The Elements of Plane TrigonometryJ. Weale, 1854 - 119 páginas |
Dentro del libro
Resultados 1-5 de 5
Página 29
... adding ( 1 ) and ( 2 ) , sin A + sin B = 2 sin ( A + B ) cos ( A - B ) ........... . ( 3 ) . By subtracting ( 2 ) from ( 1 ) , sin A - sin B = 2 sin ( 4 – B ) cos & ( A + B ) - .. ( 4 ) . cos A = cos ( A + B ) + § ( A − B ) } = cos ...
... adding ( 1 ) and ( 2 ) , sin A + sin B = 2 sin ( A + B ) cos ( A - B ) ........... . ( 3 ) . By subtracting ( 2 ) from ( 1 ) , sin A - sin B = 2 sin ( 4 – B ) cos & ( A + B ) - .. ( 4 ) . cos A = cos ( A + B ) + § ( A − B ) } = cos ...
Página 33
... adding and subtracting we have cos2 A + 2 sin A cos A + sin3 A = 1 + sin 2A , and cos2 A - 2 sin A cos A + sin2 A = 1 − sin 24 . - By extracting the square root of each of these , cos A + sin A = √ ( 1 + sin 2A ) , cos A - sin A ...
... adding and subtracting we have cos2 A + 2 sin A cos A + sin3 A = 1 + sin 2A , and cos2 A - 2 sin A cos A + sin2 A = 1 − sin 24 . - By extracting the square root of each of these , cos A + sin A = √ ( 1 + sin 2A ) , cos A - sin A ...
Página 48
... ( √3 + √2 ) . 2 By adding these equations , we have √3 1 COS x = or , 2 √2 ; x = 30 ° , or 45o . By subtracting them cos y = 1 √2 ' √3 or ; .. y = 45o , or 30o . 2 ( 1 ) Prove the following theorems , sin A 48 TRIGONOMETRY .
... ( √3 + √2 ) . 2 By adding these equations , we have √3 1 COS x = or , 2 √2 ; x = 30 ° , or 45o . By subtracting them cos y = 1 √2 ' √3 or ; .. y = 45o , or 30o . 2 ( 1 ) Prove the following theorems , sin A 48 TRIGONOMETRY .
Página 108
... Adding and subtracting ( 1 ) and ( 2 ) , cos no sin no = = ( cos o + √√− 1 sin p ) " + ( cos p −√- 1 sin p ) " ... ( - 2 - ... ( 3 ) , ( cos ¢ + √− 1 sin 4 ) " — ( cos 4 — √− 1 sin 4 ) * ...... . ( 4 ) . 2 ... Expanding by the ...
... Adding and subtracting ( 1 ) and ( 2 ) , cos no sin no = = ( cos o + √√− 1 sin p ) " + ( cos p −√- 1 sin p ) " ... ( - 2 - ... ( 3 ) , ( cos ¢ + √− 1 sin 4 ) " — ( cos 4 — √− 1 sin 4 ) * ...... . ( 4 ) . 2 ... Expanding by the ...
Página 112
... adding and subtracting , we have 1 2 cos no = x2 + > and 2 √ = 1 sin nq = x2 - = ; 1 n- .. + n x2- + n -2 + + & c . 2 n . 1 cos ( n - 4 ) p , & c .; 2 = 2 cos no + 2n.cos ( n − 2 ) p + 2 n- .. 2 " -1 ( cos ( ) " = cos no + n cos ( n − 2 ) ...
... adding and subtracting , we have 1 2 cos no = x2 + > and 2 √ = 1 sin nq = x2 - = ; 1 n- .. + n x2- + n -2 + + & c . 2 n . 1 cos ( n - 4 ) p , & c .; 2 = 2 cos no + 2n.cos ( n − 2 ) p + 2 n- .. 2 " -1 ( cos ( ) " = cos no + n cos ( n − 2 ) ...
Otras ediciones - Ver todas
Términos y frases comunes
A+ cot a+b+c AB² AC² angle ACB angle of elevation AO² AP AP Asin BOP₁ centre circumference circumscribed cos A cos cos² cos³ cosec cosine cot A cot cot A+ cot² OPA cot³ diameter distance Divide equation feet formula given height hence hexagon inscribed circle logarithms negative nth root number of sides Oa cot OA² OB² perimeter perpendicular plane triangle quadrant r² cot radii radius ratio regular polygon right angle right-angled triangle sec² sector shew sin A cos sin A sin sin-¹ sin² sin² 18 sin³ sine sine and cosine square subtend subtracted tan-¹ tan² tan³ tangent triangle ABC unity ηφ
Pasajes populares
Página 97 - From a window near the bottom of a house, which seemed to be on a level with the bottom of a steeple, I took the angle of elevation of the top of the steeple, equal to 40° ; then from another window, 18 feet directly above the former, the like angle was 37° 30'.
Página 97 - Wanting to know the height of an inaccessible tower; at the least distance from it, on the same horizontal plane, I took its angle of elevation equal to 58°; then going 300 feet directly from it, found the angle there to be only 32°: required its height, and my distance from it at the first station?
Página 86 - The area of a regular inscribed hexagon is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles.
Página 51 - It depends on the principle, that the difference of the squares of two quantities is equal to the product of the sum and difference of the quantities.
Página 63 - If from one of the angles of a rectangle a perpendicular be drawn to its diagonal, and from, the point of their intersection lines be drawn perpendicular to the sides which contain the opposite angle...
Página 87 - The square on the side of a regular pentagon inscribed in a circle is equal to the sum of the squares on the sides of the regular hexagon and decagon inscribed in the same circle.
Página 93 - To THEIR DIFFERENCE ; - So IS THE TANGENT OF HALF THE SUM OF THE OPPOSITE ANGLES; To THE TANGENT OF HALF THEIR DIFFERENCE.
Página 98 - Wanting to know my distance from an inaccessible object 0, on the other side of a river; and having no instrument for taking angles, but only a chain or cord for measuring distances; from each of two stations, A and B, which were taken at 500 yards asunder, I measured in a direct line from the object 0 100 yards, viz. AC and BD each equal to 100 yards ; also the diagonal AD measured 550 yards, and the diagonal BC 560. What then was the distance of the object 0 from each, station A and B ? . C AO...
Página 97 - Being on a horizontal plane, and wanting to know the height of a tower placed on the top of an inaccessible hill : I took the angle of elevation of the top of the hill 40°, and of the top of the tower 51°; then measuring in a line directly from it to the distance of 200 feet farther, I found the angle at the top of the tower to be 33° 45'.
Página 101 - The hypotenuse AB of a right-angled triangle ABC is trisected in the points D, E; prove that if CD, CE be joined, the sum of the squares on the sides of the triangle CDE is equal to two-thirds of the square on AB.