prove that 2(sin^6 A+cos^6 A)-3(sin^4 A+cos^4 A)+1=0​

Question

prove that 2(sin^6 A+cos^6 A)-3(sin^4 A+cos^4 A)+1=0​

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4 weeks 2021-10-29T22:21:57+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-29T22:23:14+00:00

    Answer:

    2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1

    = 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}+1

    = 2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}+1

    =2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)+1

    =2-6sin²θcos²θ-3+6sin²θcos²θ+1

    =-1+1

    =0 (Proved)

    0
    2021-10-29T22:23:15+00:00

    <body bgcolor=pink>

    \huge \purple{QUESTION}

    prove that 2(sin^6 A+cos^6 A)-3(sin^4 A+cos^4 A)+1=0

    \huge \purple{ANSWER}

    2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1

    = 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}+1

    = 2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}+1

    =2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)+1

    =2-6sin²θcos²θ-3+6sin²θcos²θ+1

    =-1+1

    =0 (Proved)

    HOPE SO IT IS HELPFUL..❣️✌️..

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