InterviewSolution
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InterviewSolution
| 1. |
2(sin⁶ &+cos⁶&)-3(sin⁴&+cos⁴&)+1=0 |
| Answer» 2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1\xa0=2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2-2sin2θcos2θ]+1The algebraic identitya3\xa0+ b3\xa0= (a+b)3\xa0- 3ab(a+b) and\xa0a2\xa0+ b2\xa0= (a+b)2\xa0- 2abare used in the above step wherea = sin2θ and b = cos2θ.writing sin2θ + cos2θ = 1, we have\xa0= 2[1-3 sin2θ cos2θ] -3[-2 sin2θcos2θ] + 1= 2-6 sin2θcos2θ -3 + 6 sin2θcos2θ + 1= -3+3=0 | |