InterviewSolution
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InterviewSolution
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`sin^2 n theta- sin^2 (n-1)theta= sin^2 theta` where `n` is constant and `n != 0,1` |
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Answer» Correct Answer - `theta=k pi, (p pi)/(n-1), ((2q+1)pi)/(2n); k, p, q, n in Z` `sin^(2)n theta-sin^(2) (n-1) theta=sin^(2) theta` or `sin(n theta-(n-1) theta) sin (n theta+(n-1) theta)= sin^(2) theta` or `sin theta sin ((2n-1) theta)=sin^(2) theta` or `sin theta=0` or `sin((2n-1) theta)= sin theta` `rArr theta=k pi`, or `(2n-1) theta=m pi +(-1)^(m) theta, k m in Z` `rArr theta= kpi`, or `(2n-1) theta=mpi+theta`, when `m` is even and `(2n-1) theta=mpi - theta`, when `m` is odd `rArr theta=kpi`, or `theta=mpi//(2n-2)`, when `m` is even and `theta=mpi//2n`, when `m` is odd `rArr theta=kpi` or `theta=p pi//(n-1)`, and `theta=(2q+1)pi//2n`, where `k, p, n q in Z` |
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