## A System of Popular TrigonometryThis historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1827 Excerpt: ...therefore the remaining angles Aco', bn'm, are equal (art. 30, Geom.). But the angle mnvl being a right one, is equal, by Abt. 8, Geom., to Cod; and consequently the triangles Cgd, d'ot, are equiangular to each other. Wherefore, by Art. 112, Geom., Cd: D'm:: Cg: D, ... Cd x D'n = D'm x Cg. By addition, ... Cd X na' + CD x D'b--. cm X Dg + v'm x Cg; that is, CD X (no! + D' ), or Cd x D'g'--Dg x cm + D'm x Cg. Now, cp is the radius, and D'g' the sine of Acd' or (acd. + Dcd'); likewise Dg is the sine, Cg the cosine of Acd, and D'm is the sine, cm-the cosine of Dcd'. Hence, R X sin (a + B) = sin A X Cot B + sin B x cos A. This, &c. Obs. 1. The above is a general formula expressing the relation between the sine of the sum of any two angles and the sines and cosines of the angles themselves. Obs. 2. I f A = B, tlien the general formula becomes R X sin 2 A = 2 sin A X Cos A, the same result as we obtained in Art. 8, by other means. Art. 8'. The rectangle under the radius and sine of the difference of two angles is equal to the difference of the two rectangles under the sine of each and the cosine of the other. Let Acd, Dcd', be the angles, represented by A and B. Then R X sin (a--B)--sin A X cos B--sin B X cos A. Dem. Draw, as before, D'm perpendicular to Cd, and mo, Dg, D'g', perpendicular to Ca; likewise mn parallel to Ca till it meets D'g' (produced so as to intersect Cd mb). By Art. 112, Geom., Cd: cm:: Dg: mo, or nti (r; mo, by Art. 20, Geom.) Therefore Cd X nti--Dg x cm (art. 100, Geom.) Again, tlie right angles Cg'5, D'rni, are equal, and the angle at b is common to the triangles The theorem is easily derived from a geometrical consideration of the figure; but the above demonstration is preferable, first, as it is more consecutive, and, s... |

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