The Elements of Plane TrigonometryJ. Weale, 1854 - 119 páginas |
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Página 34
... formula for the sin 3A ; or by the 10th proposition of the 4th book of Euclid . cos 18 ° / ( 1 - sin2 18 ° ) - - √√ / { 1- ( ~ 5 - 1 ) " } = √ = / 16-5 + 2 / 5-1 16 = = 8 15 + √5 _ √ ( 5 + √5 ) 2√2 sin 36 ° 2 sin 18 ° cos 18 ...
... formula for the sin 3A ; or by the 10th proposition of the 4th book of Euclid . cos 18 ° / ( 1 - sin2 18 ° ) - - √√ / { 1- ( ~ 5 - 1 ) " } = √ = / 16-5 + 2 / 5-1 16 = = 8 15 + √5 _ √ ( 5 + √5 ) 2√2 sin 36 ° 2 sin 18 ° cos 18 ...
Página 35
... formula , page 34 , we may find the sin 15o . - sin 15 ° = 1 { √ / ( 1 + sin 50o ) — √ / ( 1 − sin 30 ° ) } - - { { √ / ( 1 + 4 ) - { √ / ( 1-4 ) } = 1 / √3 2 12 一= √3-1 2√2 This might have been found by putting sin 150 = sin ...
... formula , page 34 , we may find the sin 15o . - sin 15 ° = 1 { √ / ( 1 + sin 50o ) — √ / ( 1 − sin 30 ° ) } - - { { √ / ( 1 + 4 ) - { √ / ( 1-4 ) } = 1 / √3 2 12 一= √3-1 2√2 This might have been found by putting sin 150 = sin ...
Página 38
... formula let A = 9o , then , 2 A = 18o , 1 - sin 9o = { √ / ( 1 + sin 18 ° ) — √ / ( 1 − sin 18o ) } = 2 1+ - - { // ( 1 + 2/5 - 1 ) - √ / ( 1 - 1/5 - 1 ) } - = ¦ { √ ( 3 + √5 ) − √ ( 5 − √5 ) } ; - .. 4 sin 9 ° = √ ( 3 ...
... formula let A = 9o , then , 2 A = 18o , 1 - sin 9o = { √ / ( 1 + sin 18 ° ) — √ / ( 1 − sin 18o ) } = 2 1+ - - { // ( 1 + 2/5 - 1 ) - √ / ( 1 - 1/5 - 1 ) } - = ¦ { √ ( 3 + √5 ) − √ ( 5 − √5 ) } ; - .. 4 sin 9 ° = √ ( 3 ...
Página 45
... formula , page 29 , = 1 - cos A 1 + cos A tan A. = tan 3A . 1 sin A + sin 5A = 2 sin 3A cos 2A , / + ) . cos A + cos 5A = 2 cos 3A cos 24 , sin A + sin 3A + sin 5A = sin 3A + 2 sin 34 cos 2A , = = sin 3A ( 1 + 2 cos 2A ) , cos A + cos ...
... formula , page 29 , = 1 - cos A 1 + cos A tan A. = tan 3A . 1 sin A + sin 5A = 2 sin 3A cos 2A , / + ) . cos A + cos 5A = 2 cos 3A cos 24 , sin A + sin 3A + sin 5A = sin 3A + 2 sin 34 cos 2A , = = sin 3A ( 1 + 2 cos 2A ) , cos A + cos ...
Página 57
... formula , cos C = a2 + b2 - c2 2ab ; which may be adapted to logarithms in the following manner : c3 = a2 + b2 — 2ab cos C ; - but cos C 1-2 sin2 C ; = .. c2 = a2 + b2 - 2ab ( 1 - 2 sin3 & C ) = a3 + b2 − 2 ab + 4ab sin3 & C - = ( a ...
... formula , cos C = a2 + b2 - c2 2ab ; which may be adapted to logarithms in the following manner : c3 = a2 + b2 — 2ab cos C ; - but cos C 1-2 sin2 C ; = .. c2 = a2 + b2 - 2ab ( 1 - 2 sin3 & C ) = a3 + b2 − 2 ab + 4ab sin3 & C - = ( a ...
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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.