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CBSE Class 11-science Answered

related to any topic in trigonometry..
Asked by archubelur2 | 25 Jul, 2010, 11:35: AM
answered-by-expert Expert Answer
Dear student,
 
We want to prove that
 
3(sinA - cosA)4 + 6(sinA + cosA)2 + 4(sin6A + cos6A) = 13
 
 
 
Here,

L.H.S. = 3(sinA - cosA)4 + 6(sinA + cosA)2 + 4(sin6A + cos6A)

           = 3(sinA – cosA)2(sinA – cosA)2 + 6(sin2A + cos2A + 2cosAsinA)

               + 4[(sin2A)3 + (cos2A)3]

          = 3(sin2A + cos2A – 2sinAcosA) (sin2A + cos2A – 2sinAcosA)

            + 6(1 + 2cosAsinA)

           + 4[(sin2A + cos2A)( sin4A - sin2A cos2A + cos4A)]

{Because sin2A + cos2A = 1 and a3 + b3 = (a + b)(a2 – ab + b2)}

         = 3(1 – 2sinAcosA) (1 – 2sinAcosA)

            + 6(1 + 2cosAsinA)

           + 4[(1)( sin4A - sin2A cos2A + cos4A)]

       = 3(1 – 4sinAcosA + 4 sin2Acos2A) + 6 + 12cosAsinA

          + 4(sin4A - sin2A cos2A + cos4A + 2sin2A cos2A - 2 sin2A cos2A)

     = 3 – 12sinAcosA + 12 sin2A cos2A + 6 + 12cosAsinA

        + 4(sin2A + cos2A)2 – 4(3 sin2A cos2A)

    = 3  + 6 – 12sinAcosA + 12 sin2A cos2A + 12cosAsinA

        + 4 – 12 sin2A cos2A

    = 3 + 6 + 4

    = 13

  = R.H.S.

Hence proved

 

Regards,

Team Topper Learning

 

 
 

L.H.S. = 3(sinA - cosA)4 + 6(sinA + cosA)2 + 4(sin6A + cos6A)

           = 3(sinA – cosA)2(sinA – cosA)2 + 6(sin2A + cos2A + 2cosAsinA)

               + 4[(sin2A)3 + (cos2A)3]

          = 3(sin2A + cos2A – 2sinAcosA) (sin2A + cos2A – 2sinAcosA)

            + 6(1 + 2cosAsinA)

           + 4[(sin2A + cos2A)( sin4A - sin2A cos2A + cos4A)]

{Because sin2A + cos2A = 1 and a3 + b3 = (a + b)(a2 – ab + b2)}

         = 3(1 – 2sinAcosA) (1 – 2sinAcosA)

            + 6(1 + 2cosAsinA)

           + 4[(1)( sin4A - sin2A cos2A + cos4A)]

       = 3(1 – 4sinAcosA + 4 sin2Acos2A) + 6 + 12cosAsinA

          + 4(sin4A - sin2A cos2A + cos4A + 2sin2A cos2A - 2 sin2A cos2A)

     = 3 – 12sinAcosA + 12 sin2A cos2A + 6 + 12cosAsinA

        + 4(sin2A + cos2A)2 – 4(3 sin2A cos2A)

    = 3  + 6 – 12sinAcosA + 12 sin2A cos2A + 12cosAsinA

        + 4 – 12 sin2A cos2A

    = 3 + 6 + 4

    = 13

Answered by | 27 Jul, 2010, 04:43: PM
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