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Let phi(z)=cz+c_0+c_1z^(-1)+c_2z^(-2)+... be an analytic function, regular and univalent for |z|>1, that maps |z|>1 conformally onto the region T preserving the point at ...

To inscribe an equilateral triangle in an ellipse, place the top polygon vertex at (0,b), then solve to find the (x,y) coordinate of the other two vertices. ...

Let a triangle have angles A, B, and C, then inequalities include sinA+sinB+sinC<=3/2sqrt(3) (1) 1<=cosA+cosB+cosC<=3/2 (2) sin(1/2A)sin(1/2B)sin(1/2C)<=1/8 (3) ...

By the definition of the trigonometric functions, cos0 = 1 (1) cot0 = infty (2) csc0 = infty (3) sec0 = 1 (4) sin0 = 0 (5) tan0 = 0. (6)

cos(pi/(15)) = 1/8(sqrt(30+6sqrt(5))+sqrt(5)-1) (1) cos((2pi)/(15)) = 1/8(sqrt(30-6sqrt(5))+sqrt(5)+1) (2) cos((4pi)/(15)) = 1/8(sqrt(30+6sqrt(5))-sqrt(5)+1) (3) ...

cos(pi/(16)) = 1/2sqrt(2+sqrt(2+sqrt(2))) (1) cos((3pi)/(16)) = 1/2sqrt(2+sqrt(2-sqrt(2))) (2) cos((5pi)/(16)) = 1/2sqrt(2-sqrt(2-sqrt(2))) (3) cos((7pi)/(16)) = ...

Values of the trigonometric functions can be expressed exactly for integer multiples of pi/20. For cosx, cos(pi/(20)) = 1/4sqrt(8+2sqrt(10+2sqrt(5))) (1) cos((3pi)/(20)) = ...

cos(pi/(24)) = 1/2sqrt(2+sqrt(2+sqrt(3))) (1) cos((5pi)/(24)) = 1/2sqrt(2+sqrt(2-sqrt(3))) (2) cos((7pi)/(24)) = 1/2sqrt(2-sqrt(2-sqrt(3))) (3) cos((11pi)/(24)) = ...

cos(pi/(30)) = 1/4sqrt(7+sqrt(5)+sqrt(6(5+sqrt(5)))) (1) cos((7pi)/(30)) = 1/4sqrt(7-sqrt(5)+sqrt(6(5-sqrt(5)))) (2) cos((11pi)/(30)) = 1/4sqrt(7+sqrt(5)-sqrt(6(5+sqrt(5)))) ...

cos(pi/(32)) = 1/2sqrt(2+sqrt(2+sqrt(2+sqrt(2)))) (1) cos((3pi)/(32)) = 1/2sqrt(2+sqrt(2+sqrt(2-sqrt(2)))) (2) cos((5pi)/(32)) = 1/2sqrt(2+sqrt(2-sqrt(2-sqrt(2)))) (3) ...