close
標題:
4maths questions sin cos tan
發問:
最佳解答:
Q1) If csc (x) = 8, then cos (x) = √(82 - 1) / 8 = √63 / 8, so,2 cos2(x/2) - 1 = √63 / 8==> cos2(x/2) = (8 + √63) / 16 = (9 + 2√63 + 7) / 32==> cos2(x/2) = [(√9 + √7) / √32]2==> cos (x/2) = (√9 + √7) / √32 or -(√9 + √7) / √32 (rej. as cos (x/2) is +ve)So, sin (x/2) = (3 - √7) / 4√2, cos (x/2) = (3 + √7) / 4√2, tan (x/2) = (3 - √7) / (3 + √7) Q2) sin?(7x)= [sin2(7x)]2= [(1/2)(1 - cos(14x))]2= 1/4 - 1/2 cos(14x) + 1/4 cos2(14x)= 1/4 - 1/2 cos(14x) + 1/4 (1/2)(1 + cos(28x)= 3/8 - 1/2 cos(14x) + 1/8 cos(28x)∴ 1st 口 is 3/8, 2nd 口 is 14, 3rd 口 is 28. Q3) 7 cos(2x) = 7 cos2(x) - 5==> 7(2 cos2(x) - 1) = 7 cos2(x) - 5==> 7 cos2(x) = 2==> cos(x) = √(2/7) or -√(2/7)==> x = 57.7°, 122.3°, 237.7°, 302.7° (corr, to 1 dp)[or x = 1.01 rad, 2.13 rad, 4.15 rad, 5.28 rad (corr to 3sf)] Q4) 6 sin(2x) + 11 sin(x) = 0==> 12 sin(x) cos(x) + 11 sin(x) = 0==> sin (x) [12 cos (x) + 11] = 0==> sin (x) = 0 or cos (x) = -11/12==> x = 156.4°, 180.0°, 336.4° (corr, to 1 dp)[or x = 2.73 rad, 3.14 rad, 5.87 rad (corr to 3sf)] 2015-03-01 10:03:43 補充： Sorry, for Q1, as 90°<180° So, cos (x) = -√(82 - 1) / 8 = -√63 / 8 2 cos2(x/2) - 1 = -√63 / 8 ==> cos2(x/2) = (8 - √63) / 16 = (9 - 2√63 + 7) / 32 ... And so the answer of Q1 is : sin (x/2) = (3 + √7) / 4√2, cos (x/2) = (3 - √7) / 4√2, tan (x/2) = (3 + √7) / (3 - √7)
其他解答:1C924F1C0172E337
4maths questions sin cos tan
發問:
此文章來自奇摩知識+如有不便請留言告知
question 1:If csc (x) = 8, for 90deg < x < 180deg, thensin (x/2) =? ; cos (x/2) =? ; tan (x/2) =? question 2:Use half angle formulas to fill in the blanks in the identity below: 口? (sin(7x))^4 = 口 -1/2cos (口 x +1/8cos( 口x)question 3:Solve 7 cos(2 x) = 7 cos^2(x) - 5 for all solutions... 顯示更多 question 1: If csc (x) = 8, for 90deg < x < 180deg, then sin (x/2) =? ; cos (x/2) =? ; tan (x/2) =? question 2: Use half angle formulas to fill in the blanks in the identity below: 口? (sin(7x))^4 = 口 -1/2cos (口 x +1/8cos( 口x) question 3: Solve 7 cos(2 x) = 7 cos^2(x) - 5 for all solutions 0最佳解答:
Q1) If csc (x) = 8, then cos (x) = √(82 - 1) / 8 = √63 / 8, so,2 cos2(x/2) - 1 = √63 / 8==> cos2(x/2) = (8 + √63) / 16 = (9 + 2√63 + 7) / 32==> cos2(x/2) = [(√9 + √7) / √32]2==> cos (x/2) = (√9 + √7) / √32 or -(√9 + √7) / √32 (rej. as cos (x/2) is +ve)So, sin (x/2) = (3 - √7) / 4√2, cos (x/2) = (3 + √7) / 4√2, tan (x/2) = (3 - √7) / (3 + √7) Q2) sin?(7x)= [sin2(7x)]2= [(1/2)(1 - cos(14x))]2= 1/4 - 1/2 cos(14x) + 1/4 cos2(14x)= 1/4 - 1/2 cos(14x) + 1/4 (1/2)(1 + cos(28x)= 3/8 - 1/2 cos(14x) + 1/8 cos(28x)∴ 1st 口 is 3/8, 2nd 口 is 14, 3rd 口 is 28. Q3) 7 cos(2x) = 7 cos2(x) - 5==> 7(2 cos2(x) - 1) = 7 cos2(x) - 5==> 7 cos2(x) = 2==> cos(x) = √(2/7) or -√(2/7)==> x = 57.7°, 122.3°, 237.7°, 302.7° (corr, to 1 dp)[or x = 1.01 rad, 2.13 rad, 4.15 rad, 5.28 rad (corr to 3sf)] Q4) 6 sin(2x) + 11 sin(x) = 0==> 12 sin(x) cos(x) + 11 sin(x) = 0==> sin (x) [12 cos (x) + 11] = 0==> sin (x) = 0 or cos (x) = -11/12==> x = 156.4°, 180.0°, 336.4° (corr, to 1 dp)[or x = 2.73 rad, 3.14 rad, 5.87 rad (corr to 3sf)] 2015-03-01 10:03:43 補充： Sorry, for Q1, as 90°<180° So, cos (x) = -√(82 - 1) / 8 = -√63 / 8 2 cos2(x/2) - 1 = -√63 / 8 ==> cos2(x/2) = (8 - √63) / 16 = (9 - 2√63 + 7) / 32 ... And so the answer of Q1 is : sin (x/2) = (3 + √7) / 4√2, cos (x/2) = (3 - √7) / 4√2, tan (x/2) = (3 + √7) / (3 - √7)
其他解答:1C924F1C0172E337
文章標籤
全站熱搜
留言列表
