InterviewSolution
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InterviewSolution
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Given that `cosx/2.cosx/4.cosx/8....=sinx/x` Then find the sum `1/2^2sec^2x/2+1/2^4sec^2x/4+...` |
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Answer» `"We have "cos""(x)/(2)cdotcos""(x)/(4)cdotcos""(x)/(8)...=(sin x)/(x)." Then find the sum "(1)/(2^(2))sec^(2)""(x)/(2)+(1)/(2^(4))sec^(2)""(x)/(4)+...` Taking log on both sides, we get `log cos""(x)/(2)+log cos""(x)/(4)+log cos""(x)/(8)...+...=log sin x-log x` Differentiating both sides with respect to x, we get `-(1)/(2)(sin""(x)/(2))/(cos""(x)/(2))-(1)/(4)(sin""(x)/(4))/(cos""(x)/(4))-(1)/(8)(sin""(x)/(8))/(cos""(x)/(8))...=(cos x)/(sin x)-(1)/(x)` `"or "-(1)/(2)tan""(x)/(2)-(1)/(4)tan""(x)/(4)-(1)/(8)tan"(x)/(8)-...=cot x -(1)/(x)` Differntiating both sides with respect to x, we get `-(1)/(2^(2))sec^(2)""(x)/(2)-(1)/(4^(2))sec^(2)""(x)/(4)-(1)/(8^(2))sec^(2)""(x)/(8)-...=-cosec^(2)x+(1)/(x^(2))` `"or "(1)/(2^(2))sec^(2)""(x)/(2)+(1)/(4^(2))sec^(2)""(x)/(4)+(1)/(8^(2))sec^(2)""(x)/(8)+...=cosec^(2)x-(1)/(x^(2))`. |
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