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हा 14 1632. RERET % sin' 22+ L 1994! U - cos - +tan o ]2 . "Show that sin -. 5805LZ +cos’e, 16.¢ |
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Answer» Via right triangles and 'sohcahtoa':t = sin⁻¹(12/13) <==> sin t = 12/13 and cos t = 5/13s = cos⁻¹(4/5) <==> cos s = 4/5 and sin s = 3/5w = tan⁻¹(63/16) <==> sin w = 63/65 and cos w = 16/65.--------------Now, repeatedly use the sum of angles identitiescos(x+y) = cos x cos y - sin x sin y and sin(x+y) = sin x cos y + cos x sin y. So, cos(sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16))= cos(sin⁻¹(12/13) + cos⁻¹(4/5)) cos⁻¹(tan⁻¹(63/16)) - sin(sin⁻¹(12/13) + cos⁻¹(4/5)) sin(tan⁻¹(63/16)) = [cos(sin⁻¹ 12/13) cos(cos⁻¹ 4/5)) - sin(sin⁻¹ 12/13) sin(cos⁻¹ 4/5))] cos⁻¹(tan⁻¹(63/16))- [sin(sin⁻¹ 12/13) cos(cos⁻¹ 4/5) + cos(sin⁻¹ 12/13) sin(cos⁻¹ 4/5)] sin(tan⁻¹(63/16)) = [(5/13)(4/5) - (12/13)(3/5)] (16/65) - [(12/13)(4/5) + (5/13)(3/5)] (63/65)= -1. Since cos(sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16)) = -1,we conclude that sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16) = cos⁻¹(-1) = π. |
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