Formulas and Theorems in Pure Mathematics |
From inside the book
Results 1-3 of 35
Page 161
... sine , and tangent . [ 610-612 625 To reduce any ratio of an angle greater than 90 ° to the ratio of an angle less ... sine and cosine all the remaining ratios may be found by ( 613-616 ) . INVERSE NOTATION . - The angle whose sine is a ...
... sine , and tangent . [ 610-612 625 To reduce any ratio of an angle greater than 90 ° to the ratio of an angle less ... sine and cosine all the remaining ratios may be found by ( 613-616 ) . INVERSE NOTATION . - The angle whose sine is a ...
Page 162
... sine , cosine , or tangent respectively , and n is any integer . Coseca , seca , cot - 1 have similar general values , by ( 610-612 ) . These formulæ are verified by taking A , in Fig . 622 , for 0 , and making n an odd or even integer ...
... sine , cosine , or tangent respectively , and n is any integer . Coseca , seca , cot - 1 have similar general values , by ( 610-612 ) . These formulæ are verified by taking A , in Fig . 622 , for 0 , and making n an odd or even integer ...
Page 187
... sine of middle part = product of tangents of adjacent parts . II . sine of middle part = product of cosines of opposite parts . In applying the rules we can take A , B , c instead of their complements , and change sine into cos , or ...
... sine of middle part = product of tangents of adjacent parts . II . sine of middle part = product of cosines of opposite parts . In applying the rules we can take A , B , c instead of their complements , and change sine into cos , or ...
Contents
other is also | 1 |
Product of Squares of Differences of same | 4 |
Prime numbers from 1 to 109 | 7 |
Copyright | |
82 other sections not shown
Common terms and phrases
A₁ axes axis b₁ becomes Binomial Theorem C₁ centre centroid chord circle coefficients column complete primitive confocal conic conjugate constant continued fraction convergent coordinates curvature curve determinant diameter differential equation divided eliminate ellipse equal expansion factors formula function given Hence homogeneous function hyperbola imaginary roots infinity integral intersection line of curvature locus method Multiply normal obtained osculating plane P₁ P₂ parabola parallel partial fractions perpendicular plane polar positive PROOF PROOF.-By PROOF.-Let PROOF.-The quadratic quadric quotient radical axis radius ratio reciprocal respectively result right line sides Similarly sin² singular solution solution Substitute surface tangent theorem tion triangle u₂ values vanish variables zero