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Tan in Surds

Julian D. A. Wiseman

Contents: values of Tan[α] expressed in surds, for α=n×3° or α=n×5⅝°, n∈ℕ: Tan[0°], Tan[3°], Tan[5.625°], Tan[6°], Tan[9°], Tan[11.25°], Tan[12°], Tan[15°], Tan[16.875°], Tan[18°], Tan[21°], Tan[22.5°], Tan[24°], Tan[27°], Tan[28.125°], Tan[30°], Tan[33°], Tan[33.75°], Tan[36°], Tan[39°], Tan[39.375°], Tan[42°], Tan[45°], Tan[48°], Tan[50.625°], Tan[51°], Tan[54°], Tan[56.25°], Tan[57°], Tan[60°], Tan[61.875°], Tan[63°], Tan[66°], Tan[67.5°], Tan[69°], Tan[72°], Tan[73.125°], Tan[75°], Tan[78°], Tan[78.75°], Tan[81°], Tan[84°], Tan[84.375°], Tan[87°], Tan[90°].

Publication history: only here. Usual disclaimer and copyright terms apply. Also see the values of Sin[] and Cos[], in surds, the values of Cosecant[] = Cosec[] = Csc[] = 1/Sin[] and Secant[] = Sec[] = 1/Cos[], in surds, and the inner radius of n/m stars, in surds.


The table shows Tan[] in surds, for angles that are integer multiples of 3° or of 5⅝° = 90°/16. The surds are shown in several formats.

Help! Further simplifications are wanted, and credit will be given to the sender.

Errors: whilst the outputs have been tested, it is possible that errors remain. Please do test things before embedding them somewhere important—and if errors or possible improvements are found, tell the author.

Tan[α] = 1/Tan[90–α]Graphical formulaLaTeXExcelCalcCenterPostScript
Tan[0°] = 1/Tan[90°] = 0Tan(0°)0000
Tan[3°] = 1/Tan[87°] ≈ 0.052407779283Tan(3°)\frac{1}{4} \left(1 + 2 \sqrt{5 - 2 \sqrt{5}} - \sqrt{5}\right) \left(\left(\sqrt{5} + 2\right) \left(2 \sqrt{3} - 3\right) - 1\right)=( 1 + 2*Sqrt(5-2*Sqrt(5)) - Sqrt(5) ) * ((Sqrt(5)+2)*(2*Sqrt(3)-3)-1) / 4( 1 + 2 Sqrt[5-2 Sqrt[Int5]] - Sqrt[Int5] ) ((Sqrt[Int5]+2) (2 Sqrt[Int3]-3)-1) / 45 sqrt dup dup -2 mul 5 add sqrt 2 mul 1 add exch sub exch 2 add 3 sqrt 2 mul 3 sub mul 1 sub mul 4 div
Tan[5⅝°] = 1/Tan[84⅜°] ≈ 0.098491403357Tan(5.625°)\frac{1}{2} \left(2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}\right) \sqrt{2 \left(2 + \sqrt{2}\right) \left(2 + \sqrt{2 + \sqrt{2}}\right)}=( 2 - Sqrt(2+Sqrt(2+Sqrt(2))) ) * Sqrt( 2 * (2+Sqrt(2)) * (2+Sqrt(2+Sqrt(2))) ) / 2( 2 - Sqrt[2+Sqrt[2+Sqrt[Int2]]] ) Sqrt[ 2 (2+Sqrt[Int2]) (2+Sqrt[2+Sqrt[Int2]]) ] / 22 sqrt 2 add dup sqrt 2 add dup 3 1 roll mul 2 mul sqrt exch sqrt 2 sub neg mul 2 div
Tan[6°] = 1/Tan[84°] ≈ 0.105104235266Tan(6°)\frac{1}{2} \left(\sqrt{10 - 2 \sqrt{5}} - \sqrt{3} \left(\sqrt{5} - 1\right)\right)=( Sqrt(10-2*Sqrt(5)) - Sqrt(3)*(Sqrt(5)-1) ) / 2( Sqrt[10-2 Sqrt[Int5]] - Sqrt[Int3] (Sqrt[Int5]-1) ) / 25 sqrt dup -2 mul 10 add sqrt exch 1 sub 3 sqrt mul sub 2 div
Tan[9°] = 1/Tan[81°] ≈ 0.158384440325Tan(9°)1 + \sqrt{5} - \sqrt{2 \sqrt{5} + 5}=1 + Sqrt(5) - Sqrt(2*Sqrt(5)+5)1 + Sqrt[Int5] - Sqrt[2 Sqrt[Int5]+5]5 sqrt dup 1 add exch 2 mul 5 add sqrt sub
Tan[11¼°] = 1/Tan[78¾°] ≈ 0.19891236738Tan(11.25°)\left(\sqrt{\sqrt{2} + 2} - 1\right) \sqrt{2} - 1=(Sqrt(Sqrt(2)+2)-1)*Sqrt(2) - 1(Sqrt[Sqrt[Int2]+2]-1) Sqrt[Int2] - 12 sqrt dup 2 add sqrt 1 sub mul 1 sub
Tan[12°] = 1/Tan[78°] ≈ 0.21255656167Tan(12°)\frac{1}{4} \left(2 \sqrt{3} - \sqrt{10 - 2 \sqrt{5}}\right) \left(3 - \sqrt{5}\right)=(2*Sqrt(3) - Sqrt(10-2*Sqrt(5))) * (3-Sqrt(5)) / 4(2 Sqrt[Int3] - Sqrt[10-2 Sqrt[Int5]]) (3-Sqrt[Int5]) / 45 sqrt neg dup 2 mul 10 add sqrt neg 3 sqrt 2 mul add exch 3 add mul 4 div
Tan[15°] = 1/Tan[75°] ≈ 0.267949192431Tan(15°)2 - \sqrt{3}=2-Sqrt(3)2-Sqrt[Int3]2 3 sqrt sub
Tan[16⅞°] = 1/Tan[73⅛°] ≈ 0.303346683607Tan(16.875°)\frac{1}{2} \left(2 - \sqrt{2 + \sqrt{2 - \sqrt{2}}}\right) \sqrt{2 \left(2 - \sqrt{2}\right) \left(2 + \sqrt{2 - \sqrt{2}}\right)}=( 2 - Sqrt(2+Sqrt(2-Sqrt(2))) ) * Sqrt( 2 * (2-Sqrt(2)) * (2+Sqrt(2-Sqrt(2))) ) / 2( 2 - Sqrt[2+Sqrt[2-Sqrt[Int2]]] ) Sqrt[ 2 (2-Sqrt[Int2]) (2+Sqrt[2-Sqrt[Int2]]) ] / 22 2 sqrt sub dup sqrt 2 add dup 3 1 roll mul 2 mul sqrt exch sqrt 2 sub neg mul 2 div
Tan[18°] = 1/Tan[72°] ≈ 0.324919696233Tan(18°)\sqrt{1 - \frac{2}{5}\sqrt{5}}=Sqrt(1-Sqrt(5)*2/5)Sqrt[1-Sqrt[Int5] 2/5]1 5 sqrt 2 mul 5 div sub sqrt
Tan[21°] = 1/Tan[69°] ≈ 0.383864035035Tan(21°)\frac{1}{4} \left(2 \sqrt{5 - 2 \sqrt{5}} + \sqrt{5} - 3\right) \left(1 + 2 \sqrt{3} - \sqrt{5}\right)=(2*Sqrt(5-2*Sqrt(5)) + Sqrt(5) - 3) * (1 + 2*Sqrt(3) - Sqrt(5)) / 4(2 Sqrt[5-2 Sqrt[Int5]] + Sqrt[Int5] - 3) (1 + 2 Sqrt[Int3] - Sqrt[Int5]) / 45 sqrt dup dup -2 mul 5 add sqrt 2 mul add 3 sub exch neg 3 sqrt 2 mul add 1 add mul 4 div
Tan[22½°] = 1/Tan[67½°] ≈ 0.414213562373Tan(22.5°)\sqrt{2} - 1=Sqrt(2)-1Sqrt[Int2]-12 sqrt 1 sub
Tan[24°] = 1/Tan[66°] ≈ 0.445228685309Tan(24°)\frac{1}{2} \left(\sqrt{22 \sqrt{5} + 50} - \sqrt{3} \left(\sqrt{5} + 3\right)\right)=( Sqrt(22*Sqrt(5)+50) - Sqrt(3)*(Sqrt(5)+3) ) / 2( Sqrt[22 Sqrt[Int5]+50] - Sqrt[Int3] (Sqrt[Int5]+3) ) / 25 sqrt dup 22 mul 50 add sqrt exch 3 add 3 sqrt mul sub 2 div
Tan[27°] = 1/Tan[63°] ≈ 0.509525449494Tan(27°)\sqrt{5} - 1 - \sqrt{5 - 2 \sqrt{5}}=Sqrt(5) - 1 - Sqrt(5-2*Sqrt(5))Sqrt[Int5] - 1 - Sqrt[5-2 Sqrt[Int5]]5 sqrt dup -2 mul 5 add sqrt sub 1 sub
Tan[28⅛°] = 1/Tan[61⅞°] ≈ 0.534511135951Tan(28.125°)\frac{1}{2} \left(2 - \sqrt{2 - \sqrt{2 - \sqrt{2}}}\right) \sqrt{2 \left(2 - \sqrt{2}\right) \left(2 - \sqrt{2 - \sqrt{2}}\right)}=( 2 - Sqrt(2-Sqrt(2-Sqrt(2))) ) * Sqrt( 2 * (2-Sqrt(2)) * (2-Sqrt(2-Sqrt(2))) ) / 2( 2 - Sqrt[2-Sqrt[2-Sqrt[Int2]]] ) Sqrt[ 2 (2-Sqrt[Int2]) (2-Sqrt[2-Sqrt[Int2]]) ] / 22 2 sqrt sub dup sqrt neg 2 add dup 3 1 roll mul 2 mul sqrt exch sqrt neg 2 add mul 2 div
Tan[30°] = 1/Tan[60°] ≈ 0.57735026919Tan(30°)\frac{\sqrt{3}}{3}=Sqrt(3)/3Sqrt[Int3]/33 sqrt 3 div
Tan[33°] = 1/Tan[57°] ≈ 0.649407593198Tan(33°)\frac{1}{2} \left(\sqrt{5 - \sqrt{5}} + \sqrt{2}\right) \sqrt{\frac{1}{5} \left(8 - \left(5 \sqrt{5} + 11\right) \left(10 \sqrt{3} - 17\right)\right)}=(Sqrt(5-Sqrt(5))+Sqrt(2)) * Sqrt( ( 8-(5*Sqrt(5)+11)*(10*Sqrt(3)-17) ) / 5 ) / 2(Sqrt[5-Sqrt[Int5]]+Sqrt[Int2]) Sqrt[ ( 8-(5 Sqrt[Int5]+11) (10 Sqrt[Int3]-17) ) / Int5 ] / 25 sqrt dup neg 5 add sqrt 2 sqrt add exch 5 mul 11 add 17 3 sqrt 10 mul sub mul 8 add 5 div sqrt mul 2 div
Tan[33¾°] = 1/Tan[56¼°] ≈ 0.668178637919Tan(33.75°)\left(\sqrt{2 - \sqrt{2}} - 1\right) \sqrt{2} + 1=(Sqrt(2-Sqrt(2))-1)*Sqrt(2) + 1(Sqrt[2-Sqrt[Int2]]-1) Sqrt[Int2] + 12 sqrt dup 2 sub neg sqrt 1 sub mul 1 add
Tan[36°] = 1/Tan[54°] ≈ 0.726542528005Tan(36°)\sqrt{5 - 2 \sqrt{5}}=Sqrt(5-2*Sqrt(5))Sqrt[5-2 Sqrt[Int5]]5 5 sqrt 2 mul sub sqrt
Tan[39°] = 1/Tan[51°] ≈ 0.809784033195Tan(39°)\frac{1}{2} \left(\sqrt{5 + \sqrt{5}} - \sqrt{2}\right) \sqrt{\frac{1}{5} \left(8 + \left(5 \sqrt{5} - 11\right) \left(10 \sqrt{3} - 17\right)\right)}=(Sqrt(5+Sqrt(5))-Sqrt(2)) * Sqrt( ( 8+(5*Sqrt(5)-11)*(10*Sqrt(3)-17) ) / 5 ) / 2(Sqrt[5+Sqrt[Int5]]-Sqrt[Int2]) Sqrt[ ( 8+(5 Sqrt[Int5]-11) (10 Sqrt[Int3]-17) ) / Int5 ] / 25 sqrt dup 5 add sqrt 2 sqrt sub exch 5 mul 11 sub 3 sqrt 10 mul 17 sub mul 8 add 5 div sqrt mul 2 div
Tan[39⅜°] = 1/Tan[50⅝°] ≈ 0.820678790829Tan(39.375°)\frac{1}{2} \left(2 - \sqrt{2 - \sqrt{2 + \sqrt{2}}}\right) \sqrt{2 \left(2 + \sqrt{2}\right) \left(2 - \sqrt{2 + \sqrt{2}}\right)}=( 2 - Sqrt(2-Sqrt(2+Sqrt(2))) ) * Sqrt( 2 * (2+Sqrt(2)) * (2-Sqrt(2+Sqrt(2))) ) / 2( 2 - Sqrt[2-Sqrt[2+Sqrt[Int2]]] ) Sqrt[ 2 (2+Sqrt[Int2]) (2-Sqrt[2+Sqrt[Int2]]) ] / 22 2 sqrt add dup sqrt neg 2 add dup 3 1 roll mul 2 mul sqrt exch sqrt neg 2 add mul 2 div
Tan[42°] = 1/Tan[48°] ≈ 0.900404044298Tan(42°)\frac{1}{4} \left(2 \sqrt{3} - \sqrt{10 - 2 \sqrt{5}}\right) \left(1 + \sqrt{5}\right)=(2*Sqrt(3)-Sqrt(10-2*Sqrt(5))) * (1+Sqrt(5)) / 4(2 Sqrt[Int3]-Sqrt[10-2 Sqrt[Int5]]) (1+Sqrt[Int5]) / 45 sqrt dup -2 mul 10 add sqrt neg 3 sqrt 2 mul add exch 1 add mul 4 div
Tan[45°] = 1Tan(45°)1111
Tan[48°] = 1/Tan[42°] ≈ 1.110612514829Tan(48°)\frac{1}{4} \left(2 \sqrt{3} + \sqrt{10 - 2 \sqrt{5}}\right) \left(3 - \sqrt{5}\right)=(2*Sqrt(3)+Sqrt(10-2*Sqrt(5))) * (3-Sqrt(5)) / 4(2 Sqrt[Int3]+Sqrt[10-2 Sqrt[Int5]]) (3-Sqrt[Int5]) / 45 sqrt neg dup 2 mul 10 add sqrt 3 sqrt 2 mul add exch 3 add mul 4 div
Tan[50⅝°] = 1/Tan[39⅜°] ≈ 1.218503525588Tan(50.625°)\frac{1}{2} \left(2 + \sqrt{2 - \sqrt{2 + \sqrt{2}}}\right) \sqrt{2 \left(2 + \sqrt{2}\right) \left(2 - \sqrt{2 + \sqrt{2}}\right)}=( 2 + Sqrt(2-Sqrt(2+Sqrt(2))) ) * Sqrt( 2 * (2+Sqrt(2)) * (2-Sqrt(2+Sqrt(2))) ) / 2( 2 + Sqrt[2-Sqrt[2+Sqrt[Int2]]] ) Sqrt[ 2 (2+Sqrt[Int2]) (2-Sqrt[2+Sqrt[Int2]]) ] / 22 sqrt 2 add dup sqrt neg 2 add dup 3 1 roll mul 2 mul sqrt exch sqrt 2 add mul 2 div
Tan[51°] = 1/Tan[39°] ≈ 1.234897156535Tan(51°)\frac{1}{2} \left(\sqrt{5 + \sqrt{5}} + \sqrt{2}\right) \sqrt{\frac{1}{5} \left(8 - \left(5 \sqrt{5} - 11\right) \left(10 \sqrt{3} + 17\right)\right)}=(Sqrt(5+Sqrt(5))+Sqrt(2)) * Sqrt( ( 8-(5*Sqrt(5)-11)*(10*Sqrt(3)+17) ) / 5 ) / 2(Sqrt[5+Sqrt[Int5]]+Sqrt[Int2]) Sqrt[ ( 8-(5 Sqrt[Int5]-11) (10 Sqrt[Int3]+17) ) / Int5 ] / 25 sqrt dup 5 add sqrt 2 sqrt add exch 5 mul 11 sub 3 sqrt 10 mul 17 add mul neg 8 add 5 div sqrt mul 2 div
Tan[54°] = 1/Tan[36°] ≈ 1.376381920471Tan(54°)\sqrt{1 + \frac{2}{5}\sqrt{5}}=Sqrt(1+Sqrt(5)*2/5)Sqrt[1+Sqrt[Int5] 2/5]5 sqrt 2 mul 5 div 1 add sqrt
Tan[56¼°] = 1/Tan[33¾°] ≈ 1.496605762665Tan(56.25°)\left(\sqrt{2 - \sqrt{2}} + 1\right) \sqrt{2} - 1=(Sqrt(2-Sqrt(2))+1)*Sqrt(2) - 1(Sqrt[2-Sqrt[Int2]]+1) Sqrt[Int2] - 12 sqrt dup 2 sub neg sqrt 1 add mul 1 sub
Tan[57°] = 1/Tan[33°] ≈ 1.539864963815Tan(57°)\frac{1}{2} \left(\sqrt{5 - \sqrt{5}} - \sqrt{2}\right) \sqrt{\frac{1}{5} \left(8 + \left(5 \sqrt{5} + 11\right) \left(10 \sqrt{3} + 17\right)\right)}=(Sqrt(5-Sqrt(5))-Sqrt(2)) * Sqrt( ( 8+(5*Sqrt(5)+11)*(10*Sqrt(3)+17) ) / 5 ) / 2(Sqrt[5-Sqrt[Int5]]-Sqrt[Int2]) Sqrt[ ( 8+(5 Sqrt[Int5]+11) (10 Sqrt[Int3]+17) ) / Int5 ] / 25 sqrt dup neg 5 add sqrt 2 sqrt sub exch 5 mul 11 add 17 3 sqrt 10 mul add mul 8 add 5 div sqrt mul 2 div
Tan[60°] = 1/Tan[30°] ≈ 1.732050807569Tan(60°)\sqrt{3}=Sqrt(3)Sqrt[Int3]3 sqrt
Tan[61⅞°] = 1/Tan[28⅛°] ≈ 1.870868411789Tan(61.875°)\frac{1}{2} \left(2 + \sqrt{2 - \sqrt{2 - \sqrt{2}}}\right) \sqrt{2 \left(2 - \sqrt{2}\right) \left(2 - \sqrt{2 - \sqrt{2}}\right)}=( 2 + Sqrt(2-Sqrt(2-Sqrt(2))) ) * Sqrt( 2 * (2-Sqrt(2)) * (2-Sqrt(2-Sqrt(2))) ) / 2( 2 + Sqrt[2-Sqrt[2-Sqrt[Int2]]] ) Sqrt[ 2 (2-Sqrt[Int2]) (2-Sqrt[2-Sqrt[Int2]]) ] / 22 2 sqrt sub dup sqrt neg 2 add dup 3 1 roll mul 2 mul sqrt exch sqrt 2 add mul 2 div
Tan[63°] = 1/Tan[27°] ≈ 1.962610505505Tan(63°)\sqrt{5 - 2 \sqrt{5}} + \sqrt{5} - 1=Sqrt(5-2*Sqrt(5)) + Sqrt(5) - 1Sqrt[5-2 Sqrt[Int5]] + Sqrt[Int5] - 15 sqrt dup -2 mul 5 add sqrt add 1 sub
Tan[66°] = 1/Tan[24°] ≈ 2.246036773904Tan(66°)\frac{1}{2} \left(\sqrt{3} \left(\sqrt{5} - 1\right) + \sqrt{10 - 2 \sqrt{5}}\right)=( Sqrt(3)*(Sqrt(5)-1) + Sqrt(10-2*Sqrt(5)) ) / 2( Sqrt[Int3] (Sqrt[Int5]-1) + Sqrt[10-2 Sqrt[Int5]] ) / 25 sqrt dup -2 mul 10 add sqrt exch 1 sub 3 sqrt mul add 2 div
Tan[67½°] = 1/Tan[22½°] ≈ 2.414213562373Tan(67.5°)\sqrt{2} + 1=Sqrt(2)+1Sqrt[Int2]+12 sqrt 1 add
Tan[69°] = 1/Tan[21°] ≈ 2.605089064694Tan(69°)\frac{1}{2} \left(\sqrt{5 + \sqrt{5}} + \sqrt{2}\right) \sqrt{\frac{1}{5} \left(\left(5 \sqrt{5} - 11\right) \left(10 \sqrt{3} - 17\right) + 8\right)}=(Sqrt(5+Sqrt(5))+Sqrt(2)) * Sqrt( ( (5*Sqrt(5)-11)*(10*Sqrt(3)-17)+8 ) / 5 ) / 2(Sqrt[5+Sqrt[Int5]]+Sqrt[Int2]) Sqrt[ ( (5 Sqrt[Int5]-11) (10 Sqrt[Int3]-17)+8 ) / Int5 ] / 25 sqrt dup 5 add sqrt 2 sqrt add exch 5 mul 11 sub 3 sqrt 10 mul 17 sub mul 8 add 5 div sqrt mul 2 div
Tan[72°] = 1/Tan[18°] ≈ 3.077683537175Tan(72°)\sqrt{2 \sqrt{5} + 5}=Sqrt(2*Sqrt(5)+5)Sqrt[2 Sqrt[Int5]+5]5 sqrt 2 mul 5 add sqrt
Tan[73⅛°] = 1/Tan[16⅞°] ≈ 3.296558208938Tan(73.125°)\frac{1}{2} \left(2 + \sqrt{2 + \sqrt{2 - \sqrt{2}}}\right) \sqrt{2 \left(2 - \sqrt{2}\right) \left(2 + \sqrt{2 - \sqrt{2}}\right)}=( 2 + Sqrt(2+Sqrt(2-Sqrt(2))) ) * Sqrt( 2 * (2-Sqrt(2)) * (2+Sqrt(2-Sqrt(2))) ) / 2( 2 + Sqrt[2+Sqrt[2-Sqrt[Int2]]] ) Sqrt[ 2 (2-Sqrt[Int2]) (2+Sqrt[2-Sqrt[Int2]]) ] / 22 2 sqrt sub dup sqrt 2 add dup 3 1 roll mul 2 mul sqrt exch sqrt 2 add mul 2 div
Tan[75°] = 1/Tan[15°] ≈ 3.732050807569Tan(75°)2 + \sqrt{3}=2+Sqrt(3)2+Sqrt[Int3]3 sqrt 2 add
Tan[78°] = 1/Tan[12°] ≈ 4.704630109478Tan(78°)\frac{1}{2} \left(\sqrt{3} \left(\sqrt{5} + 1\right) + \sqrt{2 \sqrt{5} + 10}\right)=( Sqrt(3)*(Sqrt(5)+1) + Sqrt(2*Sqrt(5)+10) ) / 2( Sqrt[Int3] (Sqrt[Int5]+1) + Sqrt[2 Sqrt[Int5]+10] ) / 25 sqrt dup 2 mul 10 add sqrt exch 1 add 3 sqrt mul add 2 div
Tan[78¾°] = 1/Tan[11¼°] ≈ 5.027339492126Tan(78.75°)\left(\sqrt{2 + \sqrt{2}} + 1\right) \sqrt{2} + 1=(Sqrt(2+Sqrt(2))+1)*Sqrt(2) + 1(Sqrt[2+Sqrt[Int2]]+1) Sqrt[Int2] + 12 sqrt dup 2 add sqrt 1 add mul 1 add
Tan[81°] = 1/Tan[9°] ≈ 6.313751514675Tan(81°)1 + \sqrt{5} + \sqrt{2 \sqrt{5} + 5}=1 + Sqrt(5) + Sqrt(2*Sqrt(5)+5)1 + Sqrt[Int5] + Sqrt[2 Sqrt[Int5]+5]5 sqrt dup 1 add exch 2 mul 5 add sqrt add
Tan[84°] = 1/Tan[6°] ≈ 9.514364454223Tan(84°)\frac{1}{2} \left(\sqrt{3} \left(\sqrt{5} + 3\right) + \sqrt{22 \sqrt{5} + 50}\right)=( Sqrt(3)*(Sqrt(5)+3) + Sqrt(22*Sqrt(5)+50) ) / 2( Sqrt[Int3] (Sqrt[Int5]+3) + Sqrt[22 Sqrt[Int5]+50] ) / 25 sqrt dup 22 mul 50 add sqrt exch 3 add 3 sqrt mul add 2 div
Tan[84⅜°] = 1/Tan[5⅝°] ≈ 10.153170387609Tan(84.375°)\frac{1}{2} \left(2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}\right) \sqrt{2 \left(2 + \sqrt{2}\right) \left(2 + \sqrt{2 + \sqrt{2}}\right)}=( 2 + Sqrt(2+Sqrt(2+Sqrt(2))) ) * Sqrt( 2 * (2+Sqrt(2)) * (2+Sqrt(2+Sqrt(2))) ) / 2( 2 + Sqrt[2+Sqrt[2+Sqrt[Int2]]] ) Sqrt[ 2 (2+Sqrt[Int2]) (2+Sqrt[2+Sqrt[Int2]]) ] / 22 sqrt 2 add dup sqrt 2 add dup 3 1 roll mul 2 mul sqrt exch sqrt 2 add mul 2 div
Tan[87°] = 1/Tan[3°] ≈ 19.081136687728Tan(87°)\frac{1}{4} \left(2 \sqrt{5 - 2 \sqrt{5}} + \sqrt{5} - 1\right) \left(\left(2 \sqrt{3} + 3\right) \left(\sqrt{5} + 2\right) + 1\right)=(2*Sqrt(5-2*Sqrt(5)) + Sqrt(5) - 1) * ((2*Sqrt(3)+3)*(Sqrt(5)+2)+1) / 4(2 Sqrt[5-2 Sqrt[Int5]] + Sqrt[Int5] - 1) ((2 Sqrt[Int3]+3) (Sqrt[Int5]+2)+1) / 45 sqrt dup dup -2 mul 5 add sqrt 2 mul add 1 sub exch 2 add 3 sqrt 2 mul 3 add mul 1 add mul 4 div
Tan[90°] = 1/Tan[0°] = ±∞Tan(90°)\pm\infty=(2^53 - 1) * (2^971)±Infinity2 23 exp 1 sub 2 104 exp mul

Julian D. A. Wiseman, December 2008


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