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0. Prove that:SIn6 |
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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) |
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