prove that: sin6 thita+cos6 thita= 1-3sin2 thita.cos2 thita
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Given: sin6ө + cos6ө =1-3sin2өcos2ө Consider LHS =sin6ө + cos6ө =(sin2ө)3 + (cos2ө)3 = (sin2ө + cos2ө)3 − 3 (sin2ө cos2ө)(sin2ө + cos2ө) [a3 +b3+ b = (a+b)3 − 3ab(a+b)] = 1 − 3sin2ө.cos2ө [sin2ө + cos2ө = 1] =RHS Hence Proved.
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