1.52137…⋅0.31119…+1.31459…=0.93798…f′(u2​)=2⋅0.31119…−1.52137…=−0.89897…u3​=1.35459…
Δu3​=∣1.35459…−0.31119…∣=1.04339…Δu3​=1.04339…
u4​=0.43805…:Δu4​=0.91653…
f(u3​)=1.35459…2−1.52137…⋅1.35459…+1.31459…=1.08866…f′(u3​)=2⋅1.35459…−1.52137…=1.18780…u4​=0.43805…
Δu4​=∣0.43805…−1.35459…∣=0.91653…Δu4​=0.91653…
u5​=1.73989…:Δu5​=1.30184…
f(u4​)=0.43805…2−1.52137…⋅0.43805…+1.31459…=0.84004…f′(u4​)=2⋅0.43805…−1.52137…=−0.64527…u5​=1.73989…
Δu5​=∣1.73989…−0.43805…∣=1.30184…Δu5​=1.30184…
u6​=0.87450…:Δu6​=0.86539…
f(u5​)=1.73989…2−1.52137…⋅1.73989…+1.31459…=1.69479…f′(u5​)=2⋅1.73989…−1.52137…=1.95841…u6​=0.87450…
Δu6​=∣0.87450…−1.73989…∣=0.86539…Δu6​=0.86539…
u7​=−2.41546…:Δu7​=3.28996…
f(u6​)=0.87450…2−1.52137…⋅0.87450…+1.31459…=0.74890…f′(u6​)=2⋅0.87450…−1.52137…=0.22763…u7​=−2.41546…
Δu7​=∣−2.41546…−0.87450…∣=3.28996…Δu7​=3.28996…
u8​=−0.71153…:Δu8​=1.70393…
f(u7​)=(−2.41546…)2−1.52137…(−2.41546…)+1.31459…=10.82387…f′(u7​)=2(−2.41546…)−1.52137…=−6.35230…u8​=−0.71153…
Δu8​=∣−0.71153…−(−2.41546…)∣=1.70393…Δu8​=1.70393…
u9​=0.27452…:Δu9​=0.98605…
f(u8​)=(−0.71153…)2−1.52137…(−0.71153…)+1.31459…=2.90337…f′(u8​)=2(−0.71153…)−1.52137…=−2.94443…u9​=0.27452…
Δu9​=∣0.27452…−(−0.71153…)∣=0.98605…Δu9​=0.98605…
u10​=1.27449…:Δu10​=0.99997…
f(u9​)=0.27452…2−1.52137…⋅0.27452…+1.31459…=0.97230…f′(u9​)=2⋅0.27452…−1.52137…=−0.97233…u10​=1.27449…
Δu10​=∣1.27449…−0.27452…∣=0.99997…Δu10​=0.99997…
Cannot find solution
The solution isu≈−1.52137…
The solutions areu=0,u≈−1.52137…
u=0,u≈−1.52137…
Verify Solutions
Find undefined (singularity) points:u=1,u=−1
Take the denominator(s) of 1−u22u​ and compare to zero
Solve 1−u2=0:u=1,u=−1
1−u2=0
Move 1to the right side
1−u2=0
Subtract 1 from both sides1−u2−1=0−1
Simplify−u2=−1
−u2=−1
Divide both sides by −1
−u2=−1
Divide both sides by −1−1−u2​=−1−1​
Simplifyu2=1
u2=1
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=1​,u=−1​
1​=1
1​
Apply radical rule: 1​=1=1
−1​=−1
−1​
Apply radical rule: 1​=11​=1=−1
u=1,u=−1
The following points are undefinedu=1,u=−1
Combine undefined points with solutions:
u=0,u≈−1.52137…
Substitute back u=tan(a)tan(a)=0,tan(a)≈−1.52137…
tan(a)=0,tan(a)≈−1.52137…
tan(a)=0:a=πn
tan(a)=0
General solutions for tan(a)=0
tan(x) periodicity table with πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​tan(x)033​​13​±∞−3​−1−33​​​​
a=0+πn
a=0+πn
Solve a=0+πn:a=πn
a=0+πn
0+πn=πna=πn
a=πn
tan(a)=−1.52137…:a=arctan(−1.52137…)+πn
tan(a)=−1.52137…
Apply trig inverse properties
tan(a)=−1.52137…
General solutions for tan(a)=−1.52137…tan(x)=−a⇒x=arctan(−a)+πna=arctan(−1.52137…)+πn
a=arctan(−1.52137…)+πn
Combine all the solutionsa=πn,a=arctan(−1.52137…)+πn
Show solutions in decimal forma=πn,a=−0.98930…+πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for tan^2(a)=((2tan(a)))/((1-tan^2(a))) ?

    The general solution for tan^2(a)=((2tan(a)))/((1-tan^2(a))) is a=pin,a=-0.98930…+pin
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