標題:
有冇人可以教我做數(有關三角恆等式)
發問:
1:答案需用平方根號表示,不用計算機a.(cos45 over sin45)+(sin30 over tan45)b.(1-sin45^2 over cos60)-tan 30^2__________________________________________________________________2.化簡:sinθ^2 + sin(90-θ) 乘(sinθ over tan... 顯示更多 1:答案需用平方根號表示,不用計算機 a.(cos45 over sin45)+(sin30 over tan45) b.(1-sin45^2 over cos60)-tan 30^2 __________________________________________________________________ 2.化簡: sinθ^2 + sin(90-θ) 乘(sinθ over tan θ) __________________________________________________________________ 3.(使用代入法) tan 25=(h over 7500-x)----------(1) tan 63=(h over x)-------------(2) h=?
1:答案需用平方根號表示,不用計算機 a.(cos45 over sin45)+(sin30 over tan45) = tan 45 + ( 1/2 over 1 ) =1 + 1/2 = 3/2 b.(1-sin45^2 over cos60)-tan 30^2 =( cos^(2) 45 over 1/2 ) - (1/平方根(3))^(2) =( (平方根(2)/2)^(2) * 2) - 1/3 =( 2/4 * 2 ) - 1/3 =1 -1/3 =2/3 2.化簡: sinθ^2 + sin(90-θ) 乘(sinθ over tan θ) 先睇右邊 sin(90-θ) 乘 sinθ over tanθ = cosθ * sinθ over (sinθ/cosθ) = cosθsinθ * cosθ/sinθ = cos^(2)θ 成條式 sinθ^2 + sin(90-θ) 乘(sinθ over tan θ) =sinθ^(2) + cos^(2)θ =1 3.(使用代入法) tan 25 = (h over 7500-x)----------(1) tan 63 = (h over x)-------------(2) 其實我想問 7500 - x 有冇括號=.=\ 我當佢有啦 from (1) tan 25=(h over 7500-x) h / tan 25 = 7500 - x 7500 - h / tan 25 = x ...(1)' from (2) tan 63 = (h over x) h / tan 63 = x ...(2)' Sub (2)' into (1)' 7500 - h / tan 25 = h / tan 63 7500 = h / tan 63 + h / tan 25 7500 = ( tan25 h + tan63 h ) / (tan25tan63) 之後按計數機 7500 = 2.428918164...h / 0.915180308... 6863.852315...= 2.428918164...h h = 2825.888668... h = 2826(Corr to nearest integer )
其他解答:
1a. cos45 / sin45 + sin30 / tan45 root2 root2 1 = ----- 除 ----- + ---- 除 1 2 2 2 = 1 + 1/2 = 3/2 1b. (1-sin45^2 over cos60)-tan 30^2 1- [(root2)/ 2 ]^2 1 = ------------------------ － [----------] ^2 1/2 root3 = 2 - 1 － 1/3 = 2/3 2. sinθ^2 + sin(90-θ) 乘(sinθ over tan θ) = sinθ^2 + cosθ乘 sinθ 除 tan θ = sinθ^2 + cosθ乘 sinθ 乘 cosθ 除 sinθ =sinθ^2 + cosθ^2 =1
有冇人可以教我做數(有關三角恆等式)
發問:
1:答案需用平方根號表示,不用計算機a.(cos45 over sin45)+(sin30 over tan45)b.(1-sin45^2 over cos60)-tan 30^2__________________________________________________________________2.化簡:sinθ^2 + sin(90-θ) 乘(sinθ over tan... 顯示更多 1:答案需用平方根號表示,不用計算機 a.(cos45 over sin45)+(sin30 over tan45) b.(1-sin45^2 over cos60)-tan 30^2 __________________________________________________________________ 2.化簡: sinθ^2 + sin(90-θ) 乘(sinθ over tan θ) __________________________________________________________________ 3.(使用代入法) tan 25=(h over 7500-x)----------(1) tan 63=(h over x)-------------(2) h=?
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最佳解答:1:答案需用平方根號表示,不用計算機 a.(cos45 over sin45)+(sin30 over tan45) = tan 45 + ( 1/2 over 1 ) =1 + 1/2 = 3/2 b.(1-sin45^2 over cos60)-tan 30^2 =( cos^(2) 45 over 1/2 ) - (1/平方根(3))^(2) =( (平方根(2)/2)^(2) * 2) - 1/3 =( 2/4 * 2 ) - 1/3 =1 -1/3 =2/3 2.化簡: sinθ^2 + sin(90-θ) 乘(sinθ over tan θ) 先睇右邊 sin(90-θ) 乘 sinθ over tanθ = cosθ * sinθ over (sinθ/cosθ) = cosθsinθ * cosθ/sinθ = cos^(2)θ 成條式 sinθ^2 + sin(90-θ) 乘(sinθ over tan θ) =sinθ^(2) + cos^(2)θ =1 3.(使用代入法) tan 25 = (h over 7500-x)----------(1) tan 63 = (h over x)-------------(2) 其實我想問 7500 - x 有冇括號=.=\ 我當佢有啦 from (1) tan 25=(h over 7500-x) h / tan 25 = 7500 - x 7500 - h / tan 25 = x ...(1)' from (2) tan 63 = (h over x) h / tan 63 = x ...(2)' Sub (2)' into (1)' 7500 - h / tan 25 = h / tan 63 7500 = h / tan 63 + h / tan 25 7500 = ( tan25 h + tan63 h ) / (tan25tan63) 之後按計數機 7500 = 2.428918164...h / 0.915180308... 6863.852315...= 2.428918164...h h = 2825.888668... h = 2826(Corr to nearest integer )
其他解答:
1a. cos45 / sin45 + sin30 / tan45 root2 root2 1 = ----- 除 ----- + ---- 除 1 2 2 2 = 1 + 1/2 = 3/2 1b. (1-sin45^2 over cos60)-tan 30^2 1- [(root2)/ 2 ]^2 1 = ------------------------ － [----------] ^2 1/2 root3 = 2 - 1 － 1/3 = 2/3 2. sinθ^2 + sin(90-θ) 乘(sinθ over tan θ) = sinθ^2 + cosθ乘 sinθ 除 tan θ = sinθ^2 + cosθ乘 sinθ 乘 cosθ 除 sinθ =sinθ^2 + cosθ^2 =1
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