 A ^Vl -sin 4. 4. Remove the ambiguous signs in Ex. 3 when A is in turn an angle of each quadrant. 5. A wall 20 feet high bears S. 59° 5' E. ; find the width of its shadow on a horizontal plane when the sun is due S. and at an altitude of 60°. 6. Solve...  Plane Trigonometry - Page 84by Elmer Adelbert Lyman, Edwin Charles Goddard - 1899 - 91 pagesFull view - About this book
 | Elmer Adelbert Lyman, Edwin Charles Goddard - Trigonometry - 1900 - 156 pages
...£ £ ,1 prove 2 sin — = ± VI + siu A ± VI — sin A, and 2 cos — = ±Vl + sin A ^Vl -sin 4. 4. Remove the ambiguous signs in Ex. 3 when A is in...+ sin 3 x — 1 + cos x + cos 2 x. 7. Prove tan-1 i + tan-1 i = ^ 8. If A = 60°, B = 45°, C = 30°, evaluate tan A + tan B + tan C tan A tan B + tan... | |
 | Elmer Adelbert Lyman - Trigonometry - 1900 - 218 pages
.... prove 2 sin — = ± Vl + sin A ± Vl — sin ^4. and 2 cos - : = ± Vl + sin A =p Vl — sin A . 4. Remove the ambiguous signs in Ex. 3 when A is in...due S. and at an altitude of 60°. 6. Solve sin x -f sin 2 x -f sin 3 x = 1 + cos x -f cos 2 x. 7. Prove tan-1 - + tan-1 ! = -• 8. If A = 60°, B =... | |
 | Edward Rutledge Robbins - Logarithms - 1909 - 184 pages
...созж=— 1. 24. tan 2 0 tan 0 — l = 0. 25. sin 4 ж — ! 26. sin20 + sind = cos20 + cos 0. 27. sin x + sin 2 x + sin 3 x = 1 + cos x + cos 2 ж. 28. sin 5 x -\- sin 3 x -\- 2 cos x = 0. 29. cos x — cos 7 a; = sin 4 ж. 30. cos 3 в + л/3... | |
 | Fletcher Durell - Plane trigonometry - 1910 - 348 pages
...Find the least possible positive value of в which will satisfy the equation 2 V3 cos2 в = sin 0. 74. Solve sin x + sin 2 x + sin 3 x = 1 + cos x + cos 2 ÍB. 75. If sin 3 x + sin 2 x = sin x, find tan x. 76. Find the length of an arc intercepted by an... | |
 | Fletcher Durell - Logarithms - 1911 - 336 pages
...the least possible positive value of 0 which will satisfy the equation 2 V§ cos2 0 — sin 0. 74. Solve sin x + sin 2 x + sin 3 x = 1 + cos x -f- cos 2 #. 75. If sin 3 x -f- sin 2 x = sin #, find tan #. 76. Find the length of an arc intercepted... | |
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