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Q-1-1-cos-2-2A-2-cos-4-A-sin-4-A-2-sin-2-A-sin-2-120-0-A-sin-2-120-0-A-3-2-




Question Number 8401 by rhm last updated on 10/Oct/16
Q .1   1+cos^2 2A=2(cos^4  A+sin^4  A)  2.  sin^2 A+sin^2 (120^0 +A)+sin^2 (120^0 −A)  = (3/2)
Q.11+cos22A=2(cos4A+sin4A)2.sin2A+sin2(1200+A)+sin2(1200A)=32
Answered by ridwan balatif last updated on 10/Oct/16
Q.1  1+cos^2 2A=2(cos^4 A+sin^4 A)  2(cos^4 A+sin^4 A)=2((cos^2 A)^2 +(sin^2 A)^2 )                                         =2((cos^2 A+sin^2 A)^2 −2sin^2 Acos^2 A)                                         =2(1−(2sinAcosA).(2/2)sinAcosA)                                         =2(1−((sin^2 2A)/2))                                         =2−sin^2 2A                                         =1+1−sin^2 2A                                         =1+cos^2 2A  ∴1+cos2A=2(cos^4 A+sin^4 A) (PROVE)
Q.11+cos22A=2(cos4A+sin4A)Missing \left or extra \right=2((cos2A+sin2A)22sin2Acos2A)=2(1(2sinAcosA).22sinAcosA)=2(1sin22A2)=2sin22A=1+1sin22A=1+cos22A1+cos2A=2(cos4A+sin4A)(PROVE)
Answered by ridwan balatif last updated on 10/Oct/16
2. sin^2 A+(sin(120^o +A))^2 +(sin(120^o −A))^2         sin^2 A+(sin120^o cosA+sinAcos120^o )^2 +(sin120^0 cosA−sinAcos120^o )^2       sin^2 A+((1/2)(√3)cosA−(1/2)sinA)^2 +((1/2)(√3)cosA+(1/2)sinA)^2       sin^2 A+{((1/2)(√3)cosA−(1/2)sinA)+((1/2)(√3)cosA+(1/2)sinA)}^2 −2((1/2)(√3)cosA−(1/2)sinA)((1/2)(√3)cosA+(1/2)sinA)      sin^2 A+((√3)cosA)^2 −2((3/4)cos^2 A−(1/4)sin^2 A)     (3/2)sin^2 A+(3/2)cos^2 A     (3/2)(sin^2 A+cos^2 A)      (3/2)  ∴sin^2 A+sin^2 (120^o +A)+sin^2 (120^o −A)=(3/2)
2.sin2A+(sin(120o+A))2+(sin(120oA))2sin2A+(sin120ocosA+sinAcos120o)2+(sin1200cosAsinAcos120o)2sin2A+(123cosA12sinA)2+(123cosA+12sinA)2sin2A+{(123cosA12sinA)+(123cosA+12sinA)}22(123cosA12sinA)(123cosA+12sinA)sin2A+(3cosA)22(34cos2A14sin2A)32sin2A+32cos2A32(sin2A+cos2A)32sin2A+sin2(120o+A)+sin2(120oA)=32

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