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12 الحساب المثلثي – الجزء الثاني

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1 ‫اﻟﻤﺜﻠﺜﻲ‬ ‫اﻟﺤﺴﺎب‬–‫اﻟﺠﺰء‬2- ‫اﻟﻤﻨﺘﻈﺮة‬ ‫اﻟﻘﺪرات‬ ‫ﻋﻠﻰ‬ ‫ﻣﺜﻠﺜﻴﺔ‬ ‫ﻣﺘﺮاﺟﺤﺔ‬ ‫أو‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬ ‫وﻗﺮاءة‬ ‫ﺗﻤﺜﻴﻞ‬ ‫ﻣﻦ‬ ‫اﻟﺘﻤﻜﻦ‬ ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫اﻟﺪرس‬‫اﻷول‬‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﺪورة‬ ‫اﻟﺴﺎﻋﺎت‬ ‫ﻋﺪد‬:15 I-‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ 1-‫اﻟﻤﻌﺎدﻟﺔ‬cosx a= ‫ﻣﺜﺎل‬1‫ﺣﻞ‬ 1 cos 2 x x∈ = ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻟﺪﻳﻨﺎ‬ 1 : 2 x∆ =‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻳﻘﻄﻊ‬ ‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﻓﻲ‬M‫و‬'M‫اﻟﺮﺋﻴﺴﻴﻴﻦ‬ ‫اﻟﻤﻨﺤﻨﻴﻴﻦ‬ ‫أﻓﺼﻮﻟﻴﻬﻤﺎ‬ ‫هﻤﺎ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ 3 π ‫و‬ 3 π -. ‫أن‬ ‫ﺑﻤﺎ‬2 3 k π π+‫ﺑﺤﻴﺚ‬k ∈‫ﻟﻠﻨﻘﻄﺔ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬M ‫و‬2 3 k π π+‫ﺑﺤﻴﺚ‬k ∈‫ﻟﻠﻨﻘﻄﺔ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬'M ‫أن‬ ‫ﻧﺴﺘﻨﺘﺞ‬ ‫ﻓﺈﻧﻨﺎ‬ 1 cos 2 x =‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬2 3 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫إذن‬2 / 2 / 3 3 S k k k k π π π π     = + ∈ ∪ − + ∈        ‫ﻣﺜﺎل‬2‫ﺣﻞ‬[ ] 1 2 ;2 cos 2 x xπ π∈ − = ‫ﻋﻠﻰ‬ ‫ﻓﻨﺤﺼﻞ‬ ‫اﻟﺴﺎﺑﻘﺔ‬ ‫اﻟﺨﻄﻮات‬ ‫ﻧﻔﺲ‬ ‫ﻧﺘﺒﻊ‬ 1 cos 2 x =‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬2 3 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻧﺤﻞ‬ ‫أﻧﻨﺎ‬ ‫وﺣﻴﺚ‬[ ]2 ;2π π− ‫ﻓﺎن‬2 2 2 3 k π π π π− ≤ + ≤‫أو‬2 2 2 3 k π π π π− ≤ − + ≤ ‫ﻟﺪﻳﻨﺎ‬2 2 2 3 k π π π π− ≤ + ≤‫ﺗﻜﺎﻓﺊ‬ 7 5 6 6 k− ≤ ≤‫ﺗﻜﺎﻓﺊ‬1k = −‫أو‬0k = ‫وﻣﻨﻪ‬ 3 x π =‫أو‬ 5 3 x π = − ‫ﻟﺪﻳﻨﺎ‬2 2 2 3 k π π π π− ≤ − + ≤‫ﺗﻜﺎﻓﺊ‬ 5 7 6 6 k− ≤ ≤‫ﺗﻜﺎﻓﺊ‬1k =‫أو‬0k = ‫وﻣﻨﻪ‬ 3 x π = −‫أو‬ 5 3 x π = ‫إذن‬ 5 5 ; ; ; 3 3 3 3 S π π π π− −  =     ‫ﺧﻼﺻﺔ‬*‫اﻟﻤﻌﺎدﻟﺔ‬cosx a=‫آﺎن‬ ‫إذا‬ ‫ﺣﻼ‬ ‫ﺗﻘﺒﻞ‬ ‫ﻻ‬1 1a a− ∨≺ *cos 1x x∈ =‫وﻓ‬ ‫إذا‬‫آﺎن‬ ‫إذا‬ ‫ﻘﻂ‬/ 2k x kπ∈ = *cos 1x x∈ = −‫آﺎن‬ ‫إذا‬ ‫وﻓﻘﻂ‬ ‫إذا‬/ 2k x kπ π∈ = + *‫آﺎن‬ ‫إذا‬1 1a− ≺ ≺‫ﻋﻨﺼﺮ‬ ‫ﻳﻮﺟﺪ‬ ‫ﻓﺎن‬α‫ﻣﻦ‬] [0;π‫ﺣﻴﺚ‬cos aα = ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬ ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬cosx a=‫ﻓﻲ‬‫هﻲ‬2x kα π= +‫أو‬2x kα π= − +‫ﺣﻴﺚ‬k ∈ { } { }2 / 2 /S k k k kα π α π= + ∈ ∪ − + ∈ ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
2 ‫ﺗﻤﺮﻳﻦ‬‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬ ( ) ] ] [ [ 2 3 3 cos cos 2 ;3 cos 2 3 4 2 ;2 2cos 3cos 1 0 x x x x x x x x π π π π π π     ∈ + = ∈ − − = −        ∈ + + = ‫اﻟﺤﻞ‬ *‫ﻧﺤﻞ‬( )cos cos 2 3 x x x π  ∈ + =    ( )cos cos 2 3 x x π  + =    ‫ﺗﻜﺎﻓﺊ‬2 2 3 x x k π π= + +‫أو‬2 2 3 x x k π π= − − +‫ﺣﻴ‬‫ﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬3 2 3 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬ 2 9 3 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫إذن‬ 2 2 / / 3 9 3 S k k k k π π π π     = + ∈ ∪ − + ∈        *‫ﻧﺤﻞ‬] ] 3 3 ;3 cos 2 4 2 x x π π π   ∈ − − = −    ‫أن‬ ‫ﻧﻌﻠﻢ‬ 3 cos 6 2 π =‫وﻣﻨﻪ‬ 5 3 cos cos 6 6 2 π π π     − = = −        ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬ 3 3 cos 2 4 2 x π  − = −    ‫ﺗﻜﺎﻓﺊ‬ 3 5 2 2 4 6 x k π π π− = +‫أو‬ 3 5 2 2 4 6 x k π π π− = − +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬ 19 2 2 12 x k π π= +‫أو‬2 2 12 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬ 19 24 x k π π= +‫أو‬ 24 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫ﺣﻴﺚ‬ ‫و‬] ];3x π π∈ −‫ﻓﺎن‬ ‫أﺟﻞ‬ ‫ﻣﻦ‬ 19 24 x k π π= +‫ﻟﺪﻳﻨﺎ‬ 19 3 24 k π π π π− + ≤≺‫أي‬ 19 1 3 24 k− + ≤≺ ‫وﻣ‬‫ﻨﻪ‬ 43 53 24 24 k− ≤≺ ‫ﺣﻴﺚ‬ ‫و‬k ∈‫ﻓﺎن‬1k = −‫أو‬0k =‫أو‬1k =‫أو‬2K = ‫إذن‬ 19 5 24 24 x π π π= − = −‫أو‬ 19 24 x π =‫أو‬ 19 43 24 24 x π π π= + =‫أو‬ 19 67 2 24 24 x π π π= + = ‫أﺟﻞ‬ ‫ﻣﻦ‬ 24 x k π π= − +‫ﻟﺪﻳﻨﺎ‬3 24 k π π π π− − + ≤≺‫أي‬ 1 1 3 24 k− − + ≤≺ ‫وﻣﻨﻪ‬ 23 73 24 24 k− ≤≺ ‫ﺣﻴﺚ‬ ‫و‬k ∈‫ﻓﺎن‬0k =‫أو‬1k =‫أو‬2K =‫أو‬3k = ‫إذن‬0 24 24 x π π π=− + ⋅ =−‫أو‬ 23 24 24 x π π π= − + =‫أو‬ 47 2 24 24 x π π π= − + =‫أو‬ 71 3 24 24 x π π π= − + = ‫إذن‬ 5 19 23 43 47 67 71 ; ; ; ; ; ; ; 24 24 24 24 24 24 24 24 S π π π π π π π π  = − −    *‫ﻧﺤﻞ‬[ [ 2 ;2 2cos 3cos 1 0x x xπ π∈ + + = ‫ﻧﻀﻊ‬cos x X=‫ﺗﺼﺒﺢ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬2 2 3 1 0X X+ + = ‫ﻟﻴﻜﻦ‬∆‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻣﻤﻴﺰ‬ ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
3 2 3 4 2 1 1∆ = − × × = ‫وﻣﻨﻪ‬ 3 1 1 4 2 X − + = = −‫أو‬ 3 1 1 4 X − − = = − ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬ 1 cos 2 x = −‫أو‬cos 1x = − ‫ﻟﺪﻳﻨﺎ‬cos 1x = −‫ﺗﻜﺎﻓﺊ‬/ 2k x kπ π∈ = + ‫و‬‫ﺣﻴﺚ‬[ [;2x π π∈‫ﻓﺎن‬2 2kπ π π π≤ + ≺‫أي‬ 1 0 2 k≤ ≺‫وﻣﻨﻪ‬0k =‫اذن‬x π= ‫ﻟﺪﻳﻨﺎ‬ 1 cos 2 x = −‫أي‬ 2 cos cos 3 x π = ‫وﻣﻨﻪ‬ 2 2 3 x k π π= +‫أو‬ 2 2 3 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫ﺣﻴﺚ‬ ‫و‬[ [;2x π π∈‫ﻓﺎن‬ ‫أﺟﻞ‬ ‫ﻣﻦ‬ 2 2 3 x k π π= − +‫ﻟﺪﻳﻨﺎ‬ 2 2 2 3 k π π π π≤ − + ≺‫أي‬ 5 4 6 3 k≤ ≺‫وﻣﻨﻪ‬1k = ‫إذن‬ 2 4 2 3 3 x π π π= − + = ‫أﺟﻞ‬ ‫ﻣﻦ‬ 2 2 3 x k π π= +‫ﻟﺪﻳﻨﺎ‬ 2 2 2 3 k π π π π≤ + ≺‫أي‬ 1 2 6 3 k≤ ≺‫ﻧﺴﺒﻲ‬ ‫ﺻﺤﻴﺢ‬ ‫ﻋﺪد‬ ‫ﻳﻮﺟﺪ‬ ‫ﻻ‬ ‫اﻷﺧﻴﺮة‬ ‫اﻟﻤﺘﻔﺎوﺗﺔ‬ ‫ﻳﺤﻘﻖ‬ ‫إذن‬ 4 ; 3 S π π   =     2-‫اﻟﻤﻌﺎدﻟﺔ‬sin x a= ‫ﻣﺜ‬‫ﺎل‬1‫ﺣﻞ‬ 3 sin 2 x x∈ = ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻟﺪﻳﻨﺎ‬ 3 : 2 y∆ =‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻳﻘﻄﻊ‬ ‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﻓﻲ‬M‫و‬'M‫اﻟﺮﺋﻴﺴﻴﻴﻦ‬ ‫اﻟﻤﻨﺤﻨﻴﻴﻦ‬ ‫أﻓﺼﻮﻟﻴﻬﻤﺎ‬ ‫هﻤﺎ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ 3 π ‫و‬ 2 3 3 π π π − =. ‫أن‬ ‫ﺑﻤﺎ‬2 3 k π π+‫ﺑﺤﻴﺚ‬k ∈‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬ ‫ﻟﻠﻨﻘﻄﺔ‬M‫و‬ 2 2 3 k π π+‫ﺑﺤﻴﺚ‬k ∈‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬ ‫ﻟﻠﻨﻘﻄﺔ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬'M‫أن‬ ‫ﻧﺴﺘﻨﺘﺞ‬ ‫ﻓﺈﻧﻨﺎ‬ 3 sin 2 x =‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬ 2 2 3 x k π π= +‫ﺣﻴﺚ‬k ∈ ‫إذن‬ 2 2 / 2 / 3 3 S k k k k π π π π     = + ∈ ∪ + ∈        ‫ﻣﺜﺎل‬2‫ﺣﻞ‬[ ] 3 2 ;3 sin 2 x xπ π∈ − = ‫ﻓﻨﺤﺼﻞ‬ ‫اﻟﺴﺎﺑﻘﺔ‬ ‫اﻟﺨﻄﻮات‬ ‫ﻧﻔﺲ‬ ‫ﻧﺘﺒﻊ‬‫ﻋﻠﻰ‬ 3 sin 2 x =‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬ 2 2 3 x k π π= +‫ﺣﻴﺚ‬k ∈ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻧﺤﻞ‬ ‫أﻧﻨﺎ‬ ‫وﺣﻴﺚ‬[ ]2 ;3π π− ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
4 ‫ﻓﺎن‬2 2 3 3 k π π π π− ≤ + ≤‫أو‬ 2 2 2 3 3 k π π π π− ≤ + ≤ ‫ﻟﺪﻳﻨﺎ‬2 2 3 3 k π π π π− ≤ + ≤‫ﺗﻜﺎﻓﺊ‬ 7 8 6 6 k− ≤ ≤ ‫ﺗﻜﺎﻓﺊ‬1k = −‫أو‬0k =‫أو‬1k = ‫وﻣﻨﻪ‬ 7 3 x π =‫أو‬ 3 x π =‫أو‬ 5 3 x π = − ‫ﻟﺪﻳﻨﺎ‬‫ﻟﺪﻳﻨﺎ‬ 2 2 2 3 3 k π π π π− ≤ + ≤‫ﺗﻜﺎﻓﺊ‬ 8 7 6 6 k− ≤ ≤ ‫ﺗﻜﺎﻓﺊ‬1k = −‫أو‬0k =‫أو‬1k = ‫وﻣﻨﻪ‬ 8 3 x π =‫أو‬ 2 3 x π =‫أو‬ 4 3 x π = − ‫إذن‬ 5 4 2 7 8 ; ; ; ; ; 3 3 3 3 3 3 S π π π π π π− −  =     ‫ﺧﻼﺻﺔ‬‫اﻟﻤﻌﺎدﻟﺔ‬sin x a=‫آﺎن‬ ‫إذا‬ ‫ﺣﻼ‬ ‫ﺗﻘﺒﻞ‬ ‫ﻻ‬1 1a a− ∨≺ / 2 sin 1 2 k x k x x π π∈ = + ⇔ ∈ = / 2 sin 1 2 k x k x x π π∈ = − + ⇔ ∈ = − ‫آﺎن‬ ‫إذا‬1 1a− ≺ ≺‫ﻋﻨﺼﺮ‬ ‫ﻳﻮﺟﺪ‬ ‫ﻓﺎن‬α‫ﻣﻦ‬; 2 2 π π  −   ‫ﺣﻴﺚ‬sin aα = ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬sin x a=‫ﻓ‬‫ﻲ‬‫هﻲ‬2x kα π= +‫أو‬2x kπ α π= − +‫ﺣﻴﺚ‬k ∈ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬{ } { }2 / 2 /S k k k kα π π α π= + ∈ ∪ − + ∈ ‫ﺗﻤﺮﻳﻦ‬‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬( )sin 2 cos 3 3 x x x π  ∈ + =    ] ] 1 ;2 sin 2 4 2 x x π π π   ∈ − − = −    ‫اﻟﺤﻞ‬-------------------------------------------------------------------------------------- ‫ﻧ‬‫ﺤﻞ‬( )sin 2 cos 3 3 x x x π  ∈ + =    ( )sin 2 cos 3 3 x x π  + =    ‫ﺗﻜﺎﻓﺊ‬sin 2 sin 3 3 2 x x π π    + = −        ‫ﺗﻜﺎﻓﺊ‬2 3 2 3 2 x x k π π π+ = − +‫أو‬2 3 2 3 2 x x k π π π π+ = − + +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬5 2 6 x k π π= +‫أو‬2 6 x k π π− = +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬ 2 30 5 x k π π= +‫أو‬( )2 6 x k π π= − + −‫ﺣﻴﺚ‬k ∈ ‫إذن‬( ) 2 / 2 / 30 5 6 S k k k k π π π π     = + ∈ ∪ − + − ∈        ‫ﻧﺤﻞ‬] ] 1 ;2 sin 2 4 2 x x π π π   ∈ − − = −    1 sin 2 4 2 x π  − = −    ‫ﺗﻜﺎﻓﺊ‬sin 2 sin 4 6 x π π    − = −        ‫ﺗﻜﺎﻓﺊ‬2 2 4 6 x k π π π− = − +‫أو‬2 2 4 6 x k π π π π− = + +‫ﺣﻴﺚ‬k ∈ ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
5 ‫ﺗﻜﺎﻓﺊ‬2 2 12 x k π π= +‫أو‬ 17 2 2 12 x k π π= +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬ 24 x k π π= +‫أو‬ 17 24 x k π π= +‫ﺣﻴﺚ‬k ∈ ‫أن‬ ‫ﺣﻴﺚ‬ ‫و‬] ];2x π π∈ −‫ﻓﺎن‬ ‫أﺟﻞ‬ ‫ﻣﻦ‬ 24 x k π π= +‫ﻟ‬‫ﺪﻳﻨﺎ‬2 24 k π π π π− + ≤≺‫أي‬ 25 47 24 24 k− ≤≺‫ﻣﻨﻪ‬ ‫و‬1k = −‫أو‬0k =‫أو‬1k = ‫إذن‬ 23 24 24 x π π π= − = −‫أو‬ 24 x π =‫أو‬ 25 24 24 x π π π= + = ‫أﺟﻞ‬ ‫ﻣﻦ‬‫ﻟﺪﻳﻨﺎ‬ 17 2 24 k π π π π− + ≤≺‫وﻣﻨﻪ‬ 41 31 24 24 k− ≤≺‫ﻣﻨﻪ‬ ‫و‬1k = −‫أو‬0k =‫أو‬1k = ‫إذن‬ 17 7 24 24 x π π π= − = −‫أو‬ 17 24 x π =‫أو‬ 17 41 24 24 x π π π= + = ‫وﻣﻨﻪ‬ 23 7 17 25 41 ; ; ; ; ; 24 24 24 24 24 24 S π π π π π π  = − −    3-‫اﻟﻤﻌﺎدﻟﺔ‬tan x a= ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬tan 1x x∈ = − ‫ﻧﻌﺘﺒﺮ‬∆‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫اﻟﻤﻤﺎس‬( )C‫أﺻﻠﻬﺎ‬ ‫ﻓﻲ‬I، ‫اﻟﻨﻘﻄﺔ‬ ‫ﻧﺄﺧﺪ‬T‫ﻣﻦ‬∆‫ﺣﻴﺚ‬1−‫أﻓﺼﻮل‬T‫ﻓﻲ‬‫اﻟﻤﺤﻮر‬∆ ‫اﻟﻤﺴﺘﻘﻴﻢ‬( )OT‫ﻳﻘﻄﻊ‬‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬( )C ‫اﻟﻨﻘﻄﺘﻴﻦ‬ ‫ﻓﻲ‬M‫و‬'M‫أن‬ ‫ﻧﻌﻠﻢ‬tan( ) 1 4 π − = − ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬ 4 π −‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬M ‫أن‬ ‫وﺑﻤﺎ‬tan( ) tanx k xπ+ =‫ﻟﻜﻞ‬/ 2 x k k π π   ∈ − + ∈    ‫ﻓﺎن‬‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬‫هﻲ‬/ 4 x k k π π − = + ∈ ‫اذن‬/ 4 S k k π π −  = + ∈    ‫ﺧﺎﺻﻴﺔ‬ tan /x a x k kα π= ⇔ = + ∈‫ﺣﻴﺚ‬α‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬tan x a=‫ﻓﻲ‬; 2 2 π π  −   ‫اﻟﻤﻌﺎدﻟﺘﻴﻦ‬ ‫ﺣﻞ‬ ‫ﺗﻤﺮﻳﻦ‬ [ ]0;3 tan 2 3x xπ∈ = tan 2 tan 3 x x x π  ∈ − = −    II-‫اﻟﻤﺘ‬‫اﻟ‬ ‫ﺮاﺟﺤﺎت‬‫ﻤﺜﻠﺜﻴﺔ‬ ‫ﻣﺜﺎل‬1 ‫ﺣﻞ‬] ] 1 ; cos 2 x xπ π∈ − ≥ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫أوﻻ‬ ‫ﻧﺤﻞ‬] ] 1 ; cos 2 x xπ π∈ − = ‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬ ‫ﺧﻄﻮات‬ ‫ﺑﺈﺗﺒﺎع‬ ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
6 ] ] 1 ; cos 2 x xπ π∈ − =‫ﺗﻜﺎﻓﺊ‬ 3 x π =‫أو‬ 3 x π = − ‫ﻟﺘﻜﻦ‬ 3 M π      ‫و‬' 3 M π  −    ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻣﻦ‬ ‫ﻧﻘﻄﺘﻴﻦ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫هﻲ‬ ‫اﻟﻤﺘﺮاﺟﺤﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻟﻠﻨﻘﻂ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬( )C‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬‫إﻟﻰ‬‫اﻟﻘﻮس‬'M IM    ‫ﻓﻲ‬] ];π π− ‫هﻲ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وهﺬﻩ‬; 3 3 S π π−  =    ‫ﻣﺜﺎل‬2‫ﺣﻞ‬[ [ 1 0;3 cos 2 x xπ∈ ≥ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫أوﻻ‬ ‫ﻧﺤﻞ‬[ [ 1 0;3 cos 2 x xπ∈ = [ [ 1 0;3 cos 2 x xπ∈ =‫ﺗﻜﺎﻓﺊ‬ 3 x π =‫أو‬ 7 3 x π =‫أو‬ 5 3 x π = 3 π ‫و‬ 7 3 π ‫اﻟﻨﻘﻄﺔ‬ ‫ﻟﻨﻔﺲ‬ ‫ﻣﻨﺤﻨﻴﻴﻦ‬ ‫أﻓﺼﻮﻟﻴﻦ‬M، ‫ﻧﻌﺘﺒﺮ‬ 5 3 π ‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬'M ‫ﻟﻠﻨﻘﻂ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫هﻲ‬ ‫اﻟﻤﺘﺮاﺟﺤﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬( )C ‫اﻟﻘﻮس‬ ‫اﻟﻰ‬ ‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬'M IM    ‫ﻓﻲ‬[ [0;3π ‫هﻲ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وهﺬﻩ‬ 5 7 0; ; 3 3 3 S π π π    = ∪       ‫ﻣﺜﺎل‬3 ‫ﺣﻞ‬[ ]0;2 tan 3x xπ∈ ≥ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻧﺤﻞ‬[ ]0;2 tan 3x xπ∈ = [ ]0;2 tan 3x xπ∈ =‫ﺗﻜﺎﻓﺊ‬ 3 x π =‫أو‬ 4 3 x π = ‫ﻧﻌﺘﺒﺮ‬ 3 π ‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬A ‫و‬ 4 3 π ‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬B ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫هﻲ‬ ‫اﻟﻤﺘﺮاﺟﺤﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻟﻠﻨﻘﻂ‬( )C‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬‫إﻟﻰ‬‫اﻟﻘﻮﺳﻴﻦ‬ ‫اﺗﺤﺎد‬AJ    ‫و‬'BJ    ‫ﻓﻲ‬[ ]0;2π ‫هﻲ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وهﺬﻩ‬ 4 3 ; ; 3 2 3 2 S π π π π    = ∪       ‫ﺗﻤﺮﻳﻦ‬ ‫ﺣﻞ‬] ] 1 ; sin 2 x xπ π − ∈ − ] ] 1 0;4 sin 2 x xπ − ∈ [ ]0;2 tan 1x xπ∈ ≺ ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
7 ‫ﺣﻠﻬﺎ‬ ‫ﻓﻲ‬ ‫ﺗﺆول‬ ‫ﻣﺘﺮاﺟﺤﺎت‬‫إﻟﻰ‬‫أ‬ ‫ﻣﺘﺮاﺟﺤﺎت‬‫ﺳﺎﺳﻴﺔ‬ ‫ﺗﻤﺮﻳﻦ‬ ‫ﺣﻞ‬ [ ] 1 ; sin 3 2 x x π π π   ∈ − − ≤    [ ]0; tan3 3x xπ∈ ] ] ( )2 ; 4cos 2 1 2 cos 2 0x x xπ π∈ − − + + ≤ ] ] 1 tan ; 0 sin 2 x x x π π + ∈ − ≥ III-‫اﻟﻤﺤﻴﻄﻴﺔ‬ ‫اﻟﺰواﻳﺎ‬–‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﺮﺑﺎﻋﻴﺎت‬ 1-‫ﺗﻌﺮﻳﻒ‬ •‫اﻟﻤﺮآﺰﻳﺔ‬ ‫اﻟﺰاوﻳﺔ‬:‫اﻟﺪاﺋﺮة‬ ‫ﻣﺮآﺰ‬ ‫رأﺳﻬﺎ‬ ‫زاوﻳﺔ‬ ‫هﻲ‬ •‫اﻟﻤﺤﻴﻄﻴﺔ‬ ‫اﻟﺰاوﻳﺔ‬:‫ﺿﻠﻌﻴﻬﺎ‬ ‫ﺑﻴﻦ‬ ‫وﺗﺤﺼﺮ‬ ‫ﻟﻠﺪاﺋﺮة‬ ‫رأﺳﻬﺎ‬ ‫ﻳﻨﺘﻤﻲ‬ ‫زاوﻳﺔ‬ ‫هﻲ‬‫ﻣﻦ‬ ‫ﻗﻮﺳﺎ‬‫هﺬﻩ‬‫اﻟﺪاﺋﺮة‬ 2-‫ﺧﺎﺻﻴ‬‫ﺎت‬ ‫ﻧﺸ‬‫ﺎط‬1 ‫ﻟﺘﻜﻦ‬( )C‫ﻣﺮآﺰهﺎ‬ ‫داﺋﺮة‬O‫ﻧﻌﺘﺒﺮ‬A‫و‬B‫ﻣﻦ‬ ‫ﻣﺨﺘﻠﻔﺘﻴﻦ‬ ‫ﻧﻘﻄﺘﻴﻦ‬( )C‫ﻗﻄﺮﻳﺎ‬ ‫ﻣﺘﻘﺎﺑﻠﺘﻴﻦ‬ ‫ﻏﻴﺮ‬ ‫و‬M‫ﻧﻘﻄ‬‫ﺔ‬‫ﻣﻦ‬( )C‫ﺑﺤﻴﺚ‬AOB‫و‬AMB‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮان‬AB    1-‫أن‬ ‫ﺑﻴﻦ‬2AOB AMB=‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟﺤﺎﻻت‬ ‫ﻓﻲ‬ ‫أ‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫ب‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫ﻏﻴﺮ‬ ‫ﻧﻘﻄﺔ‬ ‫اﻋﺘﺒﺎر‬ ‫ﻳﻤﻜﻦ‬N‫ﻣﻦ‬( )C‫ﺣﻴﺚ‬N‫و‬O‫و‬M‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫أ‬ ‫ﺑﺎﺳﺘﻌﻤﺎل‬ ‫و‬/‫اﻟﻤﻄﻠﻮب‬ ‫ﺑﻴﻦ‬ ‫ﻣﺮﺗﻴﻦ‬ 2-‫ﻧﻌﺘﺒﺮ‬( )AT‫ﻟﻠﺪاﺋﺮة‬ ‫اﻟﻤﻤﺎس‬( )C.‫اﻟﺰاوﻳﺔ‬BAT‫اﻻﻟﺰاوﻳﺔ‬ ‫ﺗﺤﺼﺮﻩ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮ‬ ‫ﻣﺤﻴﻄﻴﺔ‬ ‫اﻟﻤﺮآﺰﻳﺔ‬AOB ‫أن‬ ‫ﺑﻴﻦ‬2AOB TAB= ‫اﻟﺤﻞ‬---------------------------- 1-‫أ‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫اﻟﻤﺜﻠﺚ‬OBM‫اﻟﺮأس‬ ‫ﻓﻲ‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬O ‫وﻣﻨﻪ‬2BOM BMOπ= − ‫ﺣﻴﺚ‬ ‫و‬BOM AOBπ= −‫ﻷن‬M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫ﻓﺎن‬2AOB BMO= ‫اذن‬2AOB AMB= ‫ب‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫ﻏﻴﺮ‬ N‫ﻣﻦ‬( )C‫ﺣﻴﺚ‬N‫و‬O‫و‬M‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫أ‬ ‫ﺣﺴﺐ‬/‫ﻟﺪﻳﻨﺎ‬2NOB NMB= ‫ﻟﺪﻳﻨﺎ‬OAM‫اﻟﺮأس‬ ‫ﻓﻲ‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬ ‫ﻣﺜﻠﺚ‬O ‫ﻣﻨﻪ‬ ‫و‬2AOM AMOπ= − ‫ﻟﺪﻳﻨﺎ‬( )AOB NOB AOMπ= − + ‫وﻣﻨﻪ‬( )2 2AOB NMB AMOπ π= − + − ( )2AOB AMO NMB= − ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
8 ‫إذن‬2AOB AMB= 2/‫أن‬ ‫ﺑﻴﻦ‬2AOB TAB= ( )AT‫ﻟﻠﺪاﺋﺮة‬ ‫اﻟﻤﻤﺎس‬( )C‫وﻣﻨﻪ‬ 2 OAB BAT π = − ‫ﻟﺪﻳﻨﺎ‬OAB‫ﻓﻲ‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬‫اﻟﺮأس‬O ‫وﻣﻨﻪ‬2OAB OABπ= − ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬2 2 OAB BAT π π   = − −    ‫إذن‬2AOB TAB= ‫ﺧﺎﺻﻴﺔ‬1 ‫ﻣﺤﻴ‬ ‫زاوﻳﺔ‬ ‫ﻗﻴﺎس‬ ‫ﺿﻌﻒ‬ ‫هﻮ‬ ‫داﺋﺮة‬ ‫ﻓﻲ‬ ‫ﻣﺮآﺰﻳﺔ‬ ‫زاوﻳﺔ‬ ‫ﻗﻴﺎس‬‫ﻧﻔ‬ ‫ﺗﺤﺼﺮ‬ ‫ﻄﻴﺔ‬‫هﺬﻩ‬ ‫ﺗﺤﺼﺮﻩ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻮس‬ ‫ﺲ‬ ‫اﻟﻤﺮآﺰﻳﺔ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻧﺸﺎط‬2 ‫ﻟﺘﻜﻦ‬A‫و‬B‫و‬C‫و‬D‫داﺋﺮة‬ ‫ﻣﻦ‬ ‫ﻣﺨﺘﻠﻔﺔ‬ ‫ﻧﻘﻂ‬( )C‫ﻣﺮآﺰهﺎ‬O ‫أن‬ ‫ﺑﻴﻦ‬ABC ADC π+ =‫أو‬ABC ADC= ‫ﺧﺎﺻﻴﺔ‬2 A‫و‬B‫و‬C‫داﺋﺮة‬ ‫ﻣﻦ‬ ‫ﻧﻘﻂ‬ ‫ﺛﻼث‬( )C‫و‬D‫اﻟﻤﺴﺘﻮى‬ ‫ﻣﻦ‬ ‫ﻣﺨﺘﻠﻔﺔ‬ ‫ﻧﻘﻂ‬ ‫ﺗﻜﻮن‬D‫اﻟﺪاﺋﺮة‬ ‫ﻣﻦ‬( )C‫آﺎن‬ ‫إذا‬ ‫ﻓﻘﻂ‬ ‫و‬ ‫إذا‬ABC ADC π+ =‫أو‬ABC ADC= 3-‫ﻣﺜﻠﺚ‬ ‫ﻓﻲ‬ ‫اﻟﺠﻴﺐ‬ ‫ﻋﻼﻗﺎت‬ ‫ﻧﺸﺎط‬3 ‫ﻟﻴﻜﻦ‬ABC‫و‬ ‫ﻣﺜﻠﺜﺎ‬R‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬‫ﺑﺎﻟﻤﺜﻠﺚ‬ ‫اﻟﻤﺤﻴﻄﺔ‬ABC ‫أن‬ ‫ﺑﻴﻦ‬2 sinsin sin BC AC AB R CA B = = =‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟﺤﺎﻻت‬ ‫ﻓﻲ‬ ‫أ‬/ABC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬A ‫ب‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫ﺟﻤﻴﻊ‬ABC‫ﺣﺎدة‬ ‫ج‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫إﺣﺪى‬ABC‫ﻣﻨﻔﺮﺟﺔ‬ ‫اﻟﺠﻮاب‬ ‫أ‬/ABC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬A sin sin 1 2 A π = =‫وﻣﻨﻪ‬2 sin BC BC R A = = sin 2 AC BC B BC R = =‫وﻣﻨﻪ‬2 sin AC R B = sin 2 AB AB C BC R = =‫وﻣﻨﻪ‬2 sin AB R C = ‫إذن‬2 sinsin sin BC AC AB R CA B = = = ‫ب‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫ﺟﻤﻴﻊ‬ABC‫ﺣﺎدة‬ ‫ﻧﻌﺘﺒﺮ‬D‫ﻧﻘﻄﺔ‬‫ﻣﻘﺎﺑﻠﺔ‬‫ﻣﻊ‬ ‫ﻗﻄﺮﻳﺎ‬C DBC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬B ‫ﻟﺪﻳﻨﺎ‬D A≡‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮان‬ ‫ﻣﺤﻴﻄﻴﺘﺎن‬ ‫زاوﻳﺘﺎن‬ sin 2 BC BC D DC R = =‫وﻣﻨﻪ‬2 sin BC R D =‫إذن‬2 sin BC R A = ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
9 ‫ﻟﺪﻳﻨﺎ‬DAC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬A ‫و‬CDA B≡‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮان‬ ‫ﻣﺤﻴﻄﻴﺘﺎن‬ ‫زاوﻳﺘﺎن‬ sin 2 AC AC CDA DC R = =‫وﻣﻨﻪ‬sin 2 AC B R =‫إذن‬2 sin AC R B = ‫ﻧﻘﻄﺔ‬ ‫ﻧﻌﺘﺒﺮ‬ ‫ﺑﺎﻟﻤﺜﻞ‬‫ﻣﻘﺎﺑﻠﺔ‬‫ﻣﻊ‬ ‫ﻗﻄﺮﻳﺎ‬A‫ﻧﺒﻴﻦ‬ ‫و‬2 sin AB R C = ‫إذن‬2 sinsin sin BC AC AB R CA B = = = ‫ج‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫إﺣﺪى‬ABC‫ﻣﻨﻔﺮﺟﺔ‬ ‫ﻟﻨﻔﺘﺮض‬‫أن‬A‫ﻣﻨﻔﺮﺟﺔ‬ ‫ﻧﻌﺘﺒﺮ‬D‫ﻣﻊ‬ ‫ﻗﻄﺮﻳﺎ‬ ‫ﻣﻘﺎﺑﻠﺔ‬ ‫ﻧﻘﻄﺔ‬C ˆA‫و‬ˆD‫وﻣﻦ‬ ‫ﻣﺘﻜﺎﻣﻠﺘﺎن‬sin sinD A= sin 2 BC BC D DC R = =‫وﻣﻨﻪ‬2 sin BC R D =‫إذن‬2 sin BC R A = ‫اﻟﺰاوﻳﺘﺎن‬B‫و‬C‫ﺣﺎدﺗﺎن‬ ‫ب‬ ‫ﺣﺴﺐ‬/‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬2 sin AB R C =‫و‬2 sin AC R B = ‫إذن‬2 sinsin sin BC AC AB R CA B = = = ‫ﺧﺎﺻﻴﺔ‬ ‫ﻟﻴﻜﻦ‬ABC‫ﻣﺜﻠﺜﺎ‬‫و‬R‫اﻟﻤﺤﻴﻄﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬‫ﺑﻪ‬ 2 sinsin sin BC AC AB R CA B = = = 4-‫اﻟﻤﺜﻠﺚ‬ ‫ﻓﻲ‬ ‫ﻋﻼﻗﺎت‬)‫اﻟﻤﺴﺎﺣﺔ‬-‫اﻟﻤﺤﻴﻂ‬( ‫ﻧﺸﺎط‬ ‫ﻟﻴﻜﻦ‬ABC‫و‬ ‫ﻣﺜﻠﺜﺎ‬H‫ﻟـ‬ ‫اﻟﻌﻤﻮدي‬ ‫اﻟﻤﺴﻘﻂ‬A‫ﻋﻠﻰ‬( )BC‫و‬S‫ﻣﺴﺎﺣﺘﻪ‬ 1-‫أن‬ ‫ﺑﻴﻦ‬( )1 sin 2 S BC AC C= × × 2-‫ﻟﻴﻜﻦ‬r‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬‫ﺑﺎﻟﻤﺜﻠﺚ‬ ‫اﻟﻤﺤﺎﻃﺔ‬ABC‫و‬O‫ﻣﺮآﺰهﺎ‬ ‫أ‬/‫ﻣﺴﺎﺣﺔ‬ ‫أﺣﺴﺐ‬AOC‫ﺑﺪﻻﻟﺔ‬r‫و‬AC ‫ب‬/‫أن‬ ‫ﺑﻴﻦ‬ 1 2 S p r= ×‫ﺣﻴﺚ‬p‫اﻟﻤﺜﻠﺚ‬ ‫ﻣﺤﻴﻂ‬ABC ‫ﺧﺎﺻﻴﺔ‬ ‫ﻟﻴﻜﻦ‬ABC‫و‬ ‫ﻣﺜﻠﺜﺎ‬r‫و‬ ‫ﺑﻪ‬ ‫اﻟﻤﺤﺎﻃﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬S‫ﻣﺴﺎﺣﺘﻪ‬p‫ﻣﺤﻴﻄﻪ‬ ( )1 sin 2 S BC AC C= × × 1 2 S p r= × ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com

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12 الحساب المثلثي – الجزء الثاني

  • 1.
    1 ‫اﻟﻤﺜﻠﺜﻲ‬ ‫اﻟﺤﺴﺎب‬–‫اﻟﺠﺰء‬2- ‫اﻟﻤﻨﺘﻈﺮة‬ ‫اﻟﻘﺪرات‬ ‫ﻋﻠﻰ‬‫ﻣﺜﻠﺜﻴﺔ‬ ‫ﻣﺘﺮاﺟﺤﺔ‬ ‫أو‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬ ‫وﻗﺮاءة‬ ‫ﺗﻤﺜﻴﻞ‬ ‫ﻣﻦ‬ ‫اﻟﺘﻤﻜﻦ‬ ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫اﻟﺪرس‬‫اﻷول‬‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﺪورة‬ ‫اﻟﺴﺎﻋﺎت‬ ‫ﻋﺪد‬:15 I-‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ 1-‫اﻟﻤﻌﺎدﻟﺔ‬cosx a= ‫ﻣﺜﺎل‬1‫ﺣﻞ‬ 1 cos 2 x x∈ = ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻟﺪﻳﻨﺎ‬ 1 : 2 x∆ =‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻳﻘﻄﻊ‬ ‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﻓﻲ‬M‫و‬'M‫اﻟﺮﺋﻴﺴﻴﻴﻦ‬ ‫اﻟﻤﻨﺤﻨﻴﻴﻦ‬ ‫أﻓﺼﻮﻟﻴﻬﻤﺎ‬ ‫هﻤﺎ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ 3 π ‫و‬ 3 π -. ‫أن‬ ‫ﺑﻤﺎ‬2 3 k π π+‫ﺑﺤﻴﺚ‬k ∈‫ﻟﻠﻨﻘﻄﺔ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬M ‫و‬2 3 k π π+‫ﺑﺤﻴﺚ‬k ∈‫ﻟﻠﻨﻘﻄﺔ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬'M ‫أن‬ ‫ﻧﺴﺘﻨﺘﺞ‬ ‫ﻓﺈﻧﻨﺎ‬ 1 cos 2 x =‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬2 3 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫إذن‬2 / 2 / 3 3 S k k k k π π π π     = + ∈ ∪ − + ∈        ‫ﻣﺜﺎل‬2‫ﺣﻞ‬[ ] 1 2 ;2 cos 2 x xπ π∈ − = ‫ﻋﻠﻰ‬ ‫ﻓﻨﺤﺼﻞ‬ ‫اﻟﺴﺎﺑﻘﺔ‬ ‫اﻟﺨﻄﻮات‬ ‫ﻧﻔﺲ‬ ‫ﻧﺘﺒﻊ‬ 1 cos 2 x =‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬2 3 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻧﺤﻞ‬ ‫أﻧﻨﺎ‬ ‫وﺣﻴﺚ‬[ ]2 ;2π π− ‫ﻓﺎن‬2 2 2 3 k π π π π− ≤ + ≤‫أو‬2 2 2 3 k π π π π− ≤ − + ≤ ‫ﻟﺪﻳﻨﺎ‬2 2 2 3 k π π π π− ≤ + ≤‫ﺗﻜﺎﻓﺊ‬ 7 5 6 6 k− ≤ ≤‫ﺗﻜﺎﻓﺊ‬1k = −‫أو‬0k = ‫وﻣﻨﻪ‬ 3 x π =‫أو‬ 5 3 x π = − ‫ﻟﺪﻳﻨﺎ‬2 2 2 3 k π π π π− ≤ − + ≤‫ﺗﻜﺎﻓﺊ‬ 5 7 6 6 k− ≤ ≤‫ﺗﻜﺎﻓﺊ‬1k =‫أو‬0k = ‫وﻣﻨﻪ‬ 3 x π = −‫أو‬ 5 3 x π = ‫إذن‬ 5 5 ; ; ; 3 3 3 3 S π π π π− −  =     ‫ﺧﻼﺻﺔ‬*‫اﻟﻤﻌﺎدﻟﺔ‬cosx a=‫آﺎن‬ ‫إذا‬ ‫ﺣﻼ‬ ‫ﺗﻘﺒﻞ‬ ‫ﻻ‬1 1a a− ∨≺ *cos 1x x∈ =‫وﻓ‬ ‫إذا‬‫آﺎن‬ ‫إذا‬ ‫ﻘﻂ‬/ 2k x kπ∈ = *cos 1x x∈ = −‫آﺎن‬ ‫إذا‬ ‫وﻓﻘﻂ‬ ‫إذا‬/ 2k x kπ π∈ = + *‫آﺎن‬ ‫إذا‬1 1a− ≺ ≺‫ﻋﻨﺼﺮ‬ ‫ﻳﻮﺟﺪ‬ ‫ﻓﺎن‬α‫ﻣﻦ‬] [0;π‫ﺣﻴﺚ‬cos aα = ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬ ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬cosx a=‫ﻓﻲ‬‫هﻲ‬2x kα π= +‫أو‬2x kα π= − +‫ﺣﻴﺚ‬k ∈ { } { }2 / 2 /S k k k kα π α π= + ∈ ∪ − + ∈ ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
  • 2.
    2 ‫ﺗﻤﺮﻳﻦ‬‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬ ( )] ] [ [ 2 3 3 cos cos 2 ;3 cos 2 3 4 2 ;2 2cos 3cos 1 0 x x x x x x x x π π π π π π     ∈ + = ∈ − − = −        ∈ + + = ‫اﻟﺤﻞ‬ *‫ﻧﺤﻞ‬( )cos cos 2 3 x x x π  ∈ + =    ( )cos cos 2 3 x x π  + =    ‫ﺗﻜﺎﻓﺊ‬2 2 3 x x k π π= + +‫أو‬2 2 3 x x k π π= − − +‫ﺣﻴ‬‫ﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬3 2 3 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬ 2 9 3 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫إذن‬ 2 2 / / 3 9 3 S k k k k π π π π     = + ∈ ∪ − + ∈        *‫ﻧﺤﻞ‬] ] 3 3 ;3 cos 2 4 2 x x π π π   ∈ − − = −    ‫أن‬ ‫ﻧﻌﻠﻢ‬ 3 cos 6 2 π =‫وﻣﻨﻪ‬ 5 3 cos cos 6 6 2 π π π     − = = −        ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬ 3 3 cos 2 4 2 x π  − = −    ‫ﺗﻜﺎﻓﺊ‬ 3 5 2 2 4 6 x k π π π− = +‫أو‬ 3 5 2 2 4 6 x k π π π− = − +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬ 19 2 2 12 x k π π= +‫أو‬2 2 12 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬ 19 24 x k π π= +‫أو‬ 24 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫ﺣﻴﺚ‬ ‫و‬] ];3x π π∈ −‫ﻓﺎن‬ ‫أﺟﻞ‬ ‫ﻣﻦ‬ 19 24 x k π π= +‫ﻟﺪﻳﻨﺎ‬ 19 3 24 k π π π π− + ≤≺‫أي‬ 19 1 3 24 k− + ≤≺ ‫وﻣ‬‫ﻨﻪ‬ 43 53 24 24 k− ≤≺ ‫ﺣﻴﺚ‬ ‫و‬k ∈‫ﻓﺎن‬1k = −‫أو‬0k =‫أو‬1k =‫أو‬2K = ‫إذن‬ 19 5 24 24 x π π π= − = −‫أو‬ 19 24 x π =‫أو‬ 19 43 24 24 x π π π= + =‫أو‬ 19 67 2 24 24 x π π π= + = ‫أﺟﻞ‬ ‫ﻣﻦ‬ 24 x k π π= − +‫ﻟﺪﻳﻨﺎ‬3 24 k π π π π− − + ≤≺‫أي‬ 1 1 3 24 k− − + ≤≺ ‫وﻣﻨﻪ‬ 23 73 24 24 k− ≤≺ ‫ﺣﻴﺚ‬ ‫و‬k ∈‫ﻓﺎن‬0k =‫أو‬1k =‫أو‬2K =‫أو‬3k = ‫إذن‬0 24 24 x π π π=− + ⋅ =−‫أو‬ 23 24 24 x π π π= − + =‫أو‬ 47 2 24 24 x π π π= − + =‫أو‬ 71 3 24 24 x π π π= − + = ‫إذن‬ 5 19 23 43 47 67 71 ; ; ; ; ; ; ; 24 24 24 24 24 24 24 24 S π π π π π π π π  = − −    *‫ﻧﺤﻞ‬[ [ 2 ;2 2cos 3cos 1 0x x xπ π∈ + + = ‫ﻧﻀﻊ‬cos x X=‫ﺗﺼﺒﺢ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬2 2 3 1 0X X+ + = ‫ﻟﻴﻜﻦ‬∆‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻣﻤﻴﺰ‬ ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
  • 3.
    3 2 3 4 21 1∆ = − × × = ‫وﻣﻨﻪ‬ 3 1 1 4 2 X − + = = −‫أو‬ 3 1 1 4 X − − = = − ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬ 1 cos 2 x = −‫أو‬cos 1x = − ‫ﻟﺪﻳﻨﺎ‬cos 1x = −‫ﺗﻜﺎﻓﺊ‬/ 2k x kπ π∈ = + ‫و‬‫ﺣﻴﺚ‬[ [;2x π π∈‫ﻓﺎن‬2 2kπ π π π≤ + ≺‫أي‬ 1 0 2 k≤ ≺‫وﻣﻨﻪ‬0k =‫اذن‬x π= ‫ﻟﺪﻳﻨﺎ‬ 1 cos 2 x = −‫أي‬ 2 cos cos 3 x π = ‫وﻣﻨﻪ‬ 2 2 3 x k π π= +‫أو‬ 2 2 3 x k π π= − +‫ﺣﻴﺚ‬k ∈ ‫ﺣﻴﺚ‬ ‫و‬[ [;2x π π∈‫ﻓﺎن‬ ‫أﺟﻞ‬ ‫ﻣﻦ‬ 2 2 3 x k π π= − +‫ﻟﺪﻳﻨﺎ‬ 2 2 2 3 k π π π π≤ − + ≺‫أي‬ 5 4 6 3 k≤ ≺‫وﻣﻨﻪ‬1k = ‫إذن‬ 2 4 2 3 3 x π π π= − + = ‫أﺟﻞ‬ ‫ﻣﻦ‬ 2 2 3 x k π π= +‫ﻟﺪﻳﻨﺎ‬ 2 2 2 3 k π π π π≤ + ≺‫أي‬ 1 2 6 3 k≤ ≺‫ﻧﺴﺒﻲ‬ ‫ﺻﺤﻴﺢ‬ ‫ﻋﺪد‬ ‫ﻳﻮﺟﺪ‬ ‫ﻻ‬ ‫اﻷﺧﻴﺮة‬ ‫اﻟﻤﺘﻔﺎوﺗﺔ‬ ‫ﻳﺤﻘﻖ‬ ‫إذن‬ 4 ; 3 S π π   =     2-‫اﻟﻤﻌﺎدﻟﺔ‬sin x a= ‫ﻣﺜ‬‫ﺎل‬1‫ﺣﻞ‬ 3 sin 2 x x∈ = ‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻟﺪﻳﻨﺎ‬ 3 : 2 y∆ =‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻳﻘﻄﻊ‬ ‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﻓﻲ‬M‫و‬'M‫اﻟﺮﺋﻴﺴﻴﻴﻦ‬ ‫اﻟﻤﻨﺤﻨﻴﻴﻦ‬ ‫أﻓﺼﻮﻟﻴﻬﻤﺎ‬ ‫هﻤﺎ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ 3 π ‫و‬ 2 3 3 π π π − =. ‫أن‬ ‫ﺑﻤﺎ‬2 3 k π π+‫ﺑﺤﻴﺚ‬k ∈‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬ ‫ﻟﻠﻨﻘﻄﺔ‬M‫و‬ 2 2 3 k π π+‫ﺑﺤﻴﺚ‬k ∈‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬ ‫ﻟﻠﻨﻘﻄﺔ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬'M‫أن‬ ‫ﻧﺴﺘﻨﺘﺞ‬ ‫ﻓﺈﻧﻨﺎ‬ 3 sin 2 x =‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬ 2 2 3 x k π π= +‫ﺣﻴﺚ‬k ∈ ‫إذن‬ 2 2 / 2 / 3 3 S k k k k π π π π     = + ∈ ∪ + ∈        ‫ﻣﺜﺎل‬2‫ﺣﻞ‬[ ] 3 2 ;3 sin 2 x xπ π∈ − = ‫ﻓﻨﺤﺼﻞ‬ ‫اﻟﺴﺎﺑﻘﺔ‬ ‫اﻟﺨﻄﻮات‬ ‫ﻧﻔﺲ‬ ‫ﻧﺘﺒﻊ‬‫ﻋﻠﻰ‬ 3 sin 2 x =‫ﺗﻜﺎﻓﺊ‬2 3 x k π π= +‫أو‬ 2 2 3 x k π π= +‫ﺣﻴﺚ‬k ∈ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻧﺤﻞ‬ ‫أﻧﻨﺎ‬ ‫وﺣﻴﺚ‬[ ]2 ;3π π− ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
  • 4.
    4 ‫ﻓﺎن‬2 2 3 3 k π ππ π− ≤ + ≤‫أو‬ 2 2 2 3 3 k π π π π− ≤ + ≤ ‫ﻟﺪﻳﻨﺎ‬2 2 3 3 k π π π π− ≤ + ≤‫ﺗﻜﺎﻓﺊ‬ 7 8 6 6 k− ≤ ≤ ‫ﺗﻜﺎﻓﺊ‬1k = −‫أو‬0k =‫أو‬1k = ‫وﻣﻨﻪ‬ 7 3 x π =‫أو‬ 3 x π =‫أو‬ 5 3 x π = − ‫ﻟﺪﻳﻨﺎ‬‫ﻟﺪﻳﻨﺎ‬ 2 2 2 3 3 k π π π π− ≤ + ≤‫ﺗﻜﺎﻓﺊ‬ 8 7 6 6 k− ≤ ≤ ‫ﺗﻜﺎﻓﺊ‬1k = −‫أو‬0k =‫أو‬1k = ‫وﻣﻨﻪ‬ 8 3 x π =‫أو‬ 2 3 x π =‫أو‬ 4 3 x π = − ‫إذن‬ 5 4 2 7 8 ; ; ; ; ; 3 3 3 3 3 3 S π π π π π π− −  =     ‫ﺧﻼﺻﺔ‬‫اﻟﻤﻌﺎدﻟﺔ‬sin x a=‫آﺎن‬ ‫إذا‬ ‫ﺣﻼ‬ ‫ﺗﻘﺒﻞ‬ ‫ﻻ‬1 1a a− ∨≺ / 2 sin 1 2 k x k x x π π∈ = + ⇔ ∈ = / 2 sin 1 2 k x k x x π π∈ = − + ⇔ ∈ = − ‫آﺎن‬ ‫إذا‬1 1a− ≺ ≺‫ﻋﻨﺼﺮ‬ ‫ﻳﻮﺟﺪ‬ ‫ﻓﺎن‬α‫ﻣﻦ‬; 2 2 π π  −   ‫ﺣﻴﺚ‬sin aα = ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬sin x a=‫ﻓ‬‫ﻲ‬‫هﻲ‬2x kα π= +‫أو‬2x kπ α π= − +‫ﺣﻴﺚ‬k ∈ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬{ } { }2 / 2 /S k k k kα π π α π= + ∈ ∪ − + ∈ ‫ﺗﻤﺮﻳﻦ‬‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬( )sin 2 cos 3 3 x x x π  ∈ + =    ] ] 1 ;2 sin 2 4 2 x x π π π   ∈ − − = −    ‫اﻟﺤﻞ‬-------------------------------------------------------------------------------------- ‫ﻧ‬‫ﺤﻞ‬( )sin 2 cos 3 3 x x x π  ∈ + =    ( )sin 2 cos 3 3 x x π  + =    ‫ﺗﻜﺎﻓﺊ‬sin 2 sin 3 3 2 x x π π    + = −        ‫ﺗﻜﺎﻓﺊ‬2 3 2 3 2 x x k π π π+ = − +‫أو‬2 3 2 3 2 x x k π π π π+ = − + +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬5 2 6 x k π π= +‫أو‬2 6 x k π π− = +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬ 2 30 5 x k π π= +‫أو‬( )2 6 x k π π= − + −‫ﺣﻴﺚ‬k ∈ ‫إذن‬( ) 2 / 2 / 30 5 6 S k k k k π π π π     = + ∈ ∪ − + − ∈        ‫ﻧﺤﻞ‬] ] 1 ;2 sin 2 4 2 x x π π π   ∈ − − = −    1 sin 2 4 2 x π  − = −    ‫ﺗﻜﺎﻓﺊ‬sin 2 sin 4 6 x π π    − = −        ‫ﺗﻜﺎﻓﺊ‬2 2 4 6 x k π π π− = − +‫أو‬2 2 4 6 x k π π π π− = + +‫ﺣﻴﺚ‬k ∈ ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
  • 5.
    5 ‫ﺗﻜﺎﻓﺊ‬2 2 12 x k π π=+‫أو‬ 17 2 2 12 x k π π= +‫ﺣﻴﺚ‬k ∈ ‫ﺗﻜﺎﻓﺊ‬ 24 x k π π= +‫أو‬ 17 24 x k π π= +‫ﺣﻴﺚ‬k ∈ ‫أن‬ ‫ﺣﻴﺚ‬ ‫و‬] ];2x π π∈ −‫ﻓﺎن‬ ‫أﺟﻞ‬ ‫ﻣﻦ‬ 24 x k π π= +‫ﻟ‬‫ﺪﻳﻨﺎ‬2 24 k π π π π− + ≤≺‫أي‬ 25 47 24 24 k− ≤≺‫ﻣﻨﻪ‬ ‫و‬1k = −‫أو‬0k =‫أو‬1k = ‫إذن‬ 23 24 24 x π π π= − = −‫أو‬ 24 x π =‫أو‬ 25 24 24 x π π π= + = ‫أﺟﻞ‬ ‫ﻣﻦ‬‫ﻟﺪﻳﻨﺎ‬ 17 2 24 k π π π π− + ≤≺‫وﻣﻨﻪ‬ 41 31 24 24 k− ≤≺‫ﻣﻨﻪ‬ ‫و‬1k = −‫أو‬0k =‫أو‬1k = ‫إذن‬ 17 7 24 24 x π π π= − = −‫أو‬ 17 24 x π =‫أو‬ 17 41 24 24 x π π π= + = ‫وﻣﻨﻪ‬ 23 7 17 25 41 ; ; ; ; ; 24 24 24 24 24 24 S π π π π π π  = − −    3-‫اﻟﻤﻌﺎدﻟﺔ‬tan x a= ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬tan 1x x∈ = − ‫ﻧﻌﺘﺒﺮ‬∆‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫اﻟﻤﻤﺎس‬( )C‫أﺻﻠﻬﺎ‬ ‫ﻓﻲ‬I، ‫اﻟﻨﻘﻄﺔ‬ ‫ﻧﺄﺧﺪ‬T‫ﻣﻦ‬∆‫ﺣﻴﺚ‬1−‫أﻓﺼﻮل‬T‫ﻓﻲ‬‫اﻟﻤﺤﻮر‬∆ ‫اﻟﻤﺴﺘﻘﻴﻢ‬( )OT‫ﻳﻘﻄﻊ‬‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬( )C ‫اﻟﻨﻘﻄﺘﻴﻦ‬ ‫ﻓﻲ‬M‫و‬'M‫أن‬ ‫ﻧﻌﻠﻢ‬tan( ) 1 4 π − = − ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬ 4 π −‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬M ‫أن‬ ‫وﺑﻤﺎ‬tan( ) tanx k xπ+ =‫ﻟﻜﻞ‬/ 2 x k k π π   ∈ − + ∈    ‫ﻓﺎن‬‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬‫هﻲ‬/ 4 x k k π π − = + ∈ ‫اذن‬/ 4 S k k π π −  = + ∈    ‫ﺧﺎﺻﻴﺔ‬ tan /x a x k kα π= ⇔ = + ∈‫ﺣﻴﺚ‬α‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬tan x a=‫ﻓﻲ‬; 2 2 π π  −   ‫اﻟﻤﻌﺎدﻟﺘﻴﻦ‬ ‫ﺣﻞ‬ ‫ﺗﻤﺮﻳﻦ‬ [ ]0;3 tan 2 3x xπ∈ = tan 2 tan 3 x x x π  ∈ − = −    II-‫اﻟﻤﺘ‬‫اﻟ‬ ‫ﺮاﺟﺤﺎت‬‫ﻤﺜﻠﺜﻴﺔ‬ ‫ﻣﺜﺎل‬1 ‫ﺣﻞ‬] ] 1 ; cos 2 x xπ π∈ − ≥ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫أوﻻ‬ ‫ﻧﺤﻞ‬] ] 1 ; cos 2 x xπ π∈ − = ‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬ ‫ﺧﻄﻮات‬ ‫ﺑﺈﺗﺒﺎع‬ ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
  • 6.
    6 ] ] 1 ; cos 2 xxπ π∈ − =‫ﺗﻜﺎﻓﺊ‬ 3 x π =‫أو‬ 3 x π = − ‫ﻟﺘﻜﻦ‬ 3 M π      ‫و‬' 3 M π  −    ‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻣﻦ‬ ‫ﻧﻘﻄﺘﻴﻦ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫هﻲ‬ ‫اﻟﻤﺘﺮاﺟﺤﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻟﻠﻨﻘﻂ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬( )C‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬‫إﻟﻰ‬‫اﻟﻘﻮس‬'M IM    ‫ﻓﻲ‬] ];π π− ‫هﻲ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وهﺬﻩ‬; 3 3 S π π−  =    ‫ﻣﺜﺎل‬2‫ﺣﻞ‬[ [ 1 0;3 cos 2 x xπ∈ ≥ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫أوﻻ‬ ‫ﻧﺤﻞ‬[ [ 1 0;3 cos 2 x xπ∈ = [ [ 1 0;3 cos 2 x xπ∈ =‫ﺗﻜﺎﻓﺊ‬ 3 x π =‫أو‬ 7 3 x π =‫أو‬ 5 3 x π = 3 π ‫و‬ 7 3 π ‫اﻟﻨﻘﻄﺔ‬ ‫ﻟﻨﻔﺲ‬ ‫ﻣﻨﺤﻨﻴﻴﻦ‬ ‫أﻓﺼﻮﻟﻴﻦ‬M، ‫ﻧﻌﺘﺒﺮ‬ 5 3 π ‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬'M ‫ﻟﻠﻨﻘﻂ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫هﻲ‬ ‫اﻟﻤﺘﺮاﺟﺤﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬( )C ‫اﻟﻘﻮس‬ ‫اﻟﻰ‬ ‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬'M IM    ‫ﻓﻲ‬[ [0;3π ‫هﻲ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وهﺬﻩ‬ 5 7 0; ; 3 3 3 S π π π    = ∪       ‫ﻣﺜﺎل‬3 ‫ﺣﻞ‬[ ]0;2 tan 3x xπ∈ ≥ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻧﺤﻞ‬[ ]0;2 tan 3x xπ∈ = [ ]0;2 tan 3x xπ∈ =‫ﺗﻜﺎﻓﺊ‬ 3 x π =‫أو‬ 4 3 x π = ‫ﻧﻌﺘﺒﺮ‬ 3 π ‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬A ‫و‬ 4 3 π ‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬B ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫هﻲ‬ ‫اﻟﻤﺘﺮاﺟﺤﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫ﻟﻠﻨﻘﻂ‬( )C‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬‫إﻟﻰ‬‫اﻟﻘﻮﺳﻴﻦ‬ ‫اﺗﺤﺎد‬AJ    ‫و‬'BJ    ‫ﻓﻲ‬[ ]0;2π ‫هﻲ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وهﺬﻩ‬ 4 3 ; ; 3 2 3 2 S π π π π    = ∪       ‫ﺗﻤﺮﻳﻦ‬ ‫ﺣﻞ‬] ] 1 ; sin 2 x xπ π − ∈ − ] ] 1 0;4 sin 2 x xπ − ∈ [ ]0;2 tan 1x xπ∈ ≺ ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
  • 7.
    7 ‫ﺣﻠﻬﺎ‬ ‫ﻓﻲ‬ ‫ﺗﺆول‬‫ﻣﺘﺮاﺟﺤﺎت‬‫إﻟﻰ‬‫أ‬ ‫ﻣﺘﺮاﺟﺤﺎت‬‫ﺳﺎﺳﻴﺔ‬ ‫ﺗﻤﺮﻳﻦ‬ ‫ﺣﻞ‬ [ ] 1 ; sin 3 2 x x π π π   ∈ − − ≤    [ ]0; tan3 3x xπ∈ ] ] ( )2 ; 4cos 2 1 2 cos 2 0x x xπ π∈ − − + + ≤ ] ] 1 tan ; 0 sin 2 x x x π π + ∈ − ≥ III-‫اﻟﻤﺤﻴﻄﻴﺔ‬ ‫اﻟﺰواﻳﺎ‬–‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﺮﺑﺎﻋﻴﺎت‬ 1-‫ﺗﻌﺮﻳﻒ‬ •‫اﻟﻤﺮآﺰﻳﺔ‬ ‫اﻟﺰاوﻳﺔ‬:‫اﻟﺪاﺋﺮة‬ ‫ﻣﺮآﺰ‬ ‫رأﺳﻬﺎ‬ ‫زاوﻳﺔ‬ ‫هﻲ‬ •‫اﻟﻤﺤﻴﻄﻴﺔ‬ ‫اﻟﺰاوﻳﺔ‬:‫ﺿﻠﻌﻴﻬﺎ‬ ‫ﺑﻴﻦ‬ ‫وﺗﺤﺼﺮ‬ ‫ﻟﻠﺪاﺋﺮة‬ ‫رأﺳﻬﺎ‬ ‫ﻳﻨﺘﻤﻲ‬ ‫زاوﻳﺔ‬ ‫هﻲ‬‫ﻣﻦ‬ ‫ﻗﻮﺳﺎ‬‫هﺬﻩ‬‫اﻟﺪاﺋﺮة‬ 2-‫ﺧﺎﺻﻴ‬‫ﺎت‬ ‫ﻧﺸ‬‫ﺎط‬1 ‫ﻟﺘﻜﻦ‬( )C‫ﻣﺮآﺰهﺎ‬ ‫داﺋﺮة‬O‫ﻧﻌﺘﺒﺮ‬A‫و‬B‫ﻣﻦ‬ ‫ﻣﺨﺘﻠﻔﺘﻴﻦ‬ ‫ﻧﻘﻄﺘﻴﻦ‬( )C‫ﻗﻄﺮﻳﺎ‬ ‫ﻣﺘﻘﺎﺑﻠﺘﻴﻦ‬ ‫ﻏﻴﺮ‬ ‫و‬M‫ﻧﻘﻄ‬‫ﺔ‬‫ﻣﻦ‬( )C‫ﺑﺤﻴﺚ‬AOB‫و‬AMB‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮان‬AB    1-‫أن‬ ‫ﺑﻴﻦ‬2AOB AMB=‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟﺤﺎﻻت‬ ‫ﻓﻲ‬ ‫أ‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫ب‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫ﻏﻴﺮ‬ ‫ﻧﻘﻄﺔ‬ ‫اﻋﺘﺒﺎر‬ ‫ﻳﻤﻜﻦ‬N‫ﻣﻦ‬( )C‫ﺣﻴﺚ‬N‫و‬O‫و‬M‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫أ‬ ‫ﺑﺎﺳﺘﻌﻤﺎل‬ ‫و‬/‫اﻟﻤﻄﻠﻮب‬ ‫ﺑﻴﻦ‬ ‫ﻣﺮﺗﻴﻦ‬ 2-‫ﻧﻌﺘﺒﺮ‬( )AT‫ﻟﻠﺪاﺋﺮة‬ ‫اﻟﻤﻤﺎس‬( )C.‫اﻟﺰاوﻳﺔ‬BAT‫اﻻﻟﺰاوﻳﺔ‬ ‫ﺗﺤﺼﺮﻩ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮ‬ ‫ﻣﺤﻴﻄﻴﺔ‬ ‫اﻟﻤﺮآﺰﻳﺔ‬AOB ‫أن‬ ‫ﺑﻴﻦ‬2AOB TAB= ‫اﻟﺤﻞ‬---------------------------- 1-‫أ‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫اﻟﻤﺜﻠﺚ‬OBM‫اﻟﺮأس‬ ‫ﻓﻲ‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬O ‫وﻣﻨﻪ‬2BOM BMOπ= − ‫ﺣﻴﺚ‬ ‫و‬BOM AOBπ= −‫ﻷن‬M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫ﻓﺎن‬2AOB BMO= ‫اذن‬2AOB AMB= ‫ب‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫ﻏﻴﺮ‬ N‫ﻣﻦ‬( )C‫ﺣﻴﺚ‬N‫و‬O‫و‬M‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫أ‬ ‫ﺣﺴﺐ‬/‫ﻟﺪﻳﻨﺎ‬2NOB NMB= ‫ﻟﺪﻳﻨﺎ‬OAM‫اﻟﺮأس‬ ‫ﻓﻲ‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬ ‫ﻣﺜﻠﺚ‬O ‫ﻣﻨﻪ‬ ‫و‬2AOM AMOπ= − ‫ﻟﺪﻳﻨﺎ‬( )AOB NOB AOMπ= − + ‫وﻣﻨﻪ‬( )2 2AOB NMB AMOπ π= − + − ( )2AOB AMO NMB= − ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
  • 8.
    8 ‫إذن‬2AOB AMB= 2/‫أن‬ ‫ﺑﻴﻦ‬2AOBTAB= ( )AT‫ﻟﻠﺪاﺋﺮة‬ ‫اﻟﻤﻤﺎس‬( )C‫وﻣﻨﻪ‬ 2 OAB BAT π = − ‫ﻟﺪﻳﻨﺎ‬OAB‫ﻓﻲ‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬‫اﻟﺮأس‬O ‫وﻣﻨﻪ‬2OAB OABπ= − ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬2 2 OAB BAT π π   = − −    ‫إذن‬2AOB TAB= ‫ﺧﺎﺻﻴﺔ‬1 ‫ﻣﺤﻴ‬ ‫زاوﻳﺔ‬ ‫ﻗﻴﺎس‬ ‫ﺿﻌﻒ‬ ‫هﻮ‬ ‫داﺋﺮة‬ ‫ﻓﻲ‬ ‫ﻣﺮآﺰﻳﺔ‬ ‫زاوﻳﺔ‬ ‫ﻗﻴﺎس‬‫ﻧﻔ‬ ‫ﺗﺤﺼﺮ‬ ‫ﻄﻴﺔ‬‫هﺬﻩ‬ ‫ﺗﺤﺼﺮﻩ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻮس‬ ‫ﺲ‬ ‫اﻟﻤﺮآﺰﻳﺔ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻧﺸﺎط‬2 ‫ﻟﺘﻜﻦ‬A‫و‬B‫و‬C‫و‬D‫داﺋﺮة‬ ‫ﻣﻦ‬ ‫ﻣﺨﺘﻠﻔﺔ‬ ‫ﻧﻘﻂ‬( )C‫ﻣﺮآﺰهﺎ‬O ‫أن‬ ‫ﺑﻴﻦ‬ABC ADC π+ =‫أو‬ABC ADC= ‫ﺧﺎﺻﻴﺔ‬2 A‫و‬B‫و‬C‫داﺋﺮة‬ ‫ﻣﻦ‬ ‫ﻧﻘﻂ‬ ‫ﺛﻼث‬( )C‫و‬D‫اﻟﻤﺴﺘﻮى‬ ‫ﻣﻦ‬ ‫ﻣﺨﺘﻠﻔﺔ‬ ‫ﻧﻘﻂ‬ ‫ﺗﻜﻮن‬D‫اﻟﺪاﺋﺮة‬ ‫ﻣﻦ‬( )C‫آﺎن‬ ‫إذا‬ ‫ﻓﻘﻂ‬ ‫و‬ ‫إذا‬ABC ADC π+ =‫أو‬ABC ADC= 3-‫ﻣﺜﻠﺚ‬ ‫ﻓﻲ‬ ‫اﻟﺠﻴﺐ‬ ‫ﻋﻼﻗﺎت‬ ‫ﻧﺸﺎط‬3 ‫ﻟﻴﻜﻦ‬ABC‫و‬ ‫ﻣﺜﻠﺜﺎ‬R‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬‫ﺑﺎﻟﻤﺜﻠﺚ‬ ‫اﻟﻤﺤﻴﻄﺔ‬ABC ‫أن‬ ‫ﺑﻴﻦ‬2 sinsin sin BC AC AB R CA B = = =‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟﺤﺎﻻت‬ ‫ﻓﻲ‬ ‫أ‬/ABC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬A ‫ب‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫ﺟﻤﻴﻊ‬ABC‫ﺣﺎدة‬ ‫ج‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫إﺣﺪى‬ABC‫ﻣﻨﻔﺮﺟﺔ‬ ‫اﻟﺠﻮاب‬ ‫أ‬/ABC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬A sin sin 1 2 A π = =‫وﻣﻨﻪ‬2 sin BC BC R A = = sin 2 AC BC B BC R = =‫وﻣﻨﻪ‬2 sin AC R B = sin 2 AB AB C BC R = =‫وﻣﻨﻪ‬2 sin AB R C = ‫إذن‬2 sinsin sin BC AC AB R CA B = = = ‫ب‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫ﺟﻤﻴﻊ‬ABC‫ﺣﺎدة‬ ‫ﻧﻌﺘﺒﺮ‬D‫ﻧﻘﻄﺔ‬‫ﻣﻘﺎﺑﻠﺔ‬‫ﻣﻊ‬ ‫ﻗﻄﺮﻳﺎ‬C DBC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬B ‫ﻟﺪﻳﻨﺎ‬D A≡‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮان‬ ‫ﻣﺤﻴﻄﻴﺘﺎن‬ ‫زاوﻳﺘﺎن‬ sin 2 BC BC D DC R = =‫وﻣﻨﻪ‬2 sin BC R D =‫إذن‬2 sin BC R A = ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com
  • 9.
    9 ‫ﻟﺪﻳﻨﺎ‬DAC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬A ‫و‬CDAB≡‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮان‬ ‫ﻣﺤﻴﻄﻴﺘﺎن‬ ‫زاوﻳﺘﺎن‬ sin 2 AC AC CDA DC R = =‫وﻣﻨﻪ‬sin 2 AC B R =‫إذن‬2 sin AC R B = ‫ﻧﻘﻄﺔ‬ ‫ﻧﻌﺘﺒﺮ‬ ‫ﺑﺎﻟﻤﺜﻞ‬‫ﻣﻘﺎﺑﻠﺔ‬‫ﻣﻊ‬ ‫ﻗﻄﺮﻳﺎ‬A‫ﻧﺒﻴﻦ‬ ‫و‬2 sin AB R C = ‫إذن‬2 sinsin sin BC AC AB R CA B = = = ‫ج‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫إﺣﺪى‬ABC‫ﻣﻨﻔﺮﺟﺔ‬ ‫ﻟﻨﻔﺘﺮض‬‫أن‬A‫ﻣﻨﻔﺮﺟﺔ‬ ‫ﻧﻌﺘﺒﺮ‬D‫ﻣﻊ‬ ‫ﻗﻄﺮﻳﺎ‬ ‫ﻣﻘﺎﺑﻠﺔ‬ ‫ﻧﻘﻄﺔ‬C ˆA‫و‬ˆD‫وﻣﻦ‬ ‫ﻣﺘﻜﺎﻣﻠﺘﺎن‬sin sinD A= sin 2 BC BC D DC R = =‫وﻣﻨﻪ‬2 sin BC R D =‫إذن‬2 sin BC R A = ‫اﻟﺰاوﻳﺘﺎن‬B‫و‬C‫ﺣﺎدﺗﺎن‬ ‫ب‬ ‫ﺣﺴﺐ‬/‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬2 sin AB R C =‫و‬2 sin AC R B = ‫إذن‬2 sinsin sin BC AC AB R CA B = = = ‫ﺧﺎﺻﻴﺔ‬ ‫ﻟﻴﻜﻦ‬ABC‫ﻣﺜﻠﺜﺎ‬‫و‬R‫اﻟﻤﺤﻴﻄﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬‫ﺑﻪ‬ 2 sinsin sin BC AC AB R CA B = = = 4-‫اﻟﻤﺜﻠﺚ‬ ‫ﻓﻲ‬ ‫ﻋﻼﻗﺎت‬)‫اﻟﻤﺴﺎﺣﺔ‬-‫اﻟﻤﺤﻴﻂ‬( ‫ﻧﺸﺎط‬ ‫ﻟﻴﻜﻦ‬ABC‫و‬ ‫ﻣﺜﻠﺜﺎ‬H‫ﻟـ‬ ‫اﻟﻌﻤﻮدي‬ ‫اﻟﻤﺴﻘﻂ‬A‫ﻋﻠﻰ‬( )BC‫و‬S‫ﻣﺴﺎﺣﺘﻪ‬ 1-‫أن‬ ‫ﺑﻴﻦ‬( )1 sin 2 S BC AC C= × × 2-‫ﻟﻴﻜﻦ‬r‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬‫ﺑﺎﻟﻤﺜﻠﺚ‬ ‫اﻟﻤﺤﺎﻃﺔ‬ABC‫و‬O‫ﻣﺮآﺰهﺎ‬ ‫أ‬/‫ﻣﺴﺎﺣﺔ‬ ‫أﺣﺴﺐ‬AOC‫ﺑﺪﻻﻟﺔ‬r‫و‬AC ‫ب‬/‫أن‬ ‫ﺑﻴﻦ‬ 1 2 S p r= ×‫ﺣﻴﺚ‬p‫اﻟﻤﺜﻠﺚ‬ ‫ﻣﺤﻴﻂ‬ABC ‫ﺧﺎﺻﻴﺔ‬ ‫ﻟﻴﻜﻦ‬ABC‫و‬ ‫ﻣﺜﻠﺜﺎ‬r‫و‬ ‫ﺑﻪ‬ ‫اﻟﻤﺤﺎﻃﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬S‫ﻣﺴﺎﺣﺘﻪ‬p‫ﻣﺤﻴﻄﻪ‬ ( )1 sin 2 S BC AC C= × × 1 2 S p r= × ‫اﻟدروس‬ ‫ﻣن‬ ‫ﻟﻠﻣزﯾد‬ www.doros-bac.com