S=(100-90COSA)(100-90SINA) 已知 SINA+COSA=根号2* SIN(A+π) 求S的最大值和最小值
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![S=(100-90COSA)(100-90SINA) 已知 SINA+COSA=根号2* SIN(A+π) 求S的最大值和最小值](/uploads/image/z/12665568-48-8.jpg?t=S%3D%28100-90COSA%29%28100-90SINA%29+%E5%B7%B2%E7%9F%A5+SINA%2BCOSA%3D%E6%A0%B9%E5%8F%B72%2A+SIN%28A%2B%CF%80%EF%BC%89+%E6%B1%82S%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC)
S=(100-90COSA)(100-90SINA) 已知 SINA+COSA=根号2* SIN(A+π) 求S的最大值和最小值
S=(100-90COSA)(100-90SINA) 已知 SINA+COSA=根号2* SIN(A+π) 求S的最大值和最小值
S=(100-90COSA)(100-90SINA) 已知 SINA+COSA=根号2* SIN(A+π) 求S的最大值和最小值
sinA+cosA = √2×sin(A+π) = -√2×sinA
cosA = -(√2+1)×sinA
由于|cosA|≤1,|sinA|≤1
可得:|sinA|≤√2-1
S/100=(10-9cosA)(10-9sinA)
=100-90(sinA+cosA)+81sinAcosA
=100-90(-√2×sinA) + 81sinA[-(√2+1)×sinA]
=100+90√2×sinA-81(√2+1)sin²A
=50(√2+1)-81(√2+1)[sinA-(10-5√2)/9]²
由于0<(10-5√2)/9<√2-1
所以:
当sinA = (10-5√2)/9时,S有最大值,Smax = 5000(√2+1)
当sinA = -(√2-1)时,S有最小值,Smin = 900√2+100
sinA+cosA = √2×sin(A+π) = -√2×sinA
cosA = -(√2+1)×sinA
由于|cosA|≤1,|sinA|≤1
可得:|sinA|≤√2-1
S/100=(10-9cosA)(10-9sinA)
=100-90(sinA+cosA)+81sinAcosA
=100-90(-√2×sinA) + 81sinA[-(...
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sinA+cosA = √2×sin(A+π) = -√2×sinA
cosA = -(√2+1)×sinA
由于|cosA|≤1,|sinA|≤1
可得:|sinA|≤√2-1
S/100=(10-9cosA)(10-9sinA)
=100-90(sinA+cosA)+81sinAcosA
=100-90(-√2×sinA) + 81sinA[-(√2+1)×sinA]
=100+90√2×sinA-81(√2+1)sin²A
=50(√2+1)-81(√2+1)[sinA-(10-5√2)/9]²
由于0<(10-5√2)/9<√2-1
所以:
当sinA = (10-5√2)/9时,S有最大值,Smax = 5000(√2+1)
当sinA = -(√2-1)时,S有最小值,Smin = 900√2+100
好了。sinA+cosA = √2×sin(A+π) = -√2×sinA
cosA = -(√2+1)×sinA
由于|cosA|≤1,|sinA|≤1
可得:|sinA|≤√2-1
S/100=(10-9cosA)(10-9sinA)
=100-90(sinA+cosA)+81sinAcosA
=100-90(-√2×sinA) + 81sinA[-(√2+1)×sinA]
=100+90√2×sinA-81(√2+1)sin²A
=50(√2+1)-81(√2+1)[sinA-(10-5√2)/9]²
由于0<(10-5√2)/9<√2-1
所以:
当sinA = (10-5√2)/9时,S有最大值,Smax = 5000(√2+1)
当sinA = -(√2-1)时,S有最小值,Smin = 900√2+100
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