InterviewSolution
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InterviewSolution
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`sin 3 alpha = 4 sin alpha sin(x + alpha) sin(x-alpha)`A. `n pi pm pi//4, AA n in Z`B. `n pi pm pi//3, AA n in Z`C. `n pi pm pi//9, AA n in Z`D. `n pi pm pi//12, AA n in Z` |
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Answer» Correct Answer - B We have `sin 3 alpha =4 sin alpha (sin^(2) x-sin^(2) alpha)` or `3 sin alpha -4 sin^(3) alpha =4 sin alpha sin^(2) x-4 sin^(3) alpha` or `3 sin alpha =4 sin alpha sin^(2)x` If `sin alpha ne0, sin^(2) x=3//4=(sqrt(3)//2)^(2)= sin^(2) (pi//3)`, Therefore `x= n pi pm pi//3, AA n in Z` If `sin alpha =0`, i.e., `alpha =n pi`, equation becomes an identity. |
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