標題:
SS1 Maths sin cos tan equatio
發問:
1. Given that sin B = - 2/3 and cos B > 0, find the value of cos B - tan B in surd form. 2. Simplify the following: tan (180 + B) + tan (90 - B) 更新: Ans: 1. 11 / 3√5 2. 1 / (sinB tanB)
最佳解答:
1) sin^2 B + cos^2 B = 1 (-2/3)^2 + cos^2 B = 1 cos^2 B = 1 - 4/9 = 5/9 cos B = √5 / 3 or - √5 /3 (rejected since cos B > 0) the value of cos B - tan B = cos B - (sinB)/cosB = √5 /3 - (-2/3)/ (√5 /3) = √5 /3 + 2√5 /5 = 11√5 /15 2) tan (180 + B) + tan (90 - B) = tan B + 1/tan B 2010-06-07 13:47:44 補充： = (sin B)/cosB + (cosB)/sinB = [(sinB)^2 + (cosB)^2] / (cosB sinB) = 1 / (cosB sinB) = 2 / sin2B 2010-06-07 14:12:43 補充： 樓主品學兼憂 : For Q1 : My answer 11√5 /15 = 11√5 / (3 * 5) = 11 / 3√5 = Given answer. For Q2 : I think the given answer is wrong , Set B = 45 tan (180 + B) + tan (90 - B) = tan 225 + tan 45 = 1 + 1 = 2 But the given answer 1 / (sinB tanB) = 1 / (sin45 tan45) = √2 not = 2 2010-06-07 14:12:51 補充： The answer should be 1 / (cosB sinB) 1 / (cos45 sin45) = 1 / (√2/2 * √2/2) = 2 2010-06-07 16:51:41 補充： = sinB / cosB + cosB / sinB = [(sinB)^2 + (cosB)^2] / (sinB cosB) = 1 / (sinB cosB) or = 2 / 2(sinB cosB) = 2 / sin2B
1. sin B = -2/3 (sin B) ^2 = 4/9 1 - (cos B) ^2 = 4/9 (cos B) ^2 = 5/9 cos B = (squr 5)/3 [cos B > 0] ------------- cos B - tan B =cos B - sin B/cos B =(cos B) ^2/cos B - sin B/cos B =((cos B) ^2 - sin B)/cos B =(5/9 - (-2/3)) / (squr 5)/3 =(7/9)/((squr 5)/3) =7/3(squr 5) ------------- 2. tan (180 + B) = tan B -------------- tan (90 - B) =sin (90 - B) / cos (90 - B) =cos B/sin B =1/tan B --------------- tan (180 + B) + tan (90 - B) =tan B + 1/tan B =sin B/cos B + cos B/sin B =(sin B) ^2/sin B cos B + (cos B) ^2/ sin B cos B =((sin B) ^2 + (cos B) ^2)/sin B cos B =1/sin B cos B 2010-06-07 13:57:05 補充： 第一題計錯... =(5/9 - (-2/3)) / (squr 5)/3--->=(5/9 - (-2/3)) / (squr 5)/3 =(7/9)/((squr 5)/3)------------>=(11/9)/((squr 5)/3) =7/3(squr 5)------------------->=11/3(squr 5) 11/3(squr 5) =11/3(squr 5) x squr 5/squr 5 =11(squr 5) /157638E748CCC65837
SS1 Maths sin cos tan equatio
發問:
1. Given that sin B = - 2/3 and cos B > 0, find the value of cos B - tan B in surd form. 2. Simplify the following: tan (180 + B) + tan (90 - B) 更新: Ans: 1. 11 / 3√5 2. 1 / (sinB tanB)
最佳解答:
1) sin^2 B + cos^2 B = 1 (-2/3)^2 + cos^2 B = 1 cos^2 B = 1 - 4/9 = 5/9 cos B = √5 / 3 or - √5 /3 (rejected since cos B > 0) the value of cos B - tan B = cos B - (sinB)/cosB = √5 /3 - (-2/3)/ (√5 /3) = √5 /3 + 2√5 /5 = 11√5 /15 2) tan (180 + B) + tan (90 - B) = tan B + 1/tan B 2010-06-07 13:47:44 補充： = (sin B)/cosB + (cosB)/sinB = [(sinB)^2 + (cosB)^2] / (cosB sinB) = 1 / (cosB sinB) = 2 / sin2B 2010-06-07 14:12:43 補充： 樓主品學兼憂 : For Q1 : My answer 11√5 /15 = 11√5 / (3 * 5) = 11 / 3√5 = Given answer. For Q2 : I think the given answer is wrong , Set B = 45 tan (180 + B) + tan (90 - B) = tan 225 + tan 45 = 1 + 1 = 2 But the given answer 1 / (sinB tanB) = 1 / (sin45 tan45) = √2 not = 2 2010-06-07 14:12:51 補充： The answer should be 1 / (cosB sinB) 1 / (cos45 sin45) = 1 / (√2/2 * √2/2) = 2 2010-06-07 16:51:41 補充： = sinB / cosB + cosB / sinB = [(sinB)^2 + (cosB)^2] / (sinB cosB) = 1 / (sinB cosB) or = 2 / 2(sinB cosB) = 2 / sin2B
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其他解答:1. sin B = -2/3 (sin B) ^2 = 4/9 1 - (cos B) ^2 = 4/9 (cos B) ^2 = 5/9 cos B = (squr 5)/3 [cos B > 0] ------------- cos B - tan B =cos B - sin B/cos B =(cos B) ^2/cos B - sin B/cos B =((cos B) ^2 - sin B)/cos B =(5/9 - (-2/3)) / (squr 5)/3 =(7/9)/((squr 5)/3) =7/3(squr 5) ------------- 2. tan (180 + B) = tan B -------------- tan (90 - B) =sin (90 - B) / cos (90 - B) =cos B/sin B =1/tan B --------------- tan (180 + B) + tan (90 - B) =tan B + 1/tan B =sin B/cos B + cos B/sin B =(sin B) ^2/sin B cos B + (cos B) ^2/ sin B cos B =((sin B) ^2 + (cos B) ^2)/sin B cos B =1/sin B cos B 2010-06-07 13:57:05 補充： 第一題計錯... =(5/9 - (-2/3)) / (squr 5)/3--->=(5/9 - (-2/3)) / (squr 5)/3 =(7/9)/((squr 5)/3)------------>=(11/9)/((squr 5)/3) =7/3(squr 5)------------------->=11/3(squr 5) 11/3(squr 5) =11/3(squr 5) x squr 5/squr 5 =11(squr 5) /157638E748CCC65837
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