A Popular Course of Pure and Mixed Mathematics ...: With Tables of Logarithms, and Numerous Questions for ExerciseG. B. Whittaker, 1825 - 372 σελίδες |
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Αποτελέσματα 6 - 10 από τα 100.
Σελίδα 344
... tang + A , 2. 3. 4 & c . We have therefore in general Tang n A = n tang A- n . n -- 1 . n - tang A + n . n - i . n -- 2 . n -- 3 . n -- 4 , n- 2. 3 n . tang2 A + 2 tang ' A- & c . 2. 3. 4. 5 n . n - l . n - 2 . n- -3 tang + A- & c .
... tang + A , 2. 3. 4 & c . We have therefore in general Tang n A = n tang A- n . n -- 1 . n - tang A + n . n - i . n -- 2 . n -- 3 . n -- 4 , n- 2. 3 n . tang2 A + 2 tang ' A- & c . 2. 3. 4. 5 n . n - l . n - 2 . n- -3 tang + A- & c .
Σελίδα 344
... 3. 4. 5. 6. 7 = tang a = Cos a a a + ენ + + & c . 2 2. 3. 4 2. 3. 4. 5. 6 a4 a® + + & c . COS & 2 2. 3. 4 2. 3. 4. 5. 6 cot a = . sin a a3 as a ? a -- + + & c . 2.3 2. 3. 4. 5 2. 3. 4. 5. 6. 7 - & c . 26. Let now the arc a be ...
... 3. 4. 5. 6. 7 = tang a = Cos a a a + ენ + + & c . 2 2. 3. 4 2. 3. 4. 5. 6 a4 a® + + & c . COS & 2 2. 3. 4 2. 3. 4. 5. 6 cot a = . sin a a3 as a ? a -- + + & c . 2.3 2. 3. 4. 5 2. 3. 4. 5. 6. 7 - & c . 26. Let now the arc a be ...
Σελίδα 344
... tang ) and cot 90 ° 90 ° " By m m means of these sines we might calculate the sines and tangents of For my arcs ; we have only to substitute the proper values of m . example , to calculate the sine of the arc of 30 ° , we must make m ...
... tang ) and cot 90 ° 90 ° " By m m means of these sines we might calculate the sines and tangents of For my arcs ; we have only to substitute the proper values of m . example , to calculate the sine of the arc of 30 ° , we must make m ...
Σελίδα 344
... tang A tang A 3 + 5 7 . + & c . 31. These two series give the solution of the proposed problem . Let us now apply them to the determination of the ratio of the diameter to the circumference . If we make sin A == , we shall have the ...
... tang A tang A 3 + 5 7 . + & c . 31. These two series give the solution of the proposed problem . Let us now apply them to the determination of the ratio of the diameter to the circumference . If we make sin A == , we shall have the ...
Σελίδα 345
... tang ( A + B ) = 1 = tang A + tang B Therefore tang A = 1 - tang B 1 + tang B 1 - tang A tang B 1 Let B = 3 and we shall have tang A = 1 . Hence the sum of the two arcs A and B , or the quarter of the semi - circumferace will be T 1 1 1 ...
... tang ( A + B ) = 1 = tang A + tang B Therefore tang A = 1 - tang B 1 + tang B 1 - tang A tang B 1 Let B = 3 and we shall have tang A = 1 . Hence the sum of the two arcs A and B , or the quarter of the semi - circumferace will be T 1 1 1 ...
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Συχνά εμφανιζόμενοι όροι και φράσεις
ABC is equal altitude angle ABC angle BAC axis bisected centre circle ABCD circumference co-efficient cone conic section convergency curve cylinder described diameter divided draw equal angles equation equiangular equimultiples factors fluxion fore fraction geometrical progression given straight line gnomon greater Hence hyperbola join less Let ABC magnitudes multiple opposite parabola parallel parallelogram perpendicular plane angles polygon prism produced proportional pyramid Q. E. D. PROP Q. E. D. Proposition radius rectangle rectangle contained rectilineal figure remaining angle right angles segment shewn side BC similar sine solid angle solid parallelopiped spherical triangle square of AC subtract surd tang tangent Theorem third tiple triangle ABC vertex whence Wherefore
Δημοφιλή αποσπάσματα
Σελίδα 172 - If, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than, the other two sides of the triangle, but shall contain a greater angle. Let...
Σελίδα 191 - If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Σελίδα 190 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Σελίδα 196 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.
Σελίδα 192 - If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together •with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced.
Σελίδα 177 - That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
Σελίδα 209 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle...
Σελίδα 284 - The bases of a cylinder are the circles described by the two revolving opposite sides of the parallelogram.
Σελίδα 286 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...
Σελίδα 179 - Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.